This is a script I wrote to find the n biggest files in a given directory (recursively):

import heapq
import os, os.path
import sys
import operator

def file_sizes(directory):
    for path, _, filenames in os.walk(directory):
        for name in filenames:
            full_path = os.path.join(path, name)
            yield full_path, os.path.getsize(full_path)

num_files, directory = sys.argv[1:]
num_files = int(num_files)

big_files = heapq.nlargest(
        num_files, file_sizes(directory), key=operator.itemgetter(1))
print(*("{}\t{:>}".format(*b) for b in big_files))

It can be run as, eg: bigfiles.py 5 ~.

Ignoring the complete lack of error handling, is there any obvious way to make this clearer, or at least more succinct? I am thinking about, eg, using namedtuples in file_sizes, but is there also any way to implement file_sizes in terms of a generator expression? (I'm thinking probably not without having two calls to os.path, but I'd love to be proven wrong :-)


1 Answer 1


You could replace your function with:

file_names = (os.path.join(path, name) for path, _, filenames in os.walk(directory)
        for name in filenames)

file_sizes = ((name, os.path.getsize(name)) for name in file_names)

However, I'm not sure that really helps the clarity.

I found doing this:

big_files = heapq.nlargest(
        num_files, file_names, key=os.path.getsize)
print(*("{}\t{:>}".format(b, os.path.getsize(b)) for b in big_files))

Actually runs slightly quicker then your version.

  • \$\begingroup\$ That's an interesting result. How did you time it? You wouldn't think that twice as many syscalls would speed it up! \$\endgroup\$
    – lvc
    Feb 15, 2012 at 2:47
  • \$\begingroup\$ @lvc, I used time python script.py. Its not making twice as many sys calls because I'm only making the second call for the top 5 or so files. \$\endgroup\$ Feb 15, 2012 at 3:59
  • \$\begingroup\$ Ah, indeed, it will only be num_files extra syscalls. Presumably, what's happening is that constructing multiple tuples is more expensive - if that's the case, we could expect the difference to shrink and eventually go away as num_files approaches the number of files searched. But, when I look at it, I think I do prefer key=os.path.getsize over my many tuples for clarity, anyway. \$\endgroup\$
    – lvc
    Feb 15, 2012 at 5:06

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