Question

如何打开文件列表中的每个文件，对文件执行某些操作然后转到下一个文件？

我有一个包含1000个文本文件的目录。我已经创建了一个包含所有文件名的列表，现在我想逐个文件打开。有人知道怎么做吗？

到目前为止我所拥有的：

from os import listdir 
from os.path import isfile, join

files_in_dir = [ f for f in listdir('nes') if isfile(join('nes',f)) ]
if f.endswith(".txt"):
   print(f)

Answer 1

from os import listdir 
from os.path import isfile, join

files_in_dir = [ f for f in listdir('nes') if isfile(join('nes',f)) ]
for f in files_in_dir:
    if f.endswith(".txt"):
        with open(f, 'r') as in_file:
            for line in in_file:
                # Here you have access to lines of the opened file.

Answer 2

无需创建单独的列表来保存文件名。只需直接迭代listdir()的结果。

for fname in listdir('nes'):
    fname = join('nes', fname)
    if fname.endswith('.txt') and isfile(fname): 
        with open(fname, 'r') as f:
            # do something with open file f

如何在文件列表中按文件打开文件？

2 个答案: