572 lines
17 KiB
Python
572 lines
17 KiB
Python
# SPDX-License-Identifier: MIT
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# Copyright (C) 2022 Max Bachmann
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from __future__ import annotations
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from rapidfuzz._common_py import common_affix, conv_sequences
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from rapidfuzz._utils import is_none, setupPandas
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from rapidfuzz.distance import Indel_py as Indel
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from rapidfuzz.distance._initialize_py import Editop, Editops
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def _levenshtein_maximum(s1, s2, weights):
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len1 = len(s1)
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len2 = len(s2)
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insert, delete, replace = weights
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max_dist = len1 * delete + len2 * insert
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if len1 >= len2:
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max_dist = min(max_dist, len2 * replace + (len1 - len2) * delete)
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else:
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max_dist = min(max_dist, len1 * replace + (len2 - len1) * insert)
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return max_dist
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def _uniform_generic(s1, s2, weights):
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len1 = len(s1)
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insert, delete, replace = weights
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cache = list(range(0, (len1 + 1) * delete, delete))
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for ch2 in s2:
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temp = cache[0]
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cache[0] += insert
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for i in range(len1):
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x = temp
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if s1[i] != ch2:
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x = min(cache[i] + delete, cache[i + 1] + insert, temp + replace)
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temp = cache[i + 1]
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cache[i + 1] = x
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return cache[-1]
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def _uniform_distance(s1, s2):
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if not s1:
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return len(s2)
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VP = (1 << len(s1)) - 1
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VN = 0
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currDist = len(s1)
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mask = 1 << (len(s1) - 1)
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block = {}
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block_get = block.get
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x = 1
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for ch1 in s1:
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block[ch1] = block_get(ch1, 0) | x
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x <<= 1
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for ch2 in s2:
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# Step 1: Computing D0
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PM_j = block_get(ch2, 0)
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X = PM_j
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D0 = (((X & VP) + VP) ^ VP) | X | VN
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# Step 2: Computing HP and HN
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HP = VN | ~(D0 | VP)
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HN = D0 & VP
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# Step 3: Computing the value D[m,j]
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currDist += (HP & mask) != 0
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currDist -= (HN & mask) != 0
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# Step 4: Computing Vp and VN
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HP = (HP << 1) | 1
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HN = HN << 1
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VP = HN | ~(D0 | HP)
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VN = HP & D0
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return currDist
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def distance(
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s1,
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s2,
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*,
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weights=(1, 1, 1),
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processor=None,
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score_cutoff=None,
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score_hint=None,
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):
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"""
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Calculates the minimum number of insertions, deletions, and substitutions
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required to change one sequence into the other according to Levenshtein with custom
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costs for insertion, deletion and substitution
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Parameters
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----------
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s1 : Sequence[Hashable]
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First string to compare.
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s2 : Sequence[Hashable]
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Second string to compare.
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weights : tuple[int, int, int] or None, optional
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The weights for the three operations in the form
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(insertion, deletion, substitution). Default is (1, 1, 1),
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which gives all three operations a weight of 1.
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processor : callable, optional
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Optional callable that is used to preprocess the strings before
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comparing them. Default is None, which deactivates this behaviour.
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score_cutoff : int, optional
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Maximum distance between s1 and s2, that is
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considered as a result. If the distance is bigger than score_cutoff,
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score_cutoff + 1 is returned instead. Default is None, which deactivates
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this behaviour.
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score_hint : int, optional
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Expected distance between s1 and s2. This is used to select a
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faster implementation. Default is None, which deactivates this behaviour.
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Returns
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-------
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distance : int
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distance between s1 and s2
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Raises
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------
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ValueError
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If unsupported weights are provided a ValueError is thrown
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Examples
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--------
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Find the Levenshtein distance between two strings:
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>>> from rapidfuzz.distance import Levenshtein
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>>> Levenshtein.distance("lewenstein", "levenshtein")
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2
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Setting a maximum distance allows the implementation to select
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a more efficient implementation:
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>>> Levenshtein.distance("lewenstein", "levenshtein", score_cutoff=1)
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2
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It is possible to select different weights by passing a `weight`
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tuple.
