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make substitutions more memory efficient #5

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make substitutions more memory efficient #5

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19 changes: 8 additions & 11 deletions fizzle.py
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Original file line number Diff line number Diff line change
Expand Up @@ -12,9 +12,9 @@ def dl_distance(s1,
secondHalfDiscount=False):
"""
Return DL distance between s1 and s2. Default cost of substitution, insertion, deletion and transposition is 1
substitutions is list of tuples of characters (what, substituted by what, cost),
maximal value of substitution is 2 (ie: cost deletion & insertion that would be otherwise used)
eg: substitutions=[('a','e',0.4),('i','y',0.3)]
substitutions is list of tuples of characters (what, substituted by what, cost) or dictionary
{(what, substituted by what):cost}, maximal value of substitution is 2 (ie: cost deletion & insertion that
would be otherwise used) eg: substitutions=[('a','e',0.4),('i','y',0.3)] or {('a','e'):0.4, ('i','y'):0.3}
symetric=True mean that cost of substituting A with B is same as B with A
returnMatrix=True: the matrix of distances will be returned, if returnMatrix==False, then only distance will be returned
printMatrix==True: matrix of distances will be printed
Expand All @@ -23,9 +23,6 @@ def dl_distance(s1,

if isinstance(substitutions, list):
subs_dict = {(from_, to): cost for from_, to, cost in substitutions}
if symetric:
subs_dict.update({(to, from_)
for from_, to, cost in substitutions})
substitutions = subs_dict

if nonMatchingEnds:
Expand All @@ -39,13 +36,13 @@ def dl_distance(s1,
for i in range(len(s1)):
for j in range(len(s2)):
ch1, ch2 = s1[i], s2[j]
cost = 1
if ch1 == ch2:
cost = 0
else:
if (ch1, ch2) in substitutions:
cost = substitutions[(ch1, ch2)]
else:
cost = 1
elif (ch1, ch2) in substitutions:
cost = substitutions[(ch1, ch2)]
elif symetric and (ch2, ch1) in substitutions:
cost = substitutions[(ch2, ch1)]
if secondHalfDiscount and (s1 > half1 or s2 > half2):
deletionCost, insertionCost = 0.6, 0.6
else:
Expand Down