Is there a way to improve this regular expression to search for all words that ends with t, including don't? I also want to print the whole words, not just the last t.
r"\b\w*\Wt\b|\b\w*t\b"
I had to write out 2 separate cases for ending with either t or 't. Or this is the best it could be?
I'd use \b\S*t\b. It fixes the problem of the engine having to scan a word only to fail to find the non-word character and try the other branch in your pattern. At the very least, swap the two sides of the alternation because the common-case is that the word won't have a contraction.
>>> import re
>>> s = "mitt cat bat don't foobar"
>>> re.findall(r"\b\S*t\b", s)
['mitt', 'cat', 'bat', "don't"]
It's not clear how you want to treat non-word punctuation, but consider a variant that attempts to handle this:
>>> s = "mitt cat bat. don't foobar tee t e.t."
>>> re.findall(r"\b\S*t\b", s)
['mitt', 'cat', 'bat', "don't", 't', 'e.t']
>>> re.findall(r"\b[^.,!?\s]*t\b", s)
['mitt', 'cat', 'bat', "don't", 't', 't']
Clearly, abbreviations and edge cases may need attention if that's part of your specification.
Do not rely on generic patterns if all you want is allow an apostrophe. \W matches spaces, too. \S matches any characters different from whitespace.
Use
r"\b\w+'?t\b"
See regex proof.
EXPLANATION
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\b the boundary between a word char (\w) and
something that is not a word char
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\w+ word characters (a-z, A-Z, 0-9, _) (1 or
more times (matching the most amount
possible))
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'? '\'' (optional (matching the most amount
possible))
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t 't'
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\b the boundary between a word char (\w) and
something that is not a word char