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>>> Levenshtein.distance("lewenstein", "levenshtein", weights=(1,1,2))
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3
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"""
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_ = score_hint
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if processor is not None:
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s1 = processor(s1)
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s2 = processor(s2)
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s1, s2 = conv_sequences(s1, s2)
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if weights is None or weights == (1, 1, 1):
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dist = _uniform_distance(s1, s2)
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elif weights == (1, 1, 2):
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dist = Indel.distance(s1, s2)
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else:
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dist = _uniform_generic(s1, s2, weights)
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return dist if (score_cutoff is None or dist <= score_cutoff) else score_cutoff + 1
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def similarity(
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s1,
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s2,
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*,
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weights=(1, 1, 1),
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processor=None,
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score_cutoff=None,
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score_hint=None,
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):
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"""
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Calculates the levenshtein similarity in the range [max, 0] using custom
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costs for insertion, deletion and substitution.
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This is calculated as ``max - distance``, where max is the maximal possible
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Levenshtein distance given the lengths of the sequences s1/s2 and the weights.
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Parameters
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----------
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s1 : Sequence[Hashable]
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First string to compare.
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s2 : Sequence[Hashable]
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Second string to compare.
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weights : tuple[int, int, int] or None, optional
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The weights for the three operations in the form
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(insertion, deletion, substitution). Default is (1, 1, 1),
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which gives all three operations a weight of 1.
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processor : callable, optional
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Optional callable that is used to preprocess the strings before
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comparing them. Default is None, which deactivates this behaviour.
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score_cutoff : int, optional
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Maximum distance between s1 and s2, that is
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considered as a result. If the similarity is smaller than score_cutoff,
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0 is returned instead. Default is None, which deactivates
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this behaviour.
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score_hint : int, optional
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Expected similarity between s1 and s2. This is used to select a
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faster implementation. Default is None, which deactivates this behaviour.
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Returns
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-------
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similarity : int
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similarity between s1 and s2
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Raises
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------
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ValueError
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If unsupported weights are provided a ValueError is thrown
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"""
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_ = score_hint
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if processor is not None:
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s1 = processor(s1)
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s2 = processor(s2)
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s1, s2 = conv_sequences(s1, s2)
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weights = weights or (1, 1, 1)
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maximum = _levenshtein_maximum(s1, s2, weights)
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dist = distance(s1, s2, weights=weights)
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sim = maximum - dist
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return sim if (score_cutoff is None or sim >= score_cutoff) else 0
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def normalized_distance(
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s1,
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s2,
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*,
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weights=(1, 1, 1),
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processor=None,
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score_cutoff=None,
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score_hint=None,
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):
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"""
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Calculates a normalized levenshtein distance in the range [1, 0] using custom
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costs for insertion, deletion and substitution.
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This is calculated as ``distance / max``, where max is the maximal possible
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Levenshtein distance given the lengths of the sequences s1/s2 and the weights.
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Parameters
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----------
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s1 : Sequence[Hashable]
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First string to compare.
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s2 : Sequence[Hashable]
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Second string to compare.
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weights : tuple[int, int, int] or None, optional
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The weights for the three operations in the form
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(insertion, deletion, substitution). Default is (1, 1, 1),
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which gives all three operations a weight of 1.
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processor : callable, optional
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Optional callable that is used to preprocess the strings before
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comparing them. Default is None, which deactivates this behaviour.
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score_cutoff : float, optional
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Optional argument for a score threshold as a float between 0 and 1.0.
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For norm_dist > score_cutoff 1.0 is returned instead. Default is None,
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which deactivates this behaviour.
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score_hint : float, optional
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Expected normalized distance between s1 and s2. This is used to select a
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faster implementation. Default is None, which deactivates this behaviour.
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Returns
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-------
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norm_dist : float
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normalized distance between s1 and s2 as a float between 1.0 and 0.0
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Raises
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------
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ValueError
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If unsupported weights are provided a ValueError is thrown
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"""
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_ = score_hint
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setupPandas()
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if is_none(s1) or is_none(s2):
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return 1.0
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if processor is not None:
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s1 = processor(s1)
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s2 = processor(s2)
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s1, s2 = conv_sequences(s1, s2)
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weights = weights or (1, 1, 1)
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maximum = _levenshtein_maximum(s1, s2, weights)
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dist = distance(s1, s2, weights=weights)
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norm_dist = dist / maximum if maximum else 0
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return norm_dist if (score_cutoff is None or norm_dist <= score_cutoff) else 1
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def normalized_similarity(
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s1,
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s2,
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*,
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weights=(1, 1, 1),
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processor=None,
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score_cutoff=None,
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score_hint=None,
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):
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"""
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Calculates a normalized levenshtein similarity in the range [0, 1] using custom
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costs for insertion, deletion and substitution.
