在Python中使用 "with open "打开多个文件进行读写，而不需要专门列举和列出每个文件？[重复]

Question

这个问题在这里已经有了答案:

如何在Python中打开一个文件列表 (6个答案)

封闭 11天前.

我的意思是 知道 如何做。

with open("file1.txt","w") as file1, open("file2.txt","w") as file2, open("file3.txt","w") as file3, open("file4.txt","w") as file4:

我 想要 要大约做。

with open([list of filenames],"w") as [list of file variable names]:

有什么方法可以做到这一点？

Answer 1

你可以使用 contextlib.ExitStack 作为文件处理程序的容器。它将自动关闭所有打开的文件。

例如

filenames = "file1.txt", "file2.txt", "file3.txt", "file4.txt"
with ExitStack() as fs:
    file1, file2, file3, file4 = (fs.enter_context(open(fn, "w")) for fn in filenames)
    ...
    file2.write("Some text")