Can string replace be written in list comprehension?

You can use functools.reduce:

from functools import reduce

text = reduce(lambda x, y: x.replace(y, ''), remove_list, text)
# 'Some texts  that I want to  replace'

You can do this with a regex by building a regex from an alternation of the words to remove, taking care to escape the strings so that the [ and ] in them don't get treated as special characters:

import re

text = "Some texts [remove me] that I want to [and remove me] replace"
remove_list = ["[remove me]", "[and remove me]"]

regex = re.compile('|'.join(re.escape(r) for r in remove_list))

text = regex.sub('', text)
print(text)

Output:

Some texts  that I want to  replace

Since this may result in double spaces in the result string, you can remove them with replace e.g.

text = regex.sub('', text).replace('  ', ' ')

Output:

Some texts that I want to replace

I would do this with re.sub to remove all the substrings in one pass:

>>> import re
>>> regex = '|'.join(map(re.escape, remove_list))
>>> re.sub(regex, '', text)
'Some texts  that I want to  replace'

Note that the result has two spaces instead of one where each part was removed. If you want each occurrence to leave just one space, you can use a slightly more complicated regex:

>>> re.sub(r'\s*(' + regex + r')', '', text)
'Some texts that I want to replace'

There are other ways to write similar regexes; this one will remove the space preceding a match, but you could alternatively remove the space following a match instead. Which behaviour you want will depend on your use-case.