## deletion distance between words

Question

I'm trying to find how many characters I would need to delete to make the two words the same. For instance "at", "cat" would be 1 because I could delete the c, "boat" and "got" would be 3 because I could delete the b,a and g to make it ot. I put the words into a dictionary with their count as the value. Then I iterate over the dictionary and see if that key exists in the other dictionary otherwise I add 1 to the difference. Is this a very inefficient algorithm?

But it is overestimating the number of deletions I need.

``````def deletiondistance(firstword, secondword):
dfw = {}
dsw = {}
diff = 0
for i in range(len(firstword)):
print firstword[i]
if firstword[i] in dfw:
dfw[firstword[i]]+=1
else:
dfw[firstword[i]]=1
for j in range(len(secondword)):
if secondword[j] in dsw:
dsw[secondword[j]] +=1
else:
dsw[secondword[j]]=1

for key, value in dfw.iteritems():

if key in dsw:
#print "key exists"
pass

else:
diff +=1

print "diff",diff
``````

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2016-12-22 04:12 3 Answers

## Answers to deletion distance between words ( 3 )

1. I think your goal is similar to levenshtein distance.

Levenshtein distance is a metric for measuring distance between 2 strings.

Here is a wiki-link. https://en.wikipedia.org/wiki/Levenshtein_distance

And here is pypi package for levenshtein distance. https://pypi.python.org/pypi/python-Levenshtein

2. You can use difflib for this.

Example:

``````import difflib

cases=[('cat','bat'),
('bat','cat'),
('broom', 'ballroom'),
('boat','got')]

for a,b in cases:
print('{} => {}'.format(a,b))
cnt=0
for i,s in enumerate(difflib.ndiff(a, b)):
if s[0]==' ': continue
elif s[0]=='-':
print(u'Delete "{}" from position {}'.format(s[-1],i))
elif s[0]=='+':
print(u'Add "{}" to position {}'.format(s[-1],i))
cnt+=1
print("total=",cnt,"\n")
``````

Prints:

``````cat => bat
Delete "c" from position 0
Add "b" to position 1
total= 2

bat => cat
Delete "b" from position 0
Add "c" to position 1
total= 2

broom => ballroom
Add "a" to position 1
Add "l" to position 2
Add "l" to position 3
total= 3

boat => got
Delete "b" from position 0
Add "g" to position 1
Delete "a" from position 3
total= 3
``````
3. As @Hulk mentioned this is similar to levenshtein distance. The only difference is that substitutions are not allowed but that can be rectified by using substitution cost of 2 which is the same as removing character from both strings. Example:

``````def dist(s1, s2):
cur = list(range(len(s2) + 1))
prev = [0] * (len(s2) + 1)
for i in range(len(s1)):
cur, prev = prev, cur
cur[0] = i + 1
for j in range(len(s2)):
# Substitution is same as two deletions
sub = 0 if s1[i] == s2[j] else 2
cur[j+1] = min(prev[j] + sub, cur[j] + 1, prev[j+1] + 1)

return cur[-1]

cases=[('cat','bat'),
('bat','cat'),
('broom', 'ballroom'),
('boat','got'),
('foo', 'bar'),
('foobar', '')]

for s1, s2 in cases:
print('{} & {} = {}'.format(s1, s2, dist(s1, s2)))
``````

Output:

``````cat & bat = 2
bat & cat = 2
broom & ballroom = 3
boat & got = 3
foo & bar = 6
foobar &  = 6
``````