递归遍历所有目录，直到在Python中找到某个文件 [英] Recursively go through all directories until you find a certain file in Python

问题描述

在Python中递归地遍历所有目录直到找到特定文件的最佳方法是什么？
我想浏览目录中的所有文件，并查看我要查找的文件是否在该目录中。
如果找不到，请转到父目录并重复该过程。我还想计算
找到文件之前要经过的目录和文件数。如果循环的
末尾没有文件，则返回没有文件

What's the best way in Python to recursively go through all directories until you find a certain file? I want to look through all the files in my directory and see if the file I'm looking for is in that directory. If I cannot find it, I go to the parent directory and repeat the process. I also want to count how many directories and files I go through before I find the file. If there is no file at the end of the loop return that there is no file

startdir = Users /..../ file.txt

startdir = "Users/..../file.txt"

findfile是文件名。这是我当前的循环，但是我想使用递归来使其工作。

findfile is name of file. This is my current loop but I want to make it work using recursion.

def walkfs(startdir, findfile):
    curdir = startdir
    dircnt = 0
    filecnt = 0
    for directory in startdir:
        for file in directory:
            curdir = file
            if os.path.join(file)==findfile:
                return (dircnt, filecnt, curdir)
            else:
                dircnt+=1
                filecnt+=1

推荐答案

不要重新发明目录递归轮。只需使用 os.walk（）函数，它使您可以循环遍历目录：

Don't re-invent the directory-recursion wheel. Just use the os.walk() function, which gives you a loop over a recursive traversal of directories:

def walkfs(startdir, findfile):
    dircount = 0
    filecount = 0
    for root, dirs, files in os.walk(startdir):
        if findfile in files:
            return dircount, filecount + files.index(findfile), os.path.join(root, findfile)
        dircount += 1
        filecount += len(files)
    # nothing found, return None instead of a full path for the file
    return dircount, filecount, None

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