As Paul mentioned, your regex can either cover a limited amount of cases,
or as general of cases as possible. Yours was pretty limited to specific
delimiters and 3-character extensions. Bot yours and Paul's corrections
also don't account for negative frame numbers, if you have sequences that
maybe are simulation ramp-ups. You could do something like this to go one
step farther:
rx = re.compile(r'[._](-?\d+)\.([0-9a-z]+)$', re.I)
print rx.search("foo_bar_-1234.JPG").groups()
While the delimiter is still limited to two types of characters, it now
only allows numbers and letters for the extension (of any length) and it
still insensitive.
On Thu, Jul 31, 2014 at 7:56 AM, Paul Molodowitch <[email protected]>
wrote:
> Well... string parsing is always going to be be a bit of a crapshoot.
> Especially if you're trying to come up with something like parsing out the
> framenumbers, for a very general case, there will always be things that
> break it. In general, it's best if you can enforce some sort of uniformity
> on your filenames, so you can have a certain amount of assurance that it's
> going to work as intended.
>
> However - to your specific regex - if you're looking specifically to catch
> cases that look like:
>
> filename.####.ext
> filename_####.ext.
>
> ...you're on the right track, but there's a few things I would fix.
>
> First, some general advice - ALWAYS get in the habit of prefixing your
> regex pattern strings with "r", like: r'myPattern'
> The 'r' marks it as a raw string, meaning backslashes aren't treated as
> escape sequences. Since python keeps the backslash for sequences it
> doesn't recognize, MOST of the time it won't make a difference, but it's
> best to just get in the habit now. I also like to always triple quote my
> regex patterns, for cases when I need to search for a quote mark...
>
> Second - you probably want to make sure that your expression only matches
> at the END of the string - easiest way to do that is with a "$" character
> at the end. otherwise, if your file is:
>
> f = "myawesome_filename_ver_1_frame.0001.exr"
>
> ...you'll get the wrong result. Also, I wouldn't limit your file
> extensions to only 3 characters - ".jpeg" and ".tiff" pop up pretty
> frequently. So at least allow 4 characters (and possibly just make it 1+).
> Finally, you'll probably want to start using groups - at minimum, around
> the framenumbers. So my final take would look something like:
>
> pat = re.compile*(r'''[.|_](\d+)\.(\w{3,4})$''')*
> print pat.search(f).groups()
>
> If you want to get fancy, python regular expressions allow you to name
> your groups, which I like. The main downside is that this is a python only
> extension, so if you try to copy your regex somewhere else, it won't work:
>
> pat = re.compile*(r'''[.|_](?P<frame>\d+)\.(?P<ext>\w{3,4})$''')*
> print pat.search(f).groupdict()
>
>
> - Paul
>
> PS - Is that Kurt Russel from "The Thing", or do you just happen to look
> just like him? Either way, it's awesome...
>
>
> On Wed, Jul 30, 2014 at 12:03 PM, md <[email protected]> wrote:
>
>> Hey guys,
>>
>> I am fairly new to regular expressions.
>>
>> When I want to get the delimiter between a filename and framenumber I am
>> using the following regex.
>>
>> f = myawesome_filename_v1_0001.exr"
>>
>> pat = re.compile*('[\.|\_]\d+[\.]\w{3}')*
>> x = re.search(pat, f)
>> print x.group()
>> delimiter = x.group()[0]
>> print 'delimiter : ' + delimiter
>>
>> This works specifically for the case where you might have
>> filename.####.ext or filename_####.ext.
>>
>> Question is ... is this a decent/efficient way to do this ?
>>
>>
>>
>>
>>
>>
>>
>>
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