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 ... 

I want to extract the middle two time stamp parts after the second underscore '_' and before '.txt'. So I used the following Python regex string split:

time_info = re.split('^[0-9]+_[LU]_|-|\.txt$', f) 

But this gives me two extra empty strings in the returned list:

time_info=['', '20111007T084734', '20111008T023142', ''] 

How do I get only the two time stamp information? i.e. I want:

time_info=['20111007T084734', '20111008T023142'] 

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

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+)\.')