Python 闭包未按预期工作 [英] Python closure not working as expected

问题描述

当我运行以下脚本时，两个 lambda 都在同一个文件——junk.txt 上运行 os.startfile().我希望每个 lambda 都使用创建 lambda 时设置的值f".有没有办法让它按照我的预期运行?

When I run the following script, both lambda's run os.startfile() on the same file -- junk.txt. I would expect each lambda to use the value "f" was set to when the lambda was created. Is there a way to get this to function as I expect?

import os


def main():
    files = [r'C:\_local	est.txt', r'C:\_localjunk.txt']
    funcs = []
    for f in files:
        funcs.append(lambda: os.startfile(f))
    print funcs
    funcs[0]()
    funcs[1]()


if __name__ == '__main__':
    main()

推荐答案

一种方法是:

def main():
    files = [r'C:\_local	est.txt', r'C:\_localjunk.txt']
    funcs = []
    for f in files:
        # create a new lambda and store the current `f` as default to `path`
        funcs.append(lambda path=f: os.stat(path))
    print funcs

    # calling the lambda without a parameter uses the default value
    funcs[0]() 
    funcs[1]()

否则 f 在函数被调用时被查找，所以你得到当前(循环后)值.

Otherwise f is looked up when the function is called, so you get the current (after the loop) value.

我更喜欢的方式:

def make_statfunc(f):
    return lambda: os.stat(f)

for f in files:
    # pass the current f to another function
    funcs.append(make_statfunc(f))

甚至(在python 2.5+中):

or even (in python 2.5+):

from functools import partial
for f in files:
    # create a partially applied function
    funcs.append(partial(os.stat, f))

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