# Edit Distance in Python

PythonAlgorithmEditDistance## Python Problem Overview

I'm programming a spellcheck program in Python. I have a list of valid words (the dictionary) and I need to output a list of words from this dictionary that have an edit distance of 2 from a given invalid word.

I know I need to start by generating a list with an edit distance of one from the invalid word(and then run that again on all the generated words). I have three methods, inserts(...), deletions(...) and changes(...) that should output a list of words with an edit distance of 1, where inserts outputs all valid words with one more letter than the given word, deletions outputs all valid words with one less letter, and changes outputs all valid words with one different letter.

I've checked a bunch of places but I can't seem to find an algorithm that describes this process. All the ideas I've come up with involve looping through the dictionary list multiple times, which would be extremely time consuming. If anyone could offer some insight, I'd be extremely grateful.

## Python Solutions

## Solution 1 - Python

The thing you are looking at is called an edit distance and here is a nice explanation on wiki. There are a lot of ways how to define a distance between the two words and the one that you want is called Levenshtein distance and here is a DP (dynamic programming) implementation in python.

```
def levenshteinDistance(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
distances = range(len(s1) + 1)
for i2, c2 in enumerate(s2):
distances_ = [i2+1]
for i1, c1 in enumerate(s1):
if c1 == c2:
distances_.append(distances[i1])
else:
distances_.append(1 + min((distances[i1], distances[i1 + 1], distances_[-1])))
distances = distances_
return distances[-1]
```

## Solution 2 - Python

`difflib`

in the standard library has various utilities for sequence matching, including the `get_close_matches`

method that you could use. It uses an algorithm adapted from Ratcliff and Obershelp.

**From the docs**

```
>>> from difflib import get_close_matches
>>> get_close_matches('appel', ['ape', 'apple', 'peach', 'puppy'])
['apple', 'ape']
```

## Solution 3 - Python

Here is my version for Levenshtein distance

```
def edit_distance(s1, s2):
m=len(s1)+1
n=len(s2)+1
``````
tbl = {}
for i in range(m): tbl[i,0]=i
for j in range(n): tbl[0,j]=j
for i in range(1, m):
for j in range(1, n):
cost = 0 if s1[i-1] == s2[j-1] else 1
tbl[i,j] = min(tbl[i, j-1]+1, tbl[i-1, j]+1, tbl[i-1, j-1]+cost)
return tbl[i,j]
```

print(edit_distance("Helloworld", "HalloWorld"))

## Solution 4 - Python

```
#this calculates edit distance not levenstein edit distance
word1="rice"
word2="ice"
len_1=len(word1)
len_2=len(word2)
x =[[0]*(len_2+1) for _ in range(len_1+1)]#the matrix whose last element ->edit distance
for i in range(0,len_1+1): #initialization of base case values
x[i][0]=i
for j in range(0,len_2+1):
x[0][j]=j
for i in range (1,len_1+1):
for j in range(1,len_2+1):
if word1[i-1]==word2[j-1]:
x[i][j] = x[i-1][j-1]
else :
x[i][j]= min(x[i][j-1],x[i-1][j],x[i-1][j-1])+1
print x[i][j]
```

## Solution 5 - Python

I would recommend not creating this kind of code on your own. There are libraries for that.

For instance the Levenshtein library.

```
In [2]: Levenshtein.distance("foo", "foobar")
Out[2]: 3
In [3]: Levenshtein.distance("barfoo", "foobar")
Out[3]: 6
In [4]: Levenshtein.distance("Buroucrazy", "Bureaucracy")
Out[4]: 3
In [5]: Levenshtein.distance("Misisipi", "Mississippi")
Out[5]: 3
In [6]: Levenshtein.distance("Misisipi", "Misty Mountains")
Out[6]: 11
In [7]: Levenshtein.distance("Buroucrazy", "Born Crazy")
Out[7]: 4
```

## Solution 6 - Python

Using the `SequenceMatcher`

from Python built-in `difflib`

is another way of doing it, but (as correctly pointed out in the comments), the result does not match the definition of an edit distance exactly. Bonus: it supports ignoring "junk" parts (e.g. spaces or punctuation).

```
from difflib import SequenceMatcher
a = 'kitten'
b = 'sitting'
required_edits = [ code for code in ( SequenceMatcher(a=a, b=b, autojunk=False) .get_opcodes() ) if code[0] != 'equal'
]
required_edits
# [# # (tag, i1, i2, j1, j2)# ('replace', 0, 1, 0, 1), # replace a[0:1]="k" with b[0:1]="s"
# ('replace', 4, 5, 4, 5), # replace a[4:5]="e" with b[4:5]="i"
# ('insert', 6, 6, 6, 7), # insert b[6:7]="g" after a[6:6]="n"
# ]
# the edit distance:
len(required_edits) # == 3
```

## Solution 7 - Python

Similar to Santoshi's solution above but I made three changes:

- One line initialization instead of five
- No need to define cost alone (just use int(boolean) 0 or 1)
- Instead of double for loop use product, (this last one is only cosmetic, double loop seems unavoidable)

```
from itertools import product
def edit_distance(s1,s2):
d={ **{(i,0):i for i in range(len(s1)+1)},**{(0,j):j for j in range(len(s2)+1)}}
for i, j in product(range(1,len(s1)+1), range(1,len(s2)+1)):
d[i,j]=min((s1[i-1]!=s2[j-1]) + d[i-1,j-1], d[i-1,j]+1, d[i,j-1]+1)
return d[i,j]
```

## Solution 8 - Python

Instead of going with Levenshtein distance algo use *BK tree* or *TRIE*, as these algorithms have less complexity then edit distance. A good browse over these topic will give a detailed description.

This link will help you more about spell checking.

## Solution 9 - Python

You need Minimum Edit Distance for this task.

Following is my version of MED a.k.a Levenshtein Distance.

```
def MED_character(str1,str2):
cost=0
len1=len(str1)
len2=len(str2)
#output the length of other string in case the length of any of the string is zero
if len1==0:
return len2
if len2==0:
return len1
accumulator = [[0 for x in range(len2)] for y in range(len1)] #initializing a zero matrix
# initializing the base cases
for i in range(0,len1):
accumulator[i][0] = i;
for i in range(0,len2):
accumulator[0][i] = i;
# we take the accumulator and iterate through it row by row.
for i in range(1,len1):
char1=str1[i]
for j in range(1,len2):
char2=str2[j]
cost1=0
if char1!=char2:
cost1=2 #cost for substitution
accumulator[i][j]=min(accumulator[i-1][j]+1, accumulator[i][j-1]+1, accumulator[i-1][j-1] + cost1 )
cost=accumulator[len1-1][len2-1]
return cost
```

## Solution 10 - Python

Fine tuned codes based on the version from @Santosh and should address the issue brought up by @Artur Krajewski; The biggest difference is replacing an effective 2d matrix

```
def edit_distance(s1, s2):
# add a blank character for both strings
m=len(s1)+1
n=len(s2)+1
# launch a matrix
tbl = [[0] * n for i in range(m)]
for i in range(m): tbl[i][0]=i
for j in range(n): tbl[0][j]=j
for i in range(1, m):
for j in range(1, n):
#if strings have same letters, set operation cost as 0 otherwise 1
cost = 0 if s1[i-1] == s2[j-1] else 1
#find min practice
tbl[i][j] = min(tbl[i][j-1]+1, tbl[i-1][j]+1, tbl[i-1][j-1]+cost)
return tbl
edit_distance("birthday", "Birthdayyy")
```