python - 从另一个函数访问一个函数中声明的变量

Question

如何访问变量 x对于一个函数来说是局部的 main()来自另一个函数 sub()里面叫main() ？

下面的 MEW 抛出错误 NameError: global name 'x' is not defined当脚本运行时。

如果我添加 global x, l在main()不再有错误，但我不想让这些变量成为全局变量。

MWE:

def sub():
    print "x from main: {}".format(x)
    print "l from main: {}".format(l)


def main():
    x = 5
    l = [1,2,3,4,5]
    print "z from top: {}".format(z)
    sub()


if __name__ == "__main__":
    z = 'SOME'
    main()

Edit1 我正在编辑现有代码并且不想更改函数的数量

Answer 1

将参数作为参数传递得到你想要完成的。

def sub(x, l):
    print "x from main: {}".format(x)
    print "l from main: {}".format(l)


def main(z):
    x = 5
    l = [1,2,3,4,5]
    print "z from top: {}".format(z)
    sub(x, l)


if __name__ == "__main__":
    z = 'SOME'
    main(z)

这打印:

z from top: SOME
x from main: 5
l from main: [1, 2, 3, 4, 5]

如果你不想使用这种方法，我想你可以关注 this answer 的方法，尽管对于这样的案例来说似乎很复杂。

import inspect

z = 'SOME'

def sub():
    frame = inspect.currentframe()
    locales = None
    try: locales = frame.f_back.f_locals
    finally: del frame
    if locales:
        for k, v in locales.items():
            print("{} from main: {}".format(k, v))


def main():
    x = 5
    l = [1,2,3,4,5]
    print("z from top: {}".format(z))
    sub()

main(z)

如果你能把它变成一个上下文管理器，也许这会更干净。

您还可以使用 *args 和 *kwargs

def function(*args, **kwargs):
    if args:
        for a in args:
            print "value: {} was passed in args".format(a)
    if kwargs:
        for k, v in kwargs.items():
            print "key {} was passed with value: {} in kwargs".format(k, v)


def caller():
    x = 5
    l = [1,2,3,4,5]
    function(x, l = l)

caller()
#value: 5 was passed in args
#key l was passed with value: [1, 2, 3, 4, 5] in kwargs

从这里你可以给locals赋值

函数

locals['x'] = args[0]
locals['l'] = kwargs[l]

OR 使用 dict.get 这样您就可以设置默认值:

l = kwargs.get('l', None)