126 lines
3.4 KiB
Python
126 lines
3.4 KiB
Python
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"""
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interval.py
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--------------
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Deal with 1D intervals which are defined by:
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[start position, end position]
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"""
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import numpy as np
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def check(a, b, digits):
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"""
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Check input ranges, convert them to vector form,
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and get a fixed precision integer version of them.
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Parameters
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--------------
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a : (2, ) or (2, n) float
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Start and end of a 1D interval
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b : (2, ) or (2, n) float
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Start and end of a 1D interval
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digits : int
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How many digits to consider
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Returns
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--------------
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a : (2, n) float
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Ranges as vector
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b : (2, n) float
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Ranges as vector
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a_int : (2, n) int64
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Ranges rounded to digits, as vector
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b_int : (2, n) int64
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Ranges rounded to digits, as vector
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is_1D : bool
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If True, input was single pair of ranges
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"""
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a = np.array(a, dtype=np.float64)
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b = np.array(b, dtype=np.float64)
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if a.shape != b.shape or a.shape[-1] != 2:
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raise ValueError('ranges must be identical and (2,)!')
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# if input was single interval reshape it here
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is_1D = False
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if len(a.shape) == 1:
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a = a.reshape((-1, 2))
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b = b.reshape((-1, 2))
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is_1D = True
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# make sure ranges are sorted
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a.sort(axis=1)
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b.sort(axis=1)
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# compare in fixed point as integers
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a_int = (a * 10**digits).round().astype(np.int64)
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b_int = (b * 10**digits).round().astype(np.int64)
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return a, b, a_int, b_int, is_1D
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def intersection(a, b, digits=8):
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"""
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Given a pair of ranges, merge them in to
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one range if they overlap at all
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Parameters
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--------------
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a : (2, ) float
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Start and end of a 1D interval
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b : (2, ) float
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Start and end of a 1D interval
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digits : int
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How many digits to consider
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Returns
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--------------
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intersects : bool or (n,) bool
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Indicates if the ranges overlap at all
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new_range : (2, ) or (2, 2) float
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The unioned range from the two inputs,
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or both of the original ranges if not overlapping
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"""
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# check shape and convert
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a, b, a_int, b_int, is_1D = check(a, b, digits)
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# what are the starting and ending points of the overlap
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overlap = np.zeros(a.shape, dtype=np.float64)
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# A fully overlaps B
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current = np.logical_and(a_int[:, 0] <= b_int[:, 0],
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a_int[:, 1] >= b_int[:, 1])
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overlap[current] = b[current]
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# B fully overlaps A
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current = np.logical_and(a_int[:, 0] >= b_int[:, 0],
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a_int[:, 1] <= b_int[:, 1])
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overlap[current] = a[current]
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# A starts B ends
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# A:, 0 B:, 0 A:, 1 B:, 1
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current = np.logical_and(
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np.logical_and(a_int[:, 0] <= b_int[:, 0],
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b_int[:, 0] < a_int[:, 1]),
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a_int[:, 1] < b_int[:, 1])
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overlap[current] = np.column_stack([b[current][:, 0],
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a[current][:, 1]])
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# B starts A ends
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# B:, 0 A:, 0 B:, 1 A:, 1
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current = np.logical_and(
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np.logical_and(b_int[:, 0] <= a_int[:, 0],
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a_int[:, 0] < b_int[:, 1]),
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b_int[:, 1] < a_int[:, 1])
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overlap[current] = np.column_stack([a[current][:, 0],
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b[current][:, 1]])
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# is range overlapping at all
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intersects = overlap.ptp(axis=1) > 10**-digits
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if is_1D:
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return intersects[0], overlap[0]
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return intersects, overlap
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