上下文

目前我正在研究大型数字文件。根据法律，这些文件应在未来 20 到 70 年左右的时间内保存。我的研究是关于如何使用企业搜索解决方案从这些文件中检索文档。在我可以对这些文件做任何事情之前，它们应该被复制到本地存储。大多数文件通过 USB 驱动器和内存棒到达我这里。存档的大小在 4GB 到 250GB 之间。与此同时，我正在学习编写 python 以便能够编写搜索解决方案的脚本。

问题

由于目录和文件的嵌套很深，复制文件和目录是有问题的。许多文件或目录的名称都非常长(有时超过 255 个字符，Windows 仍然无法接受)。

解决方案

编写一个 Python 脚本，递归地删除文件夹和文件名中的所有空格，将它们缩短到 255 个字符以下。同样为了保留旧名称，新名称将采用驼峰式命名。

问题

我已经编写了一个脚本来删除空格和驼峰命名的文件和文件夹名称。但它仅适用于所选目录；它不是递归的。我尝试了 os.walk 函数，但我似乎无法使用 os.walk 实际更改名称。使用 os.listdir 我可以更改名称，但我不明白如何递归执行此操作。那么如何使用 Python 递归地重命名文件和文件夹呢？

代码

import os

current_path  = os.getcwd() # testing only
print current_path
path = "k:/test3/"
os.chdir(path)
current_path  = os.getcwd()
print current_path # testing only
new_filename= ""
#filenames = os.walk(path, topdown=False)

# all print statusses should be rewritten to lines in log files.

filenames =os.listdir(path)
print filenames
for filename in filenames:
    print "\nOldname: \n" +filename
    new_filename = filename.lower().strip() 
    #make sure all filenames are in lowercase and cut whitespace on both ends.
    if " " in filename: #test for spaces in file or foldername
        fn_parts= [w.capitalize() for w in filename.split()]
        print "The parts are: "
        print  fn_parts
        new_filename="" #empty new_filename after last iteration
        new_filename=new_filename.join(fn_parts)
        print "New filename: \n"+new_filename + "\n"
        os.rename(os.path.join(path, filename), os.path.join(path, new_filename))
    else:
        new_filename=new_filename.title()
        print "New filename: \n"+new_filename + "\n"
        os.rename(os.path.join(path, filename), os.path.join(path, new_filename))

最佳答案

对此不确定但是使用 os.walk 的代码更改可能会帮助您希望如此。

import os
current_path  = os.getcwd() # testing only
print current_path
path = "k:/test3/"
os.chdir(path)
current_path  = os.getcwd()
print current_path # testing only
new_filename= ""
#filenames = os.walk(path, topdown=False)

# all print statusses should be rewritten to lines in log files.

filenames =os.listdir(path)
print filenames
for dir,subdir,listfilename in os.walk(path):

    for filename in listfilename:
        print "\nOldname: \n" +filename 
        new_filename = filename.lower().strip() 
        #make sure all filenames are in lowercase and cut whitespace on both ends.
        if " " in filename: #test for spaces in file or foldername
            fn_parts= [w.capitalize() for w in filename.split()]
            print "The parts are: "
            print  fn_parts
            new_filename="" #empty new_filename after last iteration
            new_filename=new_filename.join(fn_parts)
            print "New filename: \n"+new_filename + "\n"
            os.rename(os.path.join(dir, filename), os.path.join(path, new_filename))
        else:
            new_filename=new_filename.title()
            print "New filename: \n"+new_filename + "\n"
            os.rename(os.path.join(path, filename), os.path.join(path, new_filename))

关于python - 递归重命名文件和文件夹，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24324359/