If any strings in a list match regex

Question

I need to check if any of the strings in a list match a regex. If any do, I want to continue. The way I've always done it in the past is using list comprehension with something like:

r = re.compile('.*search.*') if [line for line in output if r.match(line)]:   do_stuff() 

Which I now realize is pretty inefficient. If the very first item in the list matches, we can skip all the rest of the comparisons and move on. I could improve this with:

r = re.compile('.*search.*') for line in output:   if r.match(line):     do_stuff()     break 

But I'm wondering if there's a more pythonic way to do this.

Answer 1

You can use the builtin any():

r = re.compile('.*search.*') if any(r.match(line) for line in output):     do_stuff() 

Passing in the lazy generator to any() will allow it to exit on the first match without having to check any farther into the iterable.

Answer 2

Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), we can also capture a witness of an any expression when a match is found and directly use it:

# pattern = re.compile('.*search.*') # items = ['hello', 'searched', 'world', 'still', 'searching'] if any((match := pattern.match(x)) for x in items):   print(match.group(0)) # 'searched' 

For each item, this:

• Applies the regex search (pattern.match(x))
• Assigns the result to a match variable (either None or a re.Match object)
• Applies the truth value of match as part of the any expression (None -> False, Match -> True)
• If match is None, then the any search loop continues
• If match has captured a group, then we exit the any expression which is considered True and the match variable can be used within the condition's body