How do you sort files numerically?

Question

I'm processing some files in a directory and need the files to be sorted numerically. I found some examples on sorting—specifically with using the lambda pattern—at wiki.python.org, and I put this together:

import re  file_names = """ayurveda_1.tif ayurveda_11.tif ayurveda_13.tif ayurveda_2.tif ayurveda_20.tif ayurveda_22.tif""".split('\n')  num_re = re.compile('_(\d{1,2})\.')  file_names.sort(     key=lambda fname: int(num_re.search(fname).group(1)) ) 

Is there a better way to do this?

Answer 1

This is called "natural sorting" or "human sorting" (as opposed to lexicographical sorting, which is the default). Ned B wrote up a quick version of one.

import re  def tryint(s):     try:         return int(s)     except:         return s  def alphanum_key(s):     """ Turn a string into a list of string and number chunks.         "z23a" -> ["z", 23, "a"]     """     return [ tryint(c) for c in re.split('([0-9]+)', s) ]  def sort_nicely(l):     """ Sort the given list in the way that humans expect.     """     l.sort(key=alphanum_key) 

It's similar to what you're doing, but perhaps a bit more generalized.

Answer 2

Just use :

tiffFiles.sort(key=lambda var:[int(x) if x.isdigit() else x for x in re.findall(r'[^0-9]|[0-9]+', var)]) 

is faster than use try/except.