Python 只列出目录中的文件？
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List only files in a directory?
提问by tkbx
Is there a way to list the files (not directories) in a directory with Python? I know I could use os.listdirand a loop of os.path.isfile()s, but if there's something simpler (like a function os.path.listfilesindiror something), it would probably be better.
有没有办法用 Python 列出目录中的文件（而不是目录）？我知道我可以使用sos.listdir循环os.path.isfile()，但如果有更简单的东西（比如函数os.path.listfilesindir或其他东西），它可能会更好。
采纳答案by Gareth Latty
This is a simple generator expression:
这是一个简单的生成器表达式：
files = (file for file in os.listdir(path)
if os.path.isfile(os.path.join(path, file)))
for file in files: # You could shorten this to one line, but it runs on a bit.
...
Or you could make a generator function if it suited you better:
或者你可以制作一个更适合你的生成器函数：
def files(path):
for file in os.listdir(path):
if os.path.isfile(os.path.join(path, file)):
yield file
Then simply:
然后简单地：
for file in files(path):
...
回答by riamse
You could try pathlib, which has a lot of other useful stuff too.
您可以尝试pathlib，它也有很多其他有用的东西。
Pathlib is an object-oriented library for interacting with filesystem paths. To get the files in the current directory, one can do:
Pathlib 是一个面向对象的库，用于与文件系统路径交互。要获取当前目录中的文件，可以执行以下操作：
from pathlib import *
files = (x for x in Path(".") if x.is_file())
for file in files:
print(str(file), "is a file!")
This is, in my opinion, more Pythonic than using os.path.
在我看来，这比使用os.path.
See also: PEP 428.
另见：PEP 428。
回答by Noam Manos
Using pathlib in Windows as follow:
在 Windows 中使用 pathlib 如下：
files = (x for x in Path("your_path") if x.is_file())
files = (x for x in Path("your_path") if x.is_file())
Generates error:
产生错误：
TypeError: 'WindowsPath' object is not iterable
类型错误：“WindowsPath”对象不可迭代
You should rather use Path.iterdir()
你应该使用Path.iterdir()
filePath = Path("your_path")
if filePath.is_dir():
files = list(x for x in filePath.iterdir() if x.is_file())
回答by johnson
files = next(os.walk('..'))[2]
回答by bold
Using pathlib, the shortest way to list only files is:
使用pathlib，仅列出文件的最短方法是：
[x for x in Path("your_path").iterdir() if x.is_file()]
with depth support if need be.
如果需要，可以提供深度支持。
回答by Neil
For the special case of working with files in the current directory, you could do it as a simple one-liner list comprehension:
对于处理当前目录中的文件的特殊情况，您可以将其作为简单的单行列表理解：
[f for f in os.listdir(os.curdir) if os.path.isfile(f)]
Otherwise in the more general case, directory paths & filenames have to be joined:
否则，在更一般的情况下，必须加入目录路径和文件名：
dirpath = '~/path_to_dir_of_interest'
files = [f for f in os.listdir(dirpath) if os.path.isfile(os.path.join(dirpath, f))]
回答by acaruci
Since Python 3.6 you can use glob with a recursive option "**". Note that glob will give you all files and directories, so you can keep only the ones that are files
从 Python 3.6 开始，您可以使用带有递归选项“**”的 glob。请注意， glob 将为您提供所有文件和目录，因此您只能保留文件和目录
files = glob.glob(join(in_path, "**/*"), recursive=True)
files = [f for f in files if os.path.isfile(f)]
回答by PetitBrezhoneg
If you use Python 3, you could use pathlib.
如果您使用 Python 3，则可以使用pathlib。
But, you have to know that if you use the is_dir()method as :
但是，您必须知道，如果您将is_dir()方法用作：
from pathlib import *
#p is directory path
#files is list of files in the form of path type
files=[x for x in p.iterdir() if x.is_file()]
empty files will be skipped by .iterdir()
空文件将被跳过 .iterdir()
The solution I found is:
我找到的解决方案是：
from pathlib import *
#p is directory path
#listing all directory's content, even empty files
contents=list(p.glob("*"))
#if element in contents isn't a folder, it's a file
#is_dir() even works for empty folders...!
files=[x for x in contents if not x.is_dir()]
