[英]Excluding directories in os.walk
I'm writing a script that descends into a directory tree (using os.walk()) and then visits each file matching a certain file extension.我正在编写一个进入目录树的脚本(使用 os.walk()),然后访问与某个文件扩展名匹配的每个文件。 However, since some of the directory trees that my tool will be used on also contain sub directories that in turn contain a LOT of useless (for the purpose of this script) stuff, I figured I'd add an option for the user to specify a list of directories to exclude from the traversal.
但是,由于我的工具将用于的一些目录树还包含子目录,而这些子目录又包含很多无用的(为了这个脚本的目的)的东西,我想我会添加一个选项供用户指定要从遍历中排除的目录列表。
This is easy enough with os.walk().使用 os.walk() 这很容易。 After all, it's up to me to decide whether I actually want to visit the respective files / dirs yielded by os.walk() or just skip them.
毕竟,由我决定是否真的要访问 os.walk() 生成的相应文件/目录,或者只是跳过它们。 The problem is that if I have, for example, a directory tree like this:
问题是,例如,如果我有这样的目录树:
root--
|
--- dirA
|
--- dirB
|
--- uselessStuff --
|
--- moreJunk
|
--- yetMoreJunk
and I want to exclude uselessStuff and all its children, os.walk() will still descend into all the (potentially thousands of) sub directories of uselessStuff , which, needless to say, slows things down a lot.我想排除uselessStuff及其所有子目录, os.walk() 仍然会下降到uselessStuff 的所有(可能有数千个)子目录中,不用说,这会大大减慢速度。 In an ideal world, I could tell os.walk() to not even bother yielding any more children of uselessStuff , but to my knowledge there is no way of doing that (is there?).
在理想的世界中,我可以告诉 os.walk() 甚至不要再产生uselessStuff 的孩子,但据我所知,没有办法做到这一点(有吗?)。
Does anyone have an idea?有没有人有想法? Maybe there's a third-party library that provides something like that?
也许有一个第三方库可以提供类似的东西?
Modifying dirs
in-place will prune the (subsequent) files and directories visited by os.walk
:就地修改
dirs
将修剪os.walk
访问的(后续)文件和目录:
# exclude = set(['New folder', 'Windows', 'Desktop'])
for root, dirs, files in os.walk(top, topdown=True):
dirs[:] = [d for d in dirs if d not in exclude]
From help(os.walk):来自帮助(os.walk):
When topdown is true, the caller can modify the dirnames list in-place (eg, via del or slice assignment), and walk will only recurse into the subdirectories whose names remain in dirnames;
当 topdown 为 true 时,调用者可以就地修改 dirnames 列表(例如,通过 del 或 slice 赋值),而 walk 只会递归到名称保留在 dirnames 中的子目录; this can be used to prune the search...
这可用于修剪搜索...
... an alternative form of @unutbu's excellent answer that reads a little more directly, given that the intent is to exclude directories, at the cost of O(n**2) vs O(n) time. ... @unutbu 优秀答案的另一种形式,读起来更直接,因为目的是排除目录,代价是 O(n**2) 与 O(n) 时间。
(Making a copy of the dirs list with list(dirs)
is required for correct execution) (正确执行需要使用
list(dirs)
制作目录列表的副本)
# exclude = set([...])
for root, dirs, files in os.walk(top, topdown=True):
[dirs.remove(d) for d in list(dirs) if d in exclude]
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