我有c++ /Obj-C背景,我刚刚发现Python(写了大约一个小时)。 我正在写一个脚本递归地读取文件夹结构中的文本文件的内容。

我的问题是,我写的代码将只工作于一个文件夹深度。我可以在代码中看到为什么(见#hardcoded path),我只是不知道如何使用Python,因为我对它的经验只是全新的。

Python代码:

import os
import sys

rootdir = sys.argv[1]

for root, subFolders, files in os.walk(rootdir):

    for folder in subFolders:
        outfileName = rootdir + "/" + folder + "/py-outfile.txt" # hardcoded path
        folderOut = open( outfileName, 'w' )
        print "outfileName is " + outfileName

        for file in files:
            filePath = rootdir + '/' + file
            f = open( filePath, 'r' )
            toWrite = f.read()
            print "Writing '" + toWrite + "' to" + filePath
            folderOut.write( toWrite )
            f.close()

        folderOut.close()

当前回答

使用os.path.join()来构造你的路径-这样更整洁:

import os
import sys
rootdir = sys.argv[1]
for root, subFolders, files in os.walk(rootdir):
    for folder in subFolders:
        outfileName = os.path.join(root,folder,"py-outfile.txt")
        folderOut = open( outfileName, 'w' )
        print "outfileName is " + outfileName
        for file in files:
            filePath = os.path.join(root,file)
            toWrite = open( filePath).read()
            print "Writing '" + toWrite + "' to" + filePath
            folderOut.write( toWrite )
        folderOut.close()

其他回答

这招对我很管用:

import glob

root_dir = "C:\\Users\\Scott\\" # Don't forget trailing (last) slashes    
for filename in glob.iglob(root_dir + '**/*.jpg', recursive=True):
     print(filename)
     # do stuff

如果你使用的是Python 3.5或更高版本,你可以在一行内完成。

import glob

# root_dir needs a trailing slash (i.e. /root/dir/)
for filename in glob.iglob(root_dir + '**/*.txt', recursive=True):
     print(filename)

正如文档中提到的

如果递归为true,模式'**'将匹配任何文件以及零个或多个目录和子目录。

如果你想要每个文件,你可以使用

import glob

for filename in glob.iglob(root_dir + '**/**', recursive=True):
     print(filename)

我发现下面的方法是最简单的

from glob import glob
import os

files = [f for f in glob('rootdir/**', recursive=True) if os.path.isfile(f)]

使用glob('some/path/**', recursive=True)获取所有文件,但也包括目录名。添加if os.path.isfile(f)条件只过滤现有文件

试试这个:

import os
import sys

for root, subdirs, files in os.walk(path):

    for file in os.listdir(root):

        filePath = os.path.join(root, file)

        if os.path.isdir(filePath):
            pass

        else:
            f = open (filePath, 'r')
            # Do Stuff

同意Dave Webb的观点。Walk将为树中的每个目录生成一个项。事实上,您不需要关心子文件夹。

这样的代码应该可以工作:

import os
import sys

rootdir = sys.argv[1]

for folder, subs, files in os.walk(rootdir):
    with open(os.path.join(folder, 'python-outfile.txt'), 'w') as dest:
        for filename in files:
            with open(os.path.join(folder, filename), 'r') as src:
                dest.write(src.read())