Stack Overflow上有许多帖子解释了如何列出目录中的所有子目录。但是,所有这些答案都允许用户获得每个子目录的完整路径,而不仅仅是子目录的名称。
我有以下代码。问题是变量subDir[0]
会输出每个子目录的完整路径,而不仅仅是子目录的名称:
import os
#Get directory where this script is located
currentDirectory = os.path.dirname(os.path.realpath(__file__))
#Traverse all sub-directories.
for subDir in os.walk(currentDirectory):
#I know none of my subdirectories will have their own subfolders
if len(subDir[1]) == 0:
print("Subdirectory name: " + subDir[0])
print("Files in subdirectory: " + str(subDir[2]))
如何才能获得每个子目录的名称?
例如,不是这样得到的:
C:\Users\myusername\Documents\Programming\Image-Database\Curated\Hype
我想要这样:
炒作
最后,我仍然需要知道每个子目录中的文件列表。
发布于 2018-07-11 03:30:57
在'\'
上拆分子目录字符串应该足够了。注意,'\'
是一个转义字符,所以我们必须重复它才能使用实际的斜杠。
import os
#Get directory where this script is located
currentDirectory = os.path.dirname(os.path.realpath(__file__))
#Traverse all sub-directories.
for subDir in os.walk(currentDirectory):
#I know none of my subdirectories will have their own subfolders
if len(subDir[1]) == 0:
print("Subdirectory name: " + subDir[0])
print("Files in subdirectory: " + str(subDir[2]))
print('Just the name of each subdirectory: {}'.format(subDir[0].split('\\')[-1]))
发布于 2018-07-11 03:29:28
使用os.path.basename
for path, dirs, files in os.walk(currentDirectory):
#I know none of my subdirectories will have their own subfolders
if len(dirs) == 0:
print("Subdirectory name:", os.path.basename(path))
print("Files in subdirectory:", ', '.join(files))
发布于 2018-07-11 03:29:58
您可以结合使用os.listdir
和os.path.isdir
来实现这一点。
枚举所需目录中的所有项目,并在打印出属于目录的项目时遍历这些项目。
import os
current_directory = os.path.dirname(os.path.realpath(__file__))
for dir in os.listdir(current_directory):
if os.path.isdir(os.path.join(current_directory, dir)):
print(dir)
https://stackoverflow.com/questions/51272691
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