105 lines
3.1 KiB
Python
105 lines
3.1 KiB
Python
# Licensed under the Apache License, Version 2.0 (the "License");
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or
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# implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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import os
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# Each edit operation is assigned different cost, such as:
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# 'w' means swap operation, the cost is 0;
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# 's' means substitution operation, the cost is 2;
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# 'a' means insertion operation, the cost is 1;
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# 'd' means deletion operation, the cost is 3;
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# The smaller cost results in the better similarity.
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COST = {'w': 0, 's': 2, 'a': 1, 'd': 3}
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def damerau_levenshtein(s1, s2, cost):
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"""Calculates the Damerau-Levenshtein distance between two strings.
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The Damerau-Levenshtein distance says the minimum number of single
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character edits (i.e. insertions, deletions, swap or substitution)
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required to change one string to the other.
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The idea is to reserve a matrix to hold the Levenshtein distances between
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all prefixes of the first string and all prefixes of the second, then we
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can compute the values in the matrix in a dynamic programming fashion. To
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avoid a large space complexity, only the last three rows in the matrix is
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needed.(row2 holds the current row, row1 holds the previous row, and row0
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the row before that.)
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More details:
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https://en.wikipedia.org/wiki/Damerau%E2%80%93Levenshtein_distance
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https://github.com/git/git/commit/8af84dadb142f7321ff0ce8690385e99da8ede2f
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"""
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if s1 == s2:
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return 0
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len1 = len(s1)
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len2 = len(s2)
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if len1 == 0:
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return len2 * cost['a']
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if len2 == 0:
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return len1 * cost['d']
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row1 = [i * cost['a'] for i in range(len2 + 1)]
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row2 = row1[:]
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row0 = row1[:]
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for i in range(len1):
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row2[0] = (i + 1) * cost['d']
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for j in range(len2):
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# substitution
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sub_cost = row1[j] + (s1[i] != s2[j]) * cost['s']
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# insertion
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ins_cost = row2[j] + cost['a']
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# deletion
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del_cost = row1[j + 1] + cost['d']
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# swap
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swp_condition = (
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(i > 0)
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and (j > 0)
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and (s1[i - 1] == s2[j])
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and (s1[i] == s2[j - 1])
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)
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# min cost
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if swp_condition:
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swp_cost = row0[j - 1] + cost['w']
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p_cost = min(sub_cost, ins_cost, del_cost, swp_cost)
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else:
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p_cost = min(sub_cost, ins_cost, del_cost)
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row2[j + 1] = p_cost
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row0, row1, row2 = row1, row2, row0
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return row1[-1]
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def terminal_width():
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"""Return terminal width in columns
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Uses `os.get_terminal_size` function
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:returns: terminal width
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:rtype: int or None
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"""
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try:
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return os.get_terminal_size().columns
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except OSError:
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return None
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