Use replace function with list of strings python?-CodePudding

I have a bunch of replace functions to modify a list of strings that I have . Originally, I had written out each list comprehension of what strings I want to be replaced.

ls = ['sentence with string a', 'sentence with string d and string a', 
      'sentence with string d', 'sentence with string b and string b again',
      'sentence with string c', 'sentence with string c and string d']

ls = [x.replace('string a' , '') for x in ls]
ls = [x.replace('string b' , '') for x in ls]
ls = [x.replace('string c' , '') for x in ls]
ls = [x.replace('string d' , '') for x in ls]

ls = ['sentence with', 'sentence with and', 
      'sentence with', 'sentence with and again',
      'sentence with', 'sentence with and']

However, since the strings I want to replace may change I want to to use a list to do all the replace functions. When I try a for loop, this ends up removing characters from the middle of the string and doesn't give me the list like before.

words = ['string a', 'string b', 'string c', 'string d']

for txt in words:
        ls = [x.replace(txt, "") for x in ls]

How can I use the list of strings to achieve the same result?

CodePudding user response：

You can use a comprehension and functools.reduce:

from functools import reduce

words = ['string a', 'string b', 'string c', 'string d']

ls = ['sentence with string a', 'sentence with string d and string a', 
      'sentence with string d', 'sentence with string b and string b again',
      'sentence with string c', 'sentence with string c and string d']


[reduce(lambda s, w: s.replace(w, ""), words, sent) for sent in ls]
# ['sentence with ', 'sentence with  and ', 'sentence with ', 
#  'sentence with  and  again', 'sentence with ', 'sentence with  and ']

CodePudding user response：

Using re:

reg = re.compile('string [abcd]')

Then you can just use that compiled regex in the comprehension:

[reg.sub('', s) for s in ls]

Result:

['sentence with ',
 'sentence with  and ',
 'sentence with ',
 'sentence with  and  again',
 'sentence with ',
 'sentence with  and ']

CodePudding user response：

Seems that Jab was faster than me, but in case you want to keep your list of words more flexible, here's a similar solution:

import re

pattern = '|'.join(words)
[re.sub(pattern , '', x).strip() for x in ls]

results in

['sentence with',
 'sentence with  and',
 'sentence with',
 'sentence with  and  again',
 'sentence with',
 'sentence with  and']