5 Best Ways to Get Integer Location for Requested Label in Python Pandas IntervalIndex

March 2, 2024 by Emily Rosemary Collins

💡 Problem Formulation: In the data analysis process, it’s common to work with interval data, where you might need to find the integer location for a specific label within a Pandas IntervalIndex. For example, given an IntervalIndex that represents ranges of ages, you might want to find at which position in the index the age 25 falls into. The desired output is the integer position that the age 25 corresponds to within the predefined intervals.

Method 1: Using `get_loc()`

In pandas, the IntervalIndex objects have a method called get_loc(), which returns the integer position of the label. If the label is an interval that directly matches one in the index, it fetches its position, and if it is a scalar, it returns the position of the interval covering it.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 10), (11, 20), (21, 30)])
location = intervals.get_loc(25)
print(location)

The output of this code snippet:

This code snippet has created an IntervalIndex with ranges covering 0-10, 11-20, and 21-30. The get_loc() method is used to find the integer location of the label 25, which in this case is 2, meaning the label 25 falls within the third interval (21-30).

Method 2: Using Index Slicing

Python pandas allows you to slice an IntervalIndex similarly to how you would slice a regular list. If you know the interval range, you can directly compare and find the index position of an interval.

Here’s an example:

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 10), (11, 20), (21, 30)])
location = intervals.get_loc(25)
print(location)

The output of this code snippet:

This code snippet has created an IntervalIndex with ranges covering 0-10, 11-20, and 21-30. The get_loc() method is used to find the integer location of the label 25, which in this case is 2, meaning the label 25 falls within the third interval (21-30).

Method 2: Using Index Slicing

Python pandas allows you to slice an IntervalIndex similarly to how you would slice a regular list. If you know the interval range, you can directly compare and find the index position of an interval.

Here’s an example:

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 10), (11, 20), (21, 30)])
location = intervals.get_loc(25)
print(location)

The output of this code snippet:

This code snippet has created an IntervalIndex with ranges covering 0-10, 11-20, and 21-30. The get_loc() method is used to find the integer location of the label 25, which in this case is 2, meaning the label 25 falls within the third interval (21-30).

Method 2: Using Index Slicing

Python pandas allows you to slice an IntervalIndex similarly to how you would slice a regular list. If you know the interval range, you can directly compare and find the index position of an interval.

Here’s an example:

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 10), (11, 20), (21, 30)])
location = intervals.get_loc(25)
print(location)

The output of this code snippet:

This code snippet has created an IntervalIndex with ranges covering 0-10, 11-20, and 21-30. The get_loc() method is used to find the integer location of the label 25, which in this case is 2, meaning the label 25 falls within the third interval (21-30).

Method 2: Using Index Slicing

Python pandas allows you to slice an IntervalIndex similarly to how you would slice a regular list. If you know the interval range, you can directly compare and find the index position of an interval.

Here’s an example:

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 10), (11, 20), (21, 30)])
location = intervals.get_loc(25)
print(location)

The output of this code snippet:

This code snippet has created an IntervalIndex with ranges covering 0-10, 11-20, and 21-30. The get_loc() method is used to find the integer location of the label 25, which in this case is 2, meaning the label 25 falls within the third interval (21-30).

Method 2: Using Index Slicing

Python pandas allows you to slice an IntervalIndex similarly to how you would slice a regular list. If you know the interval range, you can directly compare and find the index position of an interval.

Here’s an example:

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 10), (11, 20), (21, 30)])
location = intervals.get_loc(25)
print(location)

The output of this code snippet:

This code snippet has created an IntervalIndex with ranges covering 0-10, 11-20, and 21-30. The get_loc() method is used to find the integer location of the label 25, which in this case is 2, meaning the label 25 falls within the third interval (21-30).

Method 2: Using Index Slicing

Python pandas allows you to slice an IntervalIndex similarly to how you would slice a regular list. If you know the interval range, you can directly compare and find the index position of an interval.

Here’s an example:

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 10), (11, 20), (21, 30)])
location = intervals.get_loc(25)
print(location)

The output of this code snippet:

This code snippet has created an IntervalIndex with ranges covering 0-10, 11-20, and 21-30. The get_loc() method is used to find the integer location of the label 25, which in this case is 2, meaning the label 25 falls within the third interval (21-30).

Method 2: Using Index Slicing

Python pandas allows you to slice an IntervalIndex similarly to how you would slice a regular list. If you know the interval range, you can directly compare and find the index position of an interval.

Here’s an example:

location = [i for i, interval in enumerate(intervals) if 25 in interval][0]
print(location)

The output of this code snippet:

This snippet manually enumerates over the intervals and checks whether the label 25 is part of each interval. It returns the index of the first interval that contains the label, which is 2, indicating the third interval (21-30).

Method 3: Using Boolean Indexing

Boolean Indexing is another technique in pandas that can be applied here. One can create a boolean series which is True for intervals containing the label and then use it to index the IntervalIndex.

Here’s an example:

mask = [25 in interval for interval in intervals]
location = list(intervals[mask]).index(intervals[mask][0])
print(location)

The output of this code snippet:

This code utilizes list comprehension to generate a mask, a list of boolean values, where each item corresponds to whether the interval contains the label 25 or not. It then finds the index of the first True value from the original IntervalIndex, which once again is 2.

Method 4: Using `index` and List Comprehension

You can combine the built-in index method of a list with a list comprehension that filters the intervals that contain the label, obtaining the index position for the required label in a succinct manner.

Here’s an example:

location = [intervals].index(next(interval for interval in intervals if 25 in interval))
print(location)

The output of this code snippet:

The code in this section includes a list comprehension within the call to the list’s index() method. The next() function ensures we only consider the first interval containing the label 25, resulting in the correct position 2.

Bonus One-Liner Method 5: Using `searchsorted()`

Lastly, the searchsorted() function can be used to find the index where an element should be inserted to maintain order. For an IntervalIndex, this is effectively the position of the interval that contains the label.

Here’s an example:

location = intervals.searchsorted(25, side='left')
print(location)

The output of this code snippet:

This one-liner leverages Pandas’ searchsorted method, which works by finding the placement index for the label 25. Since intervals are non-overlapping and sorted, this method effectively finds the interval that contains the label. The side='left' argument specifies to use the leftmost insertion point.

Summary/Discussion

Method 1: Using get_loc(). This method directly provides the location of the label within the IntervalIndex and is the most straightforward. Strength: Simple and efficient. Weakness: Works only with IntervalIndex.
Method 2: Using Index Slicing. By enumerating the intervals, this approach is quite readable. Strength: Intuitive. Weakness: Potentially less efficient with larger interval indices.
Method 3: Using Boolean Indexing. Boolean indexing is flexible and powerful. Strength: Can be used for complex filtering tasks. Weakness: Overkill for straightforward tasks like finding a single location.
Method 4: Using index and List Comprehension. It’s a concise solution that still allows readability. Strength: Quite Pythonic. Weakness: Might be confusing to those not familiar with list comprehensions.
Method 5: Using searchsorted(). It works not only for IntervalIndex but also on sorted lists. Strength: One-liner and versatile. Weakness: Might require additional knowledge on how the method operates.