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I have a list of strings that i would like to search for a word combination. Then delete the list if the combination is not there. Is there a python list comprehension that would work?

word_list = ["Dogs love ice cream", "Cats love balls", "Ice cream", "ice cream is good with pizza", "cats hate ice cream"]

keep_words = ["Dogs", "Cats"] 

Delete_word = ["ice cream"]

Delete words that have ice cream in it but if dogs or cats is in the sentence keep it.

Desired_output = ["Dogs love ice cream", "Cats love balls", "cats hate ice cream"]

Was trying this code also tried AND and OR but cannot get the combination right.

output_list = [x for x in  word_list if "ice cream" not in x]
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4
  • do ice cream needs to be together? Like will this sentence be considered valid:keep cream over the ice? Commented Feb 6, 2018 at 20:26
  • I was thinking of it together "ice cream" but very good point Commented Feb 6, 2018 at 20:34
  • The last one is contain cat not cats! Commented Feb 6, 2018 at 20:38
  • good spot Kasramvd cats i will edit Commented Feb 6, 2018 at 20:42

3 Answers 3

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7

Here's a list comprehension solution:

[x for x in word_list if any(kw.lower() in x.lower() for kw in keep_words) 
 or all(dw.lower() not in x.lower() for dw in Delete_word)]
# ['Dogs love ice cream', 'Cats love balls', 'cats hate ice cream']

This also adds flexibility for multiple words in the delete words list.

Explanation

Iterate over the list and keep the word if either of the following are True:

  • Any of the keep words are in x
  • None of the delete words are in x

I presume from your example that you wanted this to be case insensitive, so make all comparison on the lower-cased versions of the words.

Two helpful functions are any() and all().

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1 Comment

Pault very interesting i didn't think about "any and all" --- Thank you so much for the help
5

As an optimized approach you can put your keep_word and delete_words within set and use itertools.filterfalse() to filter the list out:

In [48]: def key(x):
             words = x.lower().split()
             return keep_words.isdisjoint(words) or not delete_words.isdisjoint(words)
   ....: 

In [49]: keep_words = {"dogs", "cats"}

In [51]: delete_words = {"ice cream"}

In [52]: list(filterfalse(key ,word_list))
Out[52]: ['Dogs love ice cream', 'Cats love balls', 'cats hate ice cream']
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4 Comments

Kasramvd a very very very interesting technique l like it - thank you for the help
return keep_words.isdisjoint(words) or Delete_word.intersection(words) could be improved not to generate the intersection like this: return keep_words.isdisjoint(words) or not Delete_word.is_disjoint(words)
@Jean-FrançoisFabre That's even better!
also x.lower() can be replaced by x.casefold() so german "beta" is lowered as double "s". more powerful than a single str.lower
1 
>>> list(filter(lambda x: not any(i in x for i in Delete_word)
...                       or  any(i in x for i in keep_words), word_list))
['Dogs love ice cream', 'Cats love balls', 'Ice cream']

Modify this accordingly for a case-insensitive implementation.

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