我试图使一个脚本列出所有目录，子目录，和文件在一个给定的目录。 我试了一下:
import sys, os
root = "/home/patate/directory/"
path = os.path.join(root, "targetdirectory")
for r, d, f in os.walk(path):
for file in f:
print(os.path.join(root, file))
不幸的是，它不能正常工作。 我得到了所有文件，但没有它们的完整路径。
例如，如果dir结构体为:
/home/patate/directory/targetdirectory/123/456/789/file.txt
它将打印:
/home/patate/directory/targetdirectory/file.txt
我需要的是第一个结果。任何帮助都将不胜感激!谢谢。
当前回答
你可以看看我做的这个样品。它使用os.path.walk函数，该函数已被弃用。使用列表存储所有文件路径
root = "Your root directory"
ex = ".txt"
where_to = "Wherever you wanna write your file to"
def fileWalker(ext,dirname,names):
'''
checks files in names'''
pat = "*" + ext[0]
for f in names:
if fnmatch.fnmatch(f,pat):
ext[1].append(os.path.join(dirname,f))
def writeTo(fList):
with open(where_to,"w") as f:
for di_r in fList:
f.write(di_r + "\n")
if __name__ == '__main__':
li = []
os.path.walk(root,fileWalker,[ex,li])
writeTo(li)
其他回答
无法评论，所以在这里写下答案。这是我见过的最清晰的一句话:
import os
[os.path.join(path, name) for path, subdirs, files in os.walk(root) for name in files]
这只是一个附加功能，有了它，你可以将数据转换成CSV格式
import sys,os
try:
import pandas as pd
except:
os.system("pip3 install pandas")
root = "/home/kiran/Downloads/MainFolder" # it may have many subfolders and files inside
lst = []
from fnmatch import fnmatch
pattern = "*.csv" #I want to get only csv files
pattern = "*.*" # Note: Use this pattern to get all types of files and folders
for path, subdirs, files in os.walk(root):
for name in files:
if fnmatch(name, pattern):
lst.append((os.path.join(path, name)))
df = pd.DataFrame({"filePaths":lst})
df.to_csv("filepaths.csv")
以防万一……获取目录中的所有文件和匹配某些模式的子目录(例如*.py):
import os
from fnmatch import fnmatch
root = '/some/directory'
pattern = "*.py"
for path, subdirs, files in os.walk(root):
for name in files:
if fnmatch(name, pattern):
print(os.path.join(path, name))
如果你想在SharePoint上列出文件这就是你列出它的方式。你的路径可能会从“\teams\”部分开始
import os
root = r"\\mycompany.sharepoint.com@SSL\DavWWWRoot\teams\MyFolder\Policies and Procedures\Deal Docs\My Deals"
list = [os.path.join(path, name) for path, subdirs, files in os.walk(root) for name in files]
print(list)
简单一点的一行代码:
import os
from itertools import product, chain
chain.from_iterable([[os.sep.join(w) for w in product([i[0]], i[2])] for i in os.walk(dir)])
