如果有人有同樣的「問題」，這是我的解決方案。我爲Kyle Gorman編寫的Wagner-Fischer算法的python實現添加了一個附件。
這個插件是在_dist函數中它的權重函數和實現。
#!/usr/bin/env python
# wagnerfischer.py: Dynamic programming Levensthein distance function
# Kyle Gorman <[email protected]>
#
# Based on:
#
# Robert A. Wagner and Michael J. Fischer (1974). The string-to-string
# correction problem. Journal of the ACM 21(1):168-173.
#
# The thresholding function was inspired by BSD-licensed code from
# Babushka, a Ruby tool by Ben Hoskings and others.
#
# Unlike many other Levenshtein distance functions out there, this works
# on arbitrary comparable Python objects, not just strings.
try: # use numpy arrays if possible...
from numpy import zeros
def _zeros(*shape):
""" like this syntax better...a la MATLAB """
return zeros(shape)
except ImportError: # otherwise do this cute solution
def _zeros(*shape):
if len(shape) == 0:
return 0
car = shape[0]
cdr = shape[1:]
return [_zeros(*cdr) for i in range(car)]
def weight(A,B, weights):
if weights == True:
from numpy import matrix
from numpy import where
# cost_weight defines the matrix structure of the AOI-placement
cost_weight = matrix([["a","b","c","d","e","f"],["g","h","i","j","k","l"],
["m","n","o","p","q","r"],["s","t","u","v","w","x"]])
max_walk = 8.00 # defined as the maximum posible distance between letters in
# the cost_weight matrix
indexA = where(cost_weight==A)
indexB = where(cost_weight==B)
walk = abs(indexA[0][0]-indexB[0][0])+abs(indexA[1][0]-indexB[1][0])
w = walk/max_walk
return w
else:
return 1
def _dist(A, B, insertion, deletion, substitution, weights=True):
D = _zeros(len(A) + 1, len(B) + 1)
for i in xrange(len(A)):
D[i + 1][0] = D[i][0] + deletion * weight(A[i],B[0], weights)
for j in xrange(len(B)):
D[0][j + 1] = D[0][j] + insertion * weight(A[0],B[j], weights)
for i in xrange(len(A)): # fill out middle of matrix
for j in xrange(len(B)):
if A[i] == B[j]:
D[i + 1][j + 1] = D[i][j] # aka, it's free.
else:
D[i + 1][j + 1] = min(D[i + 1][j] + insertion * weight(A[i],B[j], weights),
D[i][j + 1] + deletion * weight(A[i],B[j], weights),
D[i][j] + substitution * weight(A[i],B[j], weights))
return D
def _dist_thresh(A, B, thresh, insertion, deletion, substitution):
D = _zeros(len(A) + 1, len(B) + 1)
for i in xrange(len(A)):
D[i + 1][0] = D[i][0] + deletion
for j in xrange(len(B)):
D[0][j + 1] = D[0][j] + insertion
for i in xrange(len(A)): # fill out middle of matrix
for j in xrange(len(B)):
if A[i] == B[j]:
D[i + 1][j + 1] = D[i][j] # aka, it's free.
else:
D[i + 1][j + 1] = min(D[i + 1][j] + insertion,
D[i][j + 1] + deletion,
D[i][j] + substitution)
if min(D[i + 1]) >= thresh:
return
return D
def _levenshtein(A, B, insertion, deletion, substitution):
return _dist(A, B, insertion, deletion, substitution)[len(A)][len(B)]
def _levenshtein_ids(A, B, insertion, deletion, substitution):
"""
Perform a backtrace to determine the optimal path. This was hard.
"""
D = _dist(A, B, insertion, deletion, substitution)
i = len(A)
j = len(B)
ins_c = 0
del_c = 0
sub_c = 0
while True:
if i > 0:
if j > 0:
if D[i - 1][j] <= D[i][j - 1]: # if ins < del
if D[i - 1][j] < D[i - 1][j - 1]: # if ins < m/s
ins_c += 1
else:
if D[i][j] != D[i - 1][j - 1]: # if not m
sub_c += 1
j -= 1
i -= 1
else:
if D[i][j - 1] <= D[i - 1][j - 1]: # if del < m/s
del_c += 1
else:
if D[i][j] != D[i - 1][j - 1]: # if not m
sub_c += 1
i -= 1
j -= 1
else: # only insert
ins_c += 1
i -= 1
elif j > 0: # only delete
del_c += 1
j -= 1
else:
return (ins_c, del_c, sub_c)
def _levenshtein_thresh(A, B, thresh, insertion, deletion, substitution):
D = _dist_thresh(A, B, thresh, insertion, deletion, substitution)
if D != None:
return D[len(A)][len(B)]
def levenshtein(A, B, thresh=None, insertion=1, deletion=1, substitution=1):
"""
Compute levenshtein distance between iterables A and B
"""
# basic checks
if len(A) == len(B) and A == B:
return 0
if len(B) > len(A):
(A, B) = (B, A)
if len(A) == 0:
return len(B)
if thresh:
if len(A) - len(B) > thresh:
return
return _levenshtein_thresh(A, B, thresh, insertion, deletion,
substitution)
else:
return _levenshtein(A, B, insertion, deletion, substitution)
def levenshtein_ids(A, B, insertion=1, deletion=1, substitution=1):
"""
Compute number of insertions deletions, and substitutions for an
optimal alignment.
There may be more than one, in which case we disfavor substitution.
"""
# basic checks
if len(A) == len(B) and A == B:
return (0, 0, 0)
if len(B) > len(A):
(A, B) = (B, A)
if len(A) == 0:
return len(B)
else:
return _levenshtein_ids(A, B, insertion, deletion, substitution)
您好，感謝有助於StackOverflow的。如果包含這個答案，將會更好地符合[指導原則]（http://stackoverflow.com/help/how-to-answer），例如，演示如何用圖書館解決問題的簡短代碼示例。乾杯! – nthall