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Regular expression negative lookbehind of non-fixed length
the answerer says, to match things only to ignore them. I want to use that example, but I want to print only the matches that are not ignored.
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3If you want only `bar`, why do you need to match `foo`? Do you want it to be before `bar` to match `bar`? – thefourtheye Sep 10 '14 at 02:43
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of course the real example is more complex, I just gave you an SSCCE – Mark Galeck Sep 10 '14 at 02:47
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Perhaps we need a slightly less simple example? I can't fathom why I'd want to match something I didn't want to keep. – Adam Smith Sep 10 '14 at 02:49
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@MarkGaleck that question contains an answer by me using Negative Lookahead. What exactly are you trying to do? – hwnd Sep 10 '14 at 03:34
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I can't explain that better. I apologize. Thank you everybody, I appreciate. This is embarrasing. – Mark Galeck Sep 10 '14 at 03:47
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`re.findall(r'\b(?!foo)bar', 'foo bar foo bar')` ?? – hwnd Sep 10 '14 at 03:51
2 Answers
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If you want to make sure that, foo should be followed by bar and if you are only interested in bar, then you can use look-behind assertion, like this
re.findall("(?<=foo )bar", "foo bar")
# ['bar']
Instead of bar, if you want to match anything followed by foo, you can do
re.findall("(?<=foo ).*", "foo google")
# ['google']
thefourtheye
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This will leave empty items in the list. But for what I assume you are asking, you can use the alternation operator in context placing what you want to exclude on the left, ( saying throw this away, it's garbage ) and place what you want to match in a capturing group on the right side to only print the captured matches.
>>> re.findall(r'foo|(bar)', 'foo bar foo bar')
['', 'bar', '', 'bar']
hwnd
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