python - enumerating numbers in strings -

python - enumerating numbers in strings -

this seems problem have straight-forward answer; sadly, not fluent in python i'm still learning, , have not been able find helpful on google.

my goal enumerate numbers in string based on how much padding number has. think best way describe example:

0-file enumerated 0-file 9-file but 000-file enumerated 000-file 999-file.

ultimately want able [number][a-z], [a-z][number], , [a-z][number].* (so file10name.so wouldn't match), think can figure part out myself regex later on.

so, question boils downwards this:

how length of 'padding' in file? how identify in string number is, can replace it? how add together padding when i'm iterating (i'm assuming zfill, i'm interested if there's improve method).

quick edit: yes, 'psudo regex' that. concept conveyed, hence why wouldn't match things "-". padding number, not 0, thats alright. both answers received far perfect. can adapt them needs. im handeling total paths, great have there other people see in future. :)

you should figure out right specification files you're trying match before coding up. pseudo-regexps gave filenames trying match ("[number][a-z] or [a-z][number]") don't include examples gave, such 0-file.

simple version

however, taking stated specification @ face value, assuming wish include uppercase latin letters well, here's simple function match [number][a-z] or [a-z][number], , homecoming appropriate prefix, suffix, , number of numeric digits.

import re def find_number_in_filename(fn): m = re.match(r"(\d+)([a-za-z]+)$", fn) if m: prefix, suffix, num_length = "", m.group(2), len(m.group(1)) homecoming prefix, suffix, num_length m = re.match(r"([a-za-z]+)(\d+)$", fn) if m: prefix, suffix, num_length = m.group(1), "", len(m.group(2)) homecoming prefix, suffix, num_length homecoming fn, "", 0 example_fn = ("000foo", "bar14", "baz0", "file10name") fn in example_fn: prefix, suffix, num_length = find_number_in_filename(fn) if num_length == 0: print "%s: not match" % fn else: print "%s -> %s[%d-digits]%s" % (fn, prefix, num_length, suffix) all_numbered_versions = [("%s%0"+str(num_length)+"d%s") % (prefix, ii, suffix) ii in range(0,10**num_length)] print "\t", all_numbered_versions[0], "through", all_numbered_versions[-1]

the output be:

000foo -> [3-digits]foo 000foo through 999foo bar14 -> bar[2-digits] bar00 through bar99 baz0 -> baz[1-digits] baz0 through baz9 file10name: not match

notice i'm using standard printf-style string format convert numbers 0-padded strings, e.g. %03d 3-digit numbers 0-padding. using newer str.format may preferable future-proofing.

handle total paths , extensions gracefully

if input includes total paths , filenames extensions (e.g. /home/someone/project/foo000.txt) , want match based on lastly piece of path only, utilize os.path.split , .splitext trick.

update: fixed missing path separator

import re import os.path def find_number_in_filename(path): # remove path , extension head, tail = os.path.split(path) head = os.path.join(head, "") # include / or \ on end of head if it's missing fn, ext = os.path.splitext(tail) m = re.match(r"(\d+)([a-za-z]+)$", fn) if m: prefix, suffix, num_length = head, m.group(2)+ext, len(m.group(1)) homecoming prefix, suffix, num_length m = re.match(r"([a-za-z]+)(\d+)$", fn) if m: prefix, suffix, num_length = head+m.group(1), ext, len(m.group(2)) homecoming prefix, suffix, num_length homecoming path, "", 0 example_paths = ("/tmp/bar14.so", "/home/someone/0000baz.txt", "/home/someone/baz00bar.zip") path in example_paths: prefix, suffix, num_length = find_number_in_filename(path) if num_length == 0: print "%s: not match" % path else: print "%s -> %s[%d-digits]%s" % (path, prefix, num_length, suffix) all_numbered_versions = [("%s%0"+str(num_length)+"d%s") % (prefix, ii, suffix) ii in range(0,10**num_length)] print "\t", all_numbered_versions[0], "through", all_numbered_versions[-1]

python string replace enumeration