Python的glob.glob如何排序?

问题:Python的glob.glob如何排序?

我编写了以下Python代码:

#!/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

现在我明白了:

/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

它以哪种方式订购?

它可能会帮助您获得我的ls -l输出:

-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

它不是按文件名或大小排序的。

相关链接:globls

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?

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


回答 0

它可能根本没有排序,并使用条目在文件系统中出现的顺序,即使用时所获得的顺序ls -U。(至少在我的机器上,这产生与列表glob匹配相同的顺序)。

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


回答 1

顺序是任意的,但您可以自己对其进行排序

如果要按名称排序:

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

按修改时间排序:

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

按大小排序:

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

等等

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.


回答 2

通过检查您的源代码,glob.glob您可以看到它内部调用os.listdir,如下所述:

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

关键字:os.listdir(path)返回一个列表,其中包含由path给出的目录中条目的名称。该列表是任意顺序的。它不包括特殊条目“。” 和“ ..”,即使它们存在于目录中。

任意秩序。:)

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. :)


回答 3

glob.glob()是os.listdir()的包装,因此底层操作系统负责传递数据。一般而言:您不能在此假设订购。基本假设是:顺序。如果需要某种排序:在应用程序级别上排序。

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.


回答 4

顺序是任意的,但是有几种方法可以对它们进行排序。其中之一如下:

#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]))

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]))

回答 5

我有一个类似的问题,glob正在以任意顺序返回文件名列表,但是我想按照文件名指示的数字顺序逐步浏览它们。这是我实现的方式:

我的文件通过glob类似以下方式返回:

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"]

我对列表进行了排序,为此我创建了一个函数:

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

该函数返回文件名的数字部分并转换为整数,然后我在列表上这样调用sort方法:

myList.sort(key=sortKeyFunc)

这样返回一个列表:

["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"]

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"]

回答 6

如果您想知道glob.glob过去在系统上做了什么而无法添加sorted调用,则该顺序在Mac HFS +文件系统上将是一致的,并且在其他Unix系统上将是遍历顺序。因此除非重新组织了底层文件系统,否则可能是确定性的,如果添加,删除,重命名,删除,移动文件等情况发生,则可能会发生重组。

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…


回答 7

在@Johan La Rooy的解决方案中,使用排序图像sorted(glob.glob('*.png'))对我不起作用,输出列表仍未按其名称排序。

但是,sorted(glob.glob('*.png'), key=os.path.getmtime)作品完美。

我有点困惑,如何按他们的名字排序在这里不起作用。

感谢@Martin Thoma发布这个重要问题,并感谢@Johan La Rooy提供有用的解决方案。

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.


回答 8

请尝试以下代码:

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

Please try this code:

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

回答 9

'''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])

这就是我处理特定案件的方式。希望对您有所帮助。

'''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.