Python 的 glob.glob 是如何排序的？

Question

I have written the following Python code:

#!/usr/bin/python
# -*- coding: utf-8 -*-

import os, glob

path = '/home/my/path'
for infile in glob.glob( os.path.join(path, '*.png') ):
    print infile

Now I get this:

/home/my/path/output0352.png
/home/my/path/output0005.png
/home/my/path/output0137.png
/home/my/path/output0202.png
/home/my/path/output0023.png
/home/my/path/output0048.png
/home/my/path/output0069.png
/home/my/path/output0246.png
/home/my/path/output0071.png
/home/my/path/output0402.png
/home/my/path/output0230.png
/home/my/path/output0182.png
/home/my/path/output0121.png
/home/my/path/output0104.png
/home/my/path/output0219.png
/home/my/path/output0226.png
/home/my/path/output0215.png
/home/my/path/output0266.png
/home/my/path/output0347.png
/home/my/path/output0295.png
/home/my/path/output0131.png
/home/my/path/output0208.png
/home/my/path/output0194.png

In which way is it ordered?

To clarify: I am not interested in ordering - I know sorted . I want to know in which order it comes by default.

It might help you to get my ls -l output:

-rw-r--r-- 1 moose moose 627669 2011-07-17 17:26 output0005.png
-rw-r--r-- 1 moose moose 596417 2011-07-17 17:26 output0023.png
-rw-r--r-- 1 moose moose 543639 2011-07-17 17:26 output0048.png
-rw-r--r-- 1 moose moose 535384 2011-07-17 17:27 output0069.png
-rw-r--r-- 1 moose moose 543216 2011-07-17 17:27 output0071.png
-rw-r--r-- 1 moose moose 561776 2011-07-17 17:27 output0104.png
-rw-r--r-- 1 moose moose 501865 2011-07-17 17:27 output0121.png
-rw-r--r-- 1 moose moose 547144 2011-07-17 17:27 output0131.png
-rw-r--r-- 1 moose moose 530596 2011-07-17 17:27 output0137.png
-rw-r--r-- 1 moose moose 532567 2011-07-17 17:27 output0182.png
-rw-r--r-- 1 moose moose 553562 2011-07-17 17:27 output0194.png
-rw-r--r-- 1 moose moose 574065 2011-07-17 17:27 output0202.png
-rw-r--r-- 1 moose moose 552197 2011-07-17 17:27 output0208.png
-rw-r--r-- 1 moose moose 559809 2011-07-17 17:27 output0215.png
-rw-r--r-- 1 moose moose 549046 2011-07-17 17:27 output0219.png
-rw-r--r-- 1 moose moose 566661 2011-07-17 17:27 output0226.png
-rw-r--r-- 1 moose moose 561678 2011-07-17 17:27 output0246.png
-rw-r--r-- 1 moose moose 525550 2011-07-17 17:27 output0266.png
-rw-r--r-- 1 moose moose 565715 2011-07-17 17:27 output0295.png
-rw-r--r-- 1 moose moose 568381 2011-07-17 17:28 output0347.png
-rw-r--r-- 1 moose moose 532768 2011-07-17 17:28 output0352.png
-rw-r--r-- 1 moose moose 535818 2011-07-17 17:28 output0402.png

It is not ordered by filename or size.

Other links: glob , ls

Answer 1

Order is arbitrary, but you can sort them yourself

If you want sorted by name:

sorted(glob.glob('*.png'))

sorted by modification time:

import os
sorted(glob.glob('*.png'), key=os.path.getmtime)

sorted by size:

import os
sorted(glob.glob('*.png'), key=os.path.getsize)

etc.

Answer 2

It is probably not sorted at all and uses the order at which entries appear in the filesystem, ie the one you get when using ls -U . (At least on my machine this produces the same order as listing glob matches).

Answer 3

By checking the source code of glob.glob you see that it internally calls os.listdir , described here:

http://docs.python.org/library/os.html?highlight=os.listdir#os.listdir

Key sentence: os.listdir(path) Return a list containing the names of the entries in the directory given by path. The list is in arbitrary order. It does not include the special entries '.' and '..' even if they are present in the directory.

