How to remove word from a string if it contains a specific character?

Question

Ok, lets say my string contains the following: "$$$FOOD WATERMELON" . How would I remove the "$$$FOOD" from the string?

And lets say I have strings in a list:

data_list = ["$$$FOOD WATERMELON", "$$$STORY I walked to the local store"]

The method that I have in my code splits the elements in the list, then iterates through the lists inside of the list and removes any element that contains "$$$" which would work fine, if it weren't for the fact that the .split() function splits every word, so the list would end up looking like this: [["WATERMELON"],["I","walked","to","the","local,"store"]] which is not optimal, because then I would have to join the elements in the lists of the list, that takes more time.

Basically, the only thing I am wondering is: how do I remove a word in a string if it contains "$$$" . So this string: word = "$$$STORY I walked to the store" would become "I walked to the store"

Sorry if this was confusing

Answer 1

You can try this:

>>> word = "$$$STORY I walked to the store"
>>> if word.startswith('$',0,word.index(' ')):
        word=word[word.index(' ')+1:]

>>> word
'I walked to the store'

Here is how it looks using the list and the function:

def check_word(word):
    if word.startswith('$',0,word.index(' ')):
        return word[word.index(' ')+1:]
data_list = ["$$$FOOD WATERMELON", "$$$STORY I walked to the local store"]
list2=[check_word(x) for x in data_list]
print(list2)

Answer 2

Try this one.

import re

data_list = ["$$$FOOD WATERMELON", "$$$STORY I walked to the local store"]
output = [re.sub(r'\s*[$]+\w+\s*', '', x) for x in data_list]

This gave me the following output:

['WATERMELON', 'I walked to the local store']

Here's regex definition:

\s
matches any whitespace character (equivalent to [\r\n\t\f\v ])
* matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below [$]
+ matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
$ matches the character $ with index 3610 (2416 or 448) literally (case sensitive)
\w
matches any word character (equivalent to [a-zA-Z0-9_])
+ matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
\s
matches any whitespace character (equivalent to [\r\n\t\f\v ])
* matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)

Answer 3

This code checks and ensures there is no substring at any word, not just the first word like some of the other answers.

Output:

['WATERMELON', 'I walked to the local store']

Code:

data_list = ["$$$FOOD WATERMELON", "$$$STORY I walked to the local store"]

for word in range(len(data_list)):
  temp1 = data_list[word].split(" ")
  finished = False
  while not (finished):
      for temp2 in range(len(temp1)):
          if "$$$" in temp1[temp2]:
              temp1.pop(temp2)
              break
          finished = True
  data_list[word] = ' '.join(temp1)

print(data_list)