**π‘ 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**Flexible and native Python approach, suitable for complex logic within the iteration.`in`

.**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**A functional programming approach for those who prefer succinct code.`map`

.