python,将所有文件从目录树的第三、第四、第五级移动到第二级
我知道我们有python,将所有文件从目录树的第三、第四、第五级移动到第二级,python,linux,Python,Linux,我知道我们有os.walk,但我不知道如何创建它 假设我在ubuntu linux机上有以下文件夹结构: Maindir (as root called by script) +- subdir-one | +-subdir-two | +-file | +-another file | +-subdir-three | +-file3 | +-file4 | +-subdir-four | +-
os.walk
,但我不知道如何创建它
假设我在ubuntu linux机上有以下文件夹结构:
Maindir (as root called by script)
+- subdir-one
| +-subdir-two
| +-file
| +-another file
| +-subdir-three
| +-file3
| +-file4
| +-subdir-four
| +- file5
| +- file6
+- subdir-two
+- subdir-three
| +-sub-subdir-two
| +-file
| +-another file
| +-subdir-three
| +-file3
| +-file4
| +-subdir-four
| +-file5
| +-file6
+-subdir-four
+-subdir-two
+-file
+-another file
+-subdir-three
+-file3
+-file4
+-subdir-four
+-file5
+-file6
我想将所有文件从子目录移动到第2级的子目录,而不是根目录
以subdir-one为例:将subdir-four中的所有文件移动到subdir-one(在本例中为file5和file6),将subdir-four中的所有文件移动到subdir-one(在本例中为file3和file4)
Subdir two没有其他Subdir,因此脚本可以跳过它
细分曲面三:将所有文件从细分曲面二、细分曲面三和细分曲面四移动到细分曲面三
我想你明白了。如果文件被覆盖没有问题,如果它们具有相同的名称,它们仍然是重复的,这是运行此清理脚本的一个原因
当所有文件从子目录移动时,这意味着子目录将为空,因此我还想删除空的子目录
2012年1月14日更新:这是jcollado给出的更改代码,但仍然不起作用。顺便说一句,我忘了提到我还需要过滤一些目录名。在目录树中找到这些目录名时,需要将其排除在处理之外
代码我稍微改变了一下:
import os, sys
def main():
try:
main_dir = sys.argv[1]
print main_dir
# Get a list of all subdirectories of main_dir
subdirs = filter(os.path.isdir,
[os.path.join(main_dir, path)
for path in os.listdir(main_dir)])
print subdirs
# For all subdirectories,
# collect all files and all subdirectories recursively
for subdir in subdirs:
files_to_move = []
subdirs_to_remove = []
for dirpath, dirnames, filenames in os.walk(subdir):
files_to_move.extend([os.path.join(dirpath, filename)
for filename in filenames])
subdirs_to_remove.extend([os.path.join(dirpath, dirname)
for dirname in dirnames])
# To move files, just rename them replacing the original directory
# with the target directory (subdir in this case)
print files_to_move
print subdirs_to_remove
for filename in files_to_move:
source = filename
destination = os.path.join(subdir, os.path.basename(filename))
print 'Destination ='+destination
if source != destination:
os.rename(source, destination)
else:
print 'Rename cancelled, source and destination were the same'
# Reverse subdirectories order to remove them
# starting from the lower level in the tree hierarchy
subdirs_to_remove.reverse()
# Remove subdirectories
for dirname in subdirs_to_remove:
#os.rmdir(dirname)
print dirname
except ValueError:
print 'Please supply the path name on the command line'
if __name__ == '__main__':
main()
我想说以下几点:
import os
main_dir = 'main'
# Get a list of all subdirectories of main_dir
subdirs = filter(os.path.isdir,
[os.path.join(main_dir, path)
for path in os.listdir(main_dir)])
# For all subdirectories,
# collect all files and all subdirectories recursively
for subdir in subdirs:
files_to_move = []
subdirs_to_remove = []
for dirpath, dirnames, filenames in os.walk(subdir):
files_to_move.extend([os.path.join(dirpath, filename)
for filename in filenames])
subdirs_to_remove.extend([os.path.join(dirpath, dirname)
for dirname in dirnames])
# To move files, just rename them replacing the original directory
# with the target directory (subdir in this case)
for filename in files_to_move:
source = filename
destination = os.path.join(subdir, os.path.basename(filename))
os.rename(source, destination)
# Reverse subdirectories order to remove them
# starting from the lower level in the tree hierarchy
subdirs_to_remove.reverse()
# Remove subdirectories
for dirname in subdirs_to_remove:
os.rmdir(dirname)
注意:您可以使用
main\u dir
作为参数将其转换为函数。只是一个建议,也许可以尝试将程序分为三个独立的级别。一个是在层次结构中向下移动两个步骤,一个是将下面的所有文件移动到“根”(现在是您在上一步中离开的位置),另一个是递归地擦除所有空dir。它必须在Python中,还是可以使用shell命令<代码>用于*/中的d;找到“$d”-depth-mindepth 2-type f-path'*/*/*/*/*'-exec mv{}“$d”\-o-类型d-空-删除;完成Hi jcollado。您发送给我的代码不完整。它不起作用。仅处理数组中的最后一个目录。我试图更改您的代码,但运气不好。我把修改后的代码放在下面的答案中,但它不起作用…嗨,大卫,谢谢你的回答。我真的很想用python来回答这个问题。shell是可能的,但您的解决方案不是动态的?如果我们有一个10层深的目录树会发生什么??