Ignore optional suffix without leading delimiter
I would like to spell the first part of a word, ignoring the optional suffix. Both the suffix and the preceding text consist of the same character class (i.e., there is no delimiter before the suffix).
My first attempt only captures the first letter:
m = re.search(r'([A-Za-z]+?)(?:Suff)?', 'textSuff')
m.groups()
>>> ('t',)
I only want to grab "text", but when I make the first element of the group greedy, it grabs the whole line.
m = re.search(r'([A-Za-z]+)(?:Suff)?', 'textSuff')
m.groups()
>>> ('textSuff',)
Is it possible to limit the suffix without a different character?
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2 answers
If your template is built from additional templates, make sure you get as few characters as possible. Thus, there must be at least a border. I am assuming word boundary \b
is a valid way (you need to match words here):
([A-Za-z]+?)(?:Suff)?\b
Watch the demo
import re
p = re.compile(r'([A-Za-z]+?)(?:Suff)?\b')
test_str = "textSuff more words tSuff"
print(re.findall(p, test_str))
Outputs:
['text', 'more', 'words', 't']
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