在 Python 中處理多個文件時如何避免嵌套的“with”語句

Question

在 Python 代碼中處理多個文件時，使用推薦的樣式可能會變得難看：

with open("foo.txt") as foo:
    with open("bar.txt", "w") as bar:
         with open("baz.txt", "w") as baz:
              # Read from foo, write different output to bar an baz

這是僅用於處理文件的三個縮進級別！ 替代方案是這個

foo = open("foo.txt")
bar = open("bar.txt", "w")
baz = open("baz.txt", "w")
# Read from foo, write different output to bar an baz
foo.close()
bar.close()
baz.close()

我有一種感覺，這些例子中的任何一個都可以重構為更優雅的東西。 有什么例子嗎？

Answer 1

Python 2.7 及更高版本允許您在一個with語句中指定多個上下文管理器：

with open("foo.txt") as foo, open("bar.txt", "w") as bar, open("baz.txt", "w") as baz:
    # Read from foo, write different output to bar an baz

該行確實很長，並且您不能使用括號將其保持在 80 個字符以下。 但是，您可以使用\\反斜杠延續：

with open("foo.txt") as foo,\
        open("bar.txt", "w") as bar,\
        open("baz.txt", "w") as baz:
    # Read from foo, write different output to bar an baz

另一種選擇是使用contextlib.ExitStack()上下文管理器（僅在 Python 3.3 及更高版本中）：

from contextlib import ExitStack

with ExitStack() as stack:
    foo = stack.enter_context(open("foo.txt"))
    bar = stack.enter_context(open("bar.txt"))
    baz = stack.enter_context(open("baz.txt"))