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This is calculated as ``1 - normalized_distance``
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Parameters
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----------
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s1 : Sequence[Hashable]
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First string to compare.
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s2 : Sequence[Hashable]
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Second string to compare.
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weights : tuple[int, int, int] or None, optional
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The weights for the three operations in the form
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(insertion, deletion, substitution). Default is (1, 1, 1),
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which gives all three operations a weight of 1.
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processor : callable, optional
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Optional callable that is used to preprocess the strings before
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comparing them. Default is None, which deactivates this behaviour.
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score_cutoff : float, optional
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Optional argument for a score threshold as a float between 0 and 1.0.
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For norm_sim < score_cutoff 0 is returned instead. Default is None,
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which deactivates this behaviour.
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score_hint : int, optional
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Expected normalized similarity between s1 and s2. This is used to select a
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faster implementation. Default is None, which deactivates this behaviour.
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Returns
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-------
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norm_sim : float
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normalized similarity between s1 and s2 as a float between 0 and 1.0
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Raises
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------
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ValueError
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If unsupported weights are provided a ValueError is thrown
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Examples
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--------
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Find the normalized Levenshtein similarity between two strings:
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>>> from rapidfuzz.distance import Levenshtein
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>>> Levenshtein.normalized_similarity("lewenstein", "levenshtein")
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0.81818181818181
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Setting a score_cutoff allows the implementation to select
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a more efficient implementation:
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>>> Levenshtein.normalized_similarity("lewenstein", "levenshtein", score_cutoff=0.85)
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0.0
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It is possible to select different weights by passing a `weight`
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tuple.
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>>> Levenshtein.normalized_similarity("lewenstein", "levenshtein", weights=(1,1,2))
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0.85714285714285
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When a different processor is used s1 and s2 do not have to be strings
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>>> Levenshtein.normalized_similarity(["lewenstein"], ["levenshtein"], processor=lambda s: s[0])
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0.81818181818181
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"""
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_ = score_hint
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setupPandas()
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if is_none(s1) or is_none(s2):
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return 0.0
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if processor is not None:
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s1 = processor(s1)
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s2 = processor(s2)
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s1, s2 = conv_sequences(s1, s2)
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weights = weights or (1, 1, 1)
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norm_dist = normalized_distance(s1, s2, weights=weights)
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norm_sim = 1.0 - norm_dist
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return norm_sim if (score_cutoff is None or norm_sim >= score_cutoff) else 0
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def _matrix(s1, s2):
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if not s1:
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return (len(s2), [], [])
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VP = (1 << len(s1)) - 1
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VN = 0
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currDist = len(s1)
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mask = 1 << (len(s1) - 1)
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block = {}
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block_get = block.get
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x = 1
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for ch1 in s1:
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block[ch1] = block_get(ch1, 0) | x
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x <<= 1
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matrix_VP = []
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matrix_VN = []
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for ch2 in s2:
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# Step 1: Computing D0
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PM_j = block_get(ch2, 0)
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X = PM_j
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D0 = (((X & VP) + VP) ^ VP) | X | VN
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# Step 2: Computing HP and HN
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HP = VN | ~(D0 | VP)
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HN = D0 & VP
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# Step 3: Computing the value D[m,j]
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currDist += (HP & mask) != 0
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currDist -= (HN & mask) != 0
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# Step 4: Computing Vp and VN
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HP = (HP << 1) | 1
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HN = HN << 1
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VP = HN | ~(D0 | HP)
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VN = HP & D0
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matrix_VP.append(VP)
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matrix_VN.append(VN)
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return (currDist, matrix_VP, matrix_VN)
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def editops(
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s1,
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s2,
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*,
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processor=None,
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score_hint=None,
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):
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"""
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Return Editops describing how to turn s1 into s2.
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Parameters
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----------
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s1 : Sequence[Hashable]
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First string to compare.
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s2 : Sequence[Hashable]
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Second string to compare.
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processor : callable, optional
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Optional callable that is used to preprocess the strings before
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comparing them. Default is None, which deactivates this behaviour.
