Python regex split without empty string

Question

I have the following file names that exhibit this pattern:

000014_L_20111007T084734-20111008T023142.txt
000014_U_20111007T084734-20111008T023142.txt
...


        

Answer 1

I'm no Python expert but maybe you could just remove the empty strings from your list?

str_list = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
time_info = filter(None, str_list)

Answer 2

>>> f='000014_L_20111007T084734-20111008T023142.txt'
>>> f[10:-4].split('-')
['0111007T084734', '20111008T023142']

or, somewhat more general:

>>> f[f.rfind('_')+1:-4].split('-')
['20111007T084734', '20111008T023142']

Answer 3

Since this came up on google and for completeness, try using re.findall as an alternative!

This does require a little re-thinking, but it still returns a list of matches like split does. This makes it a nice drop-in replacement for some existing code and gets rid of the unwanted text. Pair it with lookaheads and/or lookbehinds and you get very similar behavior.

Yes, this is a bit of a "you're asking the wrong question" answer and doesn't use re.split(). It does solve the underlying issue- your list of matches suddenly have zero-length strings in it and you don't want that.

Answer 4

Don't use re.split(), use the groups() method of regex Match/SRE_Match objects.

>>> f = '000014_L_20111007T084734-20111008T023142.txt'
>>> time_info = re.search(r'[LU]_(\w+)-(\w+)\.', f).groups()
>>> time_info
('20111007T084734', '20111008T023142')

You can even name the capturing groups and retrieve them in a dict, though you use groupdict() rather than groups() for that. (The regex pattern for such a case would be something like r'[LU]_(?P<groupA>\w+)-(?P<groupB>\w+)\.')

Answer 5

If the timestamps are always after the second _ then you can use str.split and str.strip:

>>> strs = "000014_L_20111007T084734-20111008T023142.txt"
>>> strs.strip(".txt").split("_",2)[-1].split("-")
['20111007T084734', '20111008T023142']