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