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score_hint : int, optional
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Expected distance between s1 and s2. This is used to select a
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faster implementation. Default is None, which deactivates this behaviour.
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Returns
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-------
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editops : Editops
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edit operations required to turn s1 into s2
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Notes
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-----
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The alignment is calculated using an algorithm of Heikki Hyyrö, which is
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described [8]_. It has a time complexity and memory usage of ``O([N/64] * M)``.
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References
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----------
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.. [8] Hyyrö, Heikki. "A Note on Bit-Parallel Alignment Computation."
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Stringology (2004).
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Examples
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--------
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>>> from rapidfuzz.distance import Levenshtein
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>>> for tag, src_pos, dest_pos in Levenshtein.editops("qabxcd", "abycdf"):
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... print(("%7s s1[%d] s2[%d]" % (tag, src_pos, dest_pos)))
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delete s1[1] s2[0]
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replace s1[3] s2[2]
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insert s1[6] s2[5]
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"""
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_ = score_hint
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if processor is not None:
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s1 = processor(s1)
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s2 = processor(s2)
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s1, s2 = conv_sequences(s1, s2)
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prefix_len, suffix_len = common_affix(s1, s2)
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s1 = s1[prefix_len : len(s1) - suffix_len]
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s2 = s2[prefix_len : len(s2) - suffix_len]
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dist, VP, VN = _matrix(s1, s2)
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editops = Editops([], 0, 0)
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editops._src_len = len(s1) + prefix_len + suffix_len
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editops._dest_len = len(s2) + prefix_len + suffix_len
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if dist == 0:
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return editops
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editop_list = [None] * dist
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col = len(s1)
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row = len(s2)
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while row != 0 and col != 0:
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# deletion
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if VP[row - 1] & (1 << (col - 1)):
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dist -= 1
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col -= 1
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editop_list[dist] = Editop("delete", col + prefix_len, row + prefix_len)
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else:
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row -= 1
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# insertion
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if row and (VN[row - 1] & (1 << (col - 1))):
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dist -= 1
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editop_list[dist] = Editop("insert", col + prefix_len, row + prefix_len)
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else:
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col -= 1
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# replace (Matches are not recorded)
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if s1[col] != s2[row]:
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dist -= 1
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editop_list[dist] = Editop("replace", col + prefix_len, row + prefix_len)
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while col != 0:
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dist -= 1
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col -= 1
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editop_list[dist] = Editop("delete", col + prefix_len, row + prefix_len)
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while row != 0:
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dist -= 1
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row -= 1
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editop_list[dist] = Editop("insert", col + prefix_len, row + prefix_len)
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editops._editops = editop_list
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return editops
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def opcodes(
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s1,
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s2,
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*,
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processor=None,
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score_hint=None,
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):
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"""
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Return Opcodes describing how to turn s1 into s2.
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Parameters
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----------
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s1 : Sequence[Hashable]
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First string to compare.
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s2 : Sequence[Hashable]
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Second string to compare.
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processor : callable, optional
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Optional callable that is used to preprocess the strings before
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comparing them. Default is None, which deactivates this behaviour.
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score_hint : int, optional
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Expected distance between s1 and s2. This is used to select a
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faster implementation. Default is None, which deactivates this behaviour.
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Returns
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-------
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opcodes : Opcodes
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edit operations required to turn s1 into s2
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Notes
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-----
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The alignment is calculated using an algorithm of Heikki Hyyrö, which is
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described [9]_. It has a time complexity and memory usage of ``O([N/64] * M)``.
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References
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----------
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.. [9] Hyyrö, Heikki. "A Note on Bit-Parallel Alignment Computation."
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Stringology (2004).
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Examples
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--------
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>>> from rapidfuzz.distance import Levenshtein
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>>> a = "qabxcd"
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>>> b = "abycdf"
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>>> for tag, i1, i2, j1, j2 in Levenshtein.opcodes("qabxcd", "abycdf"):
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... print(("%7s a[%d:%d] (%s) b[%d:%d] (%s)" %
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... (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2])))
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delete a[0:1] (q) b[0:0] ()
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equal a[1:3] (ab) b[0:2] (ab)
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replace a[3:4] (x) b[2:3] (y)
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equal a[4:6] (cd) b[3:5] (cd)
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insert a[6:6] () b[5:6] (f)
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"""
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return editops(s1, s2, processor=processor, score_hint=score_hint).as_opcodes()
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