Python control "with" 有条件的上下文管理器

Question

Python2.7 可以使用条件来控制 "with" 上下文管理器吗？我的场景是，如果存在一个 gzip 压缩文件，我想附加到它，如果它不存在，我想写入一个新文件。伪代码为：

with gzip.open(outfile, 'a+') if os.isfile(outfile) else with open(outfile, 'w') as outhandle:

或者...

if os.isfile(outfile):
    with gzip.open(outfile, 'a+') as outhandle:
        # do stuff
else:
    with open(outfile, 'w') as outhandle:
        # do the same stuff

我不想重复 "do stuff" 因为它们之间是一样的。但是如何使用条件来控制 with 上下文？

Answer 1

您可以尝试只为 "do stuff"

编写一个函数

def do_stuff():
    #do stuff here 

if os.isfile(outfile):
    with gzip.open(outfile, 'a+') as outhandle:
        do_stuff()
else:
    with open(outfile, 'w') as outhandle:
        do_stuff()

Answer 2

记住函数也可以赋值给变量

if os.isfile(outfile):
    open_function = gzip.open
    mode = 'a+'
else:
    open_function = open
    mode = 'w'

with open_function(outfile, mode) as outhandle:
    # do stuff

Python control "with" 有条件的上下文管理器

Python control "with" context manager with conditional

python

with-statement

conditional-statements

python-2.7