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列出文件夹中的png文件

分类: 技术问答 • 2022-05-03 20:54:54
问题描述：

我正在尝试在包含其他文件夹的文件夹中列出所有png文件的列表.这是我的代码.

I'm trying to make a list of all png files in a folder that contains other folders. Here's my code.

import os
filelist=os.listdir('images')
for fichier in filelist:
    if not(fichier.endswith(".png")):
        filelist.remove(fichier)
print(filelist)

问题是，最后一张照片显示某些子文件夹(但不是全部)逃脱了剔除...

problem is, that last print shows that some of the subfolders (but not all) escaped the culling...

['aa.png', 'Nouveau dossier - Copie', 'Nouveau dossier - Copie (3)', 'Nouveau dossier - Copie (5)', 'zz.png']

我在做什么错了?

答

摘要:永远，永远，永远修改要迭代的列表.

Summary: Never, ever, ever modify the list that you are iterating over.

相反，遍历一个副本:

import os
filelist=os.listdir('images')
for fichier in filelist[:]: # filelist[:] makes a copy of filelist.
    if not(fichier.endswith(".png")):
        filelist.remove(fichier)
print(filelist)

或者，如果您不希望制作不必要的副本，请反向进行迭代(仅当您可以保证列表中的项目是唯一的时才有效；对于文件系统，这是一个很好的假设):

Or if you don't like to make unnecessary copies, iterate in reverse (this will only work if you can guarantee that the items in the list are unique; for a file system this is a good assumption):

for fichier in reversed(filelist):
    # do stuff

请注意，您可以使用Python的glob模块来简化此操作:

Note that you can use Python's glob module to simplify this:

import glob
print(glob.glob('images/*.png'))

原因

当您遍历Python中的列表时，在后台，Python实际上正在遍历该列表的索引.您可以看到，每当您实际删除这些项目时，这都是一个巨大的问题:

The reason why

When you iterate over a list in Python, behind the scenes Python is actually iterating over the indices of the list. You can see that this is a huge problem whenever you actually remove the items:

l = [1, 2, 2, 3]:
for x in l:
    if x%2 == 0:
        l.remove(x)
    print('x == {}'.format(x))
print('l == {}'.format(l))

您可以通过此处显示的内容来跳过第二个2，并且l的值为[1, 2, 3].这是因为，每当到达和删除前两个2时，索引都是1(第二个元素).在下一次迭代中，索引为2.此时为l == [1,2,3]，因此为x == 3.如果您运行代码，则可能比该解释更为明显.

You can tell by what is printed here that the second 2 is skipped, and that l has a value of [1, 2, 3]. This is because, whenever the first 2 is reached and removed, the index is 1 (the second element). On the next iteration, the index is 2. At this point, l == [1,2,3], so x == 3. If you run the code, it will probably be more obvious than this explanation.

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