Replace operation inside a Python for loop

Question

my code below works fine.

import re

def mycontent(text: str):
    netx = text
    urls = re.findall("MyText", netx)
    for i in urls:
        netx = netx.replace(i, "MyReplaceText")
    
    return netx

I want to combine this word with a few more words. How can I do that? Thank you.

import re

def mycontent(text: str):
    netx = text
    urls = re.findall("MyText", "MyText1" tnetx)
    for i in urls:
        netx = netx.replace(i, "MyReplaceText", "MyText1Replace"")

    return netx

I tried in the form but I couldn't get results, it didn't work.

Just put all the words in a single string (`"MyReplaceTextMyReplaceText"`). If that's not what you mean, please clarify. — Dan, Dec 12 '22 at 23:25
Your code does not compile. `re.findall(pattern, string, flags=0)` takes only one `pattern` parameter. You should construct a regular expression for your pattern. — Yuri Ginsburg, Dec 12 '22 at 23:32
Why are you using the loop instead of `re.sub()` to replace all the regexp matches? — Barmar, Dec 12 '22 at 23:41
don't use `re`, just have a list of tuple pairs of find and replace strings, and loop through those applying `.replace(find_str, replace_str)` — bn_ln, Dec 12 '22 at 23:44
Does this answer your question? https://stackoverflow.com/q/6116978/6045800 — Tomerikoo, Dec 12 '22 at 23:55

score 0 · Answer 1 · answered Dec 12 '22 at 23:47

To match multiple strings, use alternatives in the regexp. Then use re.sub() with a replacement function that looks up the corresponding replacement.


def mycontent(text: str) -> str:
    replacements = {'MyText': 'MyReplacement1', 'MyText1', 'MyReplacement1'}
    return re.sub(r'MyText|MyText1', lambda m: replacements[m.group(), text]

Replace operation inside a Python for loop

1 Answers1