Understanding pandas.IntervalIndex.get_loc for Efficient Interval Navigation

Purpose

• It helps you find the position of a specific value (label) relative to the intervals in the IntervalIndex.

Arguments

• method (optional): This argument allows you to specify how to handle labels falling on interval boundaries. By default (method=None), it only considers labels that fall strictly within an interval (not on the edges). You can use other options like 'left' or 'right' to include labels on those specific edges.

• key: This is the label (value) you're searching for. It can be a single value, an interval itself, or even a list of labels.

Return Value

• The output depends on the structure of your IntervalIndex:
  • If the index is unique (no overlapping intervals), it returns an integer representing the position of the label.
  • If the index is monotonic (intervals are ordered in a specific way, either increasing or decreasing), it returns a slice object indicating the relevant interval(s).
  • In other cases (overlapping intervals or non-monotonic order), it returns a boolean mask with the same length as the IntervalIndex, indicating for each interval whether the label falls within it.

Examples

1. Finding the position of a value within an interval:

import pandas as pd

# Create an IntervalIndex
intervals = pd.IntervalIndex.from_tuples([(0, 1), (2, 3)])

# Find the location of 0.5
location = intervals.get_loc(0.5)

# Output: location will be 0 (since 0.5 falls within the first interval)

1. Finding all intervals containing a specific value (overlapping case):

intervals = pd.IntervalIndex.from_tuples([(0, 2), (1, 3)])

# Find locations for label 1.5
location = intervals.get_loc(1.5)

# Output: location will be a boolean mask [ True,  True] (both intervals contain 1.5)

• The return type adapts to the structure of your IntervalIndex for clear interpretation.
• The method argument provides flexibility in handling labels on interval boundaries.
• Use get_loc for efficient retrieval based on intervals rather than standard indexing.

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Finding multiple locations with a list

import pandas as pd

intervals = pd.IntervalIndex.from_tuples([(0, 2), (3, 5), (6, 8)])

# Find locations for multiple values
locations = intervals.get_loc([1, 7])

# Output: locations will likely be [0, -1] (1 falls within the first interval, 7 is not present)

Handling labels on boundaries (left/right)

intervals = pd.IntervalIndex.from_tuples([(0, 2), (2, 4)])

# Find location for 2 (default excludes it)
location = intervals.get_loc(2)

# Now, include labels on the left boundary with method='left'
location_left = intervals.get_loc(2, method='left')

# Output: location will be -1, location_left will be 1 (including 2 on the left)

import pandas.util.testing as tm

# Create a DataFrame with IntervalIndex
df = tm.makeIntervalFrame(periods=5)

# Get a specific row by label within the IntervalIndex
specific_row = df.loc[df.index.get_loc(2)]  # Assuming interval labeled 2 exists

# Output: specific_row will contain the row corresponding to interval labeled 2

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Looping with membership testing

This approach iterates through the IntervalIndex and checks if the target value falls within each interval using methods like contains or comparisons. It's less efficient than get_loc but might be suitable for simple cases or if you need more control over the logic.

import pandas as pd

def find_location(intervals, target):
  for i, interval in enumerate(intervals):
    if target in interval:  # or use interval.contains(target)
      return i
  return -1  # Not found

intervals = pd.IntervalIndex.from_tuples([(0, 2), (3, 5)])
target = 1.5

location = find_location(intervals, target)

# Output: location will be 0

numpy.searchsorted (for sorted IntervalIndex):

If your IntervalIndex is sorted (either ascending or descending), you can leverage numpy.searchsorted from the NumPy library. It performs a binary search to find the insertion point for the target value within the sorted intervals. However, it requires additional steps to interpret the results and might not handle overlapping intervals well.

import pandas as pd
import numpy as np

intervals = pd.IntervalIndex.from_tuples([(0, 2), (3, 5)])
target = 1.5

sorted_values = intervals.left  # Assuming sorted ascending
location = np.searchsorted(sorted_values, target)

# Further processing needed to interpret location for overlapping intervals

• Consider numpy.searchsorted only for very specific circumstances where you have a sorted IntervalIndex and need a basic binary search functionality.
• If you need more control over the logic or don't have a sorted IntervalIndex, looping with membership testing might be appropriate.
• For most cases, pandas.IntervalIndex.get_loc is the recommended option due to its efficiency and built-in handling of different IntervalIndex structures (unique, monotonic, overlapping).