Reading files in a particular order in python

Question

Lets say I have three files in a folder: file9.txt, file10.txt and file11.txt and i want to read them in this particular order. Can anyone help me with this?

Right now I am using the code

import glob, os
for infile in glob.glob(os.path.join( '*.txt')):
    print "Current File Being Processed is: " + infile

and it reads first file10.txt then file11.txt and then file9.txt.

Can someone help me how to get the right order?

Answer 1

Files on the filesystem are not sorted. You can sort the resulting filenames yourself using the sorted() function:

for infile in sorted(glob.glob('*.txt')):     print "Current File Being Processed is: " + infile 

Note that the os.path.join call in your code is a no-op; with only one argument it doesn't do anything but return that argument unaltered.

Note that your files will sort in alphabetical ordering, which puts 10 before 9. You can use a custom key function to improve the sorting:

import re numbers = re.compile(r'(\d+)') def numericalSort(value):     parts = numbers.split(value)     parts[1::2] = map(int, parts[1::2])     return parts   for infile in sorted(glob.glob('*.txt'), key=numericalSort):     print "Current File Being Processed is: " + infile 

The numericalSort function splits out any digits in a filename, turns it into an actual number, and returns the result for sorting:

>>> files = ['file9.txt', 'file10.txt', 'file11.txt', '32foo9.txt', '32foo10.txt'] >>> sorted(files) ['32foo10.txt', '32foo9.txt', 'file10.txt', 'file11.txt', 'file9.txt'] >>> sorted(files, key=numericalSort) ['32foo9.txt', '32foo10.txt', 'file9.txt', 'file10.txt', 'file11.txt'] 

Answer 2

You can wrap your glob.glob( ... ) expression inside a sorted( ... ) statement and sort the resulting list of files. Example:

for infile in sorted(glob.glob('*.txt')):

You can give sorted a comparison function or, better, use the key= ... argument to give it a custom key that is used for sorting.

Example:

There are the following files:

x/blub01.txt
x/blub02.txt
x/blub10.txt
x/blub03.txt
y/blub05.txt

The following code will produce the following output:

for filename in sorted(glob.glob('[xy]/*.txt')):
        print filename
# x/blub01.txt
# x/blub02.txt
# x/blub03.txt
# x/blub10.txt
# y/blub05.txt

Now with key function:

def key_func(x):
        return os.path.split(x)[-1]
for filename in sorted(glob.glob('[xy]/*.txt'), key=key_func):
        print filename
# x/blub01.txt
# x/blub02.txt
# x/blub03.txt
# y/blub05.txt
# x/blub10.txt

EDIT: Possibly this key function can sort your files:

pat=re.compile("(\d+)\D*$")
...
def key_func(x):
        mat=pat.search(os.path.split(x)[-1]) # match last group of digits
        if mat is None:
            return x
        return "{:>10}".format(mat.group(1)) # right align to 10 digits.

It sure can be improved, but I think you get the point. Paths without numbers will be left alone, paths with numbers will be converted to a string that is 10 digits wide and contains the number.