Python - 打开连续文件而不是实际打开每个文件

Question

如果我要阅读Python 3.2中的许多文件,比如30-40,我想将文件引用保留在列表中

(所有文件都在一个公共文件夹中)

无论如何我可以打开所有文件到列表中各自的文件句柄,而不必通过file.open()函数单独打开每个文件

Answer 1

这很简单,只需根据文件路径列表使用列表推导.或者,如果您只需要一次访问它们,请使用生成器表达式以避免一次打开所有四十个文件.

list_of_filenames = ['/foo/bar', '/baz', '/tmp/foo']
open_files = [open(f) for f in list_of_filenames]

如果要在某个目录中的所有文件上使用句柄,请使用以下os.listdir函数:

import os
open_files = [open(f) for f in os.listdir(some_path)]

我认为一个简单的,平坦的目录在这里,但注意到该os.listdir返回的路径列表的所有文件对象在指定目录下,无论是"真正"的文件或目录.因此,如果您打开的目录中有目录,则需要使用os.path.isfile以下内容过滤结果:

import os
open_files = [open(f) for f in os.listdir(some_path) if os.path.isfile(f)]

Also, os.listdir only returns the bare filename, rather than the whole path, so if the current working directory is not some_path, you'll want to make absolute paths using os.path.join.

import os
open_files = [open(os.path.join(some_path, f)) for f in os.listdir(some_path) 
              if os.path.isfile(f)]

With a generator expression:

import os
all_files = (open(f) for f in os.listdir(some_path)) # note () instead of []
for f in all_files:
    pass # do something with the open file here.

In all cases, make sure you close the files when you're done with them. If you can upgrade to Python 3.3 or higher, I recommend you use an ExitStack for one more level of convenience .