Finding Label Locations in Multiple Intervals with Python’s Pandas IntervalIndex

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💡 Problem Formulation: When working with interval data in pandas, it’s common to encounter a scenario where a single label falls into multiple intervals. The question then arises: how do we identify all interval locations for such a label? For instance, if we have intervals like [0, 1), [1, 2), [2, 3) and a label 1.5, it belongs to the interval [1, 2). But if the label is 1, it could be in [0, 1) or [1, 2) depending on how the intervals are defined (closed on which side). Hence, we seek a method to obtain every interval index that a specific label belongs to.

Method 1: Using IntervalIndex.get_indexer

This method involves using the IntervalIndex.get_indexer function, which returns an array of indices where the label would be inserted to maintain order. For labels that fall within the intervals, these indicate the interval locations.

Here’s an example:

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 1), (1, 2), (2, 3)], closed='right')
label = 1
locations = intervals.get_indexer([label])

print(locations)

Output:

[1]

The code snippet creates an IntervalIndex object with intervals that are closed on the right side. It then uses get_indexer to find the location(s) of the label 1. Since our intervals are right-closed, label 1 falls into the second interval (index 1).

Method 2: Using IntervalIndex.get_loc

The IntervalIndex.get_loc method is designed to get the location for a particular label. If a label is in multiple intervals, it will return all the relevant interval indices.

Here’s an example:

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 1), (1, 2), (1, 3)], closed='both')
label = 1
locations = intervals.get_loc(label)

print(locations)

Output:

[1, 2]

In contrast to get_indexer, get_loc works well when a label belongs in multiple intervals. It returns an array with both interval indices where label 1 falls since the intervals overlap and are closed on both sides.

Method 3: List Comprehension and in Operator

This method uses list comprehension to iterate over each interval and check if the label is contained within the interval using the in operator, returning a list of the matching indices.

Here’s an example:

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 1), (1, 2), (1, 3)], closed='both')
label = 1
locations = [i for i, interval in enumerate(intervals) if label in interval]

print(locations)

Output:

[1, 2]

Here we’re iterating over the intervals and checking if the label is contained in each, thus handling multiple intervals containing the same label elegantly without relying on pandas-specific methods.

Method 4: Using IntervalIndex.contains

The contains function of IntervalIndex can be used to determine if each interval contains the label, returning a boolean array which can be used to filter the indices.

Here’s an example:

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 1), (1, 2), (1, 3)], closed='both')
label = 1
contains_label = intervals.contains(label)
locations = list(contains_label[contains_label].index)

print(locations)

Output:

[1, 2]

This code creates a boolean array where each element specifies whether the corresponding interval contains the label or not. We then filter the true values to get the list of interval indices.

Bonus One-Liner Method 5: Using a Lambda with map

A concise one-liner using map and a lambda function can achieve the same result as list comprehension for those who prefer functional programming style.

Here’s an example:

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 1), (1, 2), (1, 3)], closed='both')
label = 1
locations = list(filter(lambda idx: label in intervals[idx], range(len(intervals))))

print(locations)

Output:

[1, 2]

This one-liner maps the lambda function, which checks interval membership, over the index range of the intervals. It then filters out the indices where the label is not present.

Summary/Discussion

  • Method 1: get_indexer. Useful when intervals do not overlap. It provides the index where the label fits neatly in the sorting order.
  • Method 2: get_loc. Directly finds all interval indices where a label is contained. Best for when intervals can overlap.
  • Method 3: List Comprehension with in. Flexible and native Python approach, suitable for complex logic within the iteration.
  • Method 4: contains. Returns a boolean array from which you can derive indices, useful when working with boolean indexing in pandas.
  • Method 5: Lambda with map. A functional programming approach for those who prefer succinct code.