How can I remove characters after underscore?-CodePudding

I need to convert a text like

photo_id 102297_skdjksd223238 text black dog in a water

photo_id 102297 text black dog in a water

by removeing the substring after underscore

inputFile = open("text.txt", "r") 
exportFile = open("result", "w")
sub_str = "_"
for line in inputFile:    
   new_line = line[:line.index(sub_str)   len(sub_str)]
   exportFile.writelines(new_line)

but couldn't access the second underscore as it removed all text after photo_id ..

CodePudding user response：

^{Note: The question was tagged regex when I wrote this:}

_[^\s]*

_ - a literal _
[^\s]* - (or \S* if supported) any character but whitespaces - zero or more times

Substitute with a blank string.

Demo

inp = 'photo_id 102297_skdjksd223238 text black dog in a water foo_baz bar'

res = re.sub(r'_[^\s]*', '', inp)

print(res)

Output

photo 102297 text black dog in a water foo bar

CodePudding user response：

You could split the first underscore from the right:

s= "photo_id 102297_skdjksd223238 text black dog in a water"
prefix, suffix = s.rsplit('_', 1)
print(f"{prefix} {suffix.split(' ', 1)[-1]}")

Out:

photo_id 102297 text black dog in a water

CodePudding user response：

You might use a pattern to capture the leading digits to make it a bit more specific, and then match the underscore followed by optional non whitespace characters.

In the replacement use the first capture group.

\b(\d )_\S*

Explanation

\b A word boundary to prevent a partial word match
(\d ) Capture group 1, match 1 digits
_\S* Match an underscore and optional non whitespace characters

See a regex101 demo.

import re
 
pattern = r"\b(\d )_\S*"
s = "photo_id 102297_skdjksd223238 text black dog in a water"
 
result = re.sub(pattern, r"\1", s)
 
if result:
    print (result)

Output

photo_id 102297 text black dog in a water

Another option including photo_id and matching until the first underscore:

\b(photo_id\s [^_\s] )_\S*

See another regex101 demo.