如何在Python

Question

有多个文件时，在Python代码多个文件时可以得到丑陋使用推荐的风格时，当避免嵌套“与”语句：

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 

这三个缩进级别只是为了有工作文件！替代方案是这样的

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() 

我有一种感觉，这些例子中的任何一个都可以被重构为更优雅的东西。任何例子？

Answer 1

的Python 2.7和最多让你在一个with语句中指定多个上下文经理：

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 

线确实长期做下去，你不能使用括号保持，下面80个字符。您可以使用，但是\反斜杠延续：

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 

另一种选择是使用contextlib.ExitStack() context manager（只在Python 3.3及以上）：

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"))