I want to match digits betwen "000" or betwen \b and "000" or "000" and \b from a string like this:
11101110001011101000000011101010111
I have tried with expressions like this:
(?<=000)\d+(?=000)
but I only get the largest occurrence
I expect to get:
1110111
1011101
0
11101010111
You can use the regex package and the .findall() method:
In [1]: s = "11101110001011101000000011101010111"
In [2]: import regex
In [3]: regex.findall(r"(?<=000|^)\d+?(?=000|$)", s)
Out[3]: ['1110111', '1011101', '0', '00011101010111']
The 000|^ and 000|$ would help to match either the 000 or the beginning and the end of a string respectively. Also note the ? after the \d+ - we are making it non-greedy.
Note that the regular re.findall() would fail with the following error in this case:
error: look-behind requires fixed-width pattern
This is because re does not support variable-length lookarounds but regex does.
(?<=000)[^0]\d+?(?=000|\b)|(?<=\b)[^0]\d+?(?=000|\b) but now it doesn't get the 0 betwen "000" and "000"(?<=000)[^0]\d+?(?=000|\b)|(?<=\b)[^0]\d+?(?=000|\b)|(?<=000)\d(?=000)you can do it with the re module like this:
re.findall(r'(?:\b|(?<=000))(\d+?)(?:000|\b)', s)
