Extract words in a paragraph that are similar to words in list

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I have the following string:

"The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"

List of words to be extracted:

["town","teddy","chicken","boy went"]

NB: town and teddy are wrongly spelt in the given sentence.

I have tried the following but I get other words that are not part of the answer:

import difflib

sent = "The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"

list1 = ["town","teddy","chicken","boy went"]

[difflib.get_close_matches(x.lower().strip(), sent.split()) for x in list1 ]

I am getting the following result:

[['twn', 'to'], ['tddy'], ['chicken.', 'picked'], ['went']]

instead of:

'twn', 'tddy', 'chicken','boy went'
2 Answers

Notice in the documentation for difflib.get_closest_matches():

difflib.get_close_matches(word, possibilities, n=3, cutoff=0.6)

Return a list of the best "good enough" matches. word is a sequence for which close matches are desired (typically a string), and possibilities is a list of sequences against which to match word (typically a list of strings).

Optional argument n (default 3) is the maximum number of close matches to return; n must be greater than 0.

Optional argument cutoff (default 0.6) is a float in the range [0, 1]. Possibilities that don’t score at least that similar to word are ignored.


At the moment, you are using the default n and cutoff arguments.

You can specify either (or both), to narrow down the returned matches.

For example, you could use a cutoff score of 0.75:

result = [difflib.get_close_matches(x.lower().strip(), sent.split(), cutoff=0.75) for x in list1]

Or, you could specify that only at most 1 match should be returned:

result = [difflib.get_close_matches(x.lower().strip(), sent.split(), n=1) for x in list1]

In either case, you could use a list comprehension to flatten the lists of lists (since difflib.get_close_matches() always returns a list):

matches = [r[0] for r in result]

Since you also want to check for close matches of bigrams, you can do so by extracting pairings of adjacent "words", and pass them to difflib.get_close_matches() as part of the possibilities argument.

Here is a full working example of this in action:

import difflib
import re

sent = "The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"

list1 = ["town", "teddy", "chicken", "boy went"]

# this extracts overlapping pairings of "words"
# i.e. ['The boy', 'boy went', 'went to', 'to twn', ...
pairs = re.findall(r'(?=(\b[^ ]+ [^ ]+\b))', sent)

# we pass the sent.split() list as before
# and concatenate the new pairs list to the end of it also
result = [difflib.get_close_matches(x.lower().strip(), sent.split() + pairs, n=1) for x in list1]

matches = [r[0] for r in result]

print(matches)
# ['twn', 'tddy', 'chicken.', 'boy went']

If you read Python documentation fordifflib.get_close_matches() https://docs.python.org/3/library/difflib.html It returns all possible best matches. Method signature: difflib.get_close_matches(word, possibilities, n=3, cutoff=0.6)

Here n is the maximum number of close matches to return. So I think you can pass this as 1.

>>> [difflib.get_close_matches(x.lower().strip(), sent.split(),1)[0] for x in list1]
['twn', 'tddy', 'chicken.', 'went']
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