在python中以特定顺序读取文件 [英] Reading files in a particular order in python

问题描述

让我们说我在一个文件夹中有三个文件:file9.txt，file10.txt和file11.txt，我想按此特定顺序阅读它们.有人可以帮我吗?

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?

现在我正在使用代码

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

它首先读取file10.txt，然后读取file11.txt，然后读取file9.txt.

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

有人可以帮助我如何获得正确的订单吗?

Can someone help me how to get the right order?

推荐答案

文件系统上的文件未排序.您可以使用 sorted()函数:

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

请注意，代码中的os.path.join调用是无操作的；仅带有一个参数，它什么也不会做，只会使该参数保持不变.

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.

请注意，您的文件将按字母顺序排序，这会将10放在9之前.您可以使用自定义键功能来改善排序:

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

numericalSort函数将文件名中的所有数字分开，将其转换为实际数字，然后返回结果进行排序:

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']

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