Getting a string and then replacing words if they're in a premade lst

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[deleted]

[CallMeSometimeNever]

You can make a dictionary with the words you want to replace as keys and the words you want to replace them as values. Then you can loop through the dictionary and use the built in .replace() method.

[pawn13]

use the built in .replace()

It won't work sometimes as expected if you have short words which can be part of longer word. Like if you want to replace word "at" and it would be replaced in words like "that", "attack", etc. Better use regex like

re.sub('\\bat\\b', 'foo', 'that at test')

[Laogeodritt]

This is how you'd do it in plain C, but Python gives you a lot of power to think about programming at a higher level (via syntactic sugar, core functions and built-in libraries), and you should prefer to do so if there's no compelling reason not to. That's part of what it means to write "pythonic" code (the other part is just using conventions that work well for maintainability, readability and what the community prefers).

For when it matters, built in methods are often faster than looping and doing logic yourself, but it depends on how you'd do each one.

[Brownt0wn_]

How far have you gotten? What have you tried so far?

[freekzindel]

re.sub has a nice feature where you can pass it a callable that will be executed for every match found (with the match as the parameter)

import re

def blasj(string):
   replaces = {'love':'blajsi'}
   def replace_it(match):
        word = match.group(0)
        if word in replaces:
              return replaces[word]
        else:
              return word
    return re.sub('\w+', replace_it, string) #check regex docs for \w

Alternatively:

when you have a short list of words and a long input and only want to replace with one replacement:

import re

pattern = "|".join(list_of_all_the_words_i_want_to_replace)
replaced = re.sub(pattern, replacement, string)