Arbitrary order . :)

Answer 4

Order is arbitrary, but there are several ways to sort them. One of them is as following:

#First, get the files:
import glob
import re
files =glob.glob1(img_folder,'*'+output_image_format)
# if you want sort files according to the digits included in the filename, you can do as following:
files = sorted(files, key=lambda x:float(re.findall("(\d+)",x)[0]))

Answer 5

glob.glob() is a wrapper around os.listdir() so the underlaying OS is in charge for delivering the data. In general: you can not make an assumption on the ordering here. The basic assumption is: no ordering. If you need some sorting: sort on the application level.

Answer 6

I had a similar issue, glob was returning a list of file names in an arbitrary order but I wanted to step through them in numerical order as indicated by the file name. This is how I achieved it:

My files were returned by glob something like:

myList = ["c:\tmp\x\123.csv", "c:\tmp\x\44.csv", "c:\tmp\x\101.csv", "c:\tmp\x\102.csv", "c:\tmp\x\12.csv"]

I sorted the list in place, to do this I created a function:

def sortKeyFunc(s):
    return int(os.path.basename(s)[:-4])

This function returns the numeric part of the file name and converts to an integer.I then called the sort method on the list as such:

myList.sort(key=sortKeyFunc)

This returned a list as such:

["c:\tmp\x\12.csv", "c:\tmp\x\44.csv", "c:\tmp\x\101.csv", "c:\tmp\x\102.csv", "c:\tmp\x\123.csv"]

Answer 7

From @Johan La Rooy's solution, sorting the images using sorted(glob.glob('*.png')) does not work for me, the output list is still not ordered by their names.

However, the sorted(glob.glob('*.png'), key=os.path.getmtime) works perfectly.

I am a bit confused how can sorting by their names does not work here.

Thank @Martin Thoma for posting this great question and @Johan La Rooy for the helpful solutions.

Answer 8

If you're wondering about what glob.glob has done on your system in the past and cannot add a sorted call, the ordering will be consistent on Mac HFS+ filesystems and will be traversal order on other Unix systems. So it will likely have been deterministic unless the underlying filesystem was reorganized which can happen if files were added, removed, renamed, deleted, moved, etc...

Answer 9

At least in Python3 you also can do this:

import os, re, glob

path = '/home/my/path'
files = glob.glob(os.path.join(path, '*.png'))
files.sort(key=lambda x:[int(c) if c.isdigit() else c for c in re.split(r'(\d+)', x)])
for infile in files:
    print(infile)

This should lexicographically order your input array of strings (eg respect numbers in strings while ordering).

Answer 10

I used the built in sorted so solve this problem:

from pathlib import Path

p = Path('/home/my/path')
sorted(list(p.glob('**/*.png')))

Answer 11

Please try this code:

sorted(glob.glob( os.path.join(path, '*.png') ),key=lambda x:float(re.findall("([0-9]+?)\.png",x)[0]))

Answer 12

'''my file name is 
"0_male_0.wav", "0_male_2.wav"... "0_male_30.wav"... 
"1_male_0.wav", "1_male_2.wav"... "1_male_30.wav"... 
"8_male_0.wav", "8_male_2.wav"... "8_male_30.wav"

when I wav.read(files) I want to read them in a sorted torder, i.e., "0_male_0.wav"
"0_male_1.wav"
"0_male_2.wav" ...
"0_male_30.wav"
"1_male_0.wav"
"1_male_1.wav"
"1_male_2.wav" ...
"1_male_30.wav"
so this is how I did it.

Just take all files start with "0_*" as an example. Others you can just put it in a loop
'''

import scipy.io.wavfile as wav
import glob 
from os.path import isfile, join

#get all the file names in file_names. THe order is totally messed up
file_names = [f for f in listdir(audio_folder_dir) if isfile(join(audio_folder_dir, f)) and '.wav' in f] 
#find files that belongs to "0_*" group
filegroup0 = glob.glob(audio_folder_dir+'/0_*')
#now you get sorted files in group '0_*' by the last number in the filename
filegroup0 = sorted(filegroup0, key=getKey)

def getKey(filename):
    file_text_name = os.path.splitext(os.path.basename(filename))  #you get the file's text name without extension
    file_last_num = os.path.basename(file_text_name[0]).split('_')  #you get three elements, the last one is the number. You want to sort it by this number
    return int(file_last_num[2])

That's how I did my particular case. Hope it's helpful.