Python：在本地范围内执行

Question

在发帖之前，我已经阅读了以下内容：模块级变量执行无法访问的问题python 执行命令但我无法让我的代码运行。这是代码：import stringarr = [0, 1, 2, 3, 4]required = 4red = ['0']alpha = string.printable[10:62]ss = ''it = len(arr) - required + 1for i in range(required):    now = alpha[i]    rd = '-'.join(red)    ss += '\t' * (i + 1) + f'for {now} in range({it}-{rd}):\n'    red.append(now)exec('def inner_reducer():\n' + ss + '\t' * (required + 1) + f'yield {red[-1]}')a = inner_reducer()print(a.__next__())print(a.__next__())print(a.__next__())print(a.__next__())arr我需要一个以和作为参数的生成器，而不是直接在全局范围内编写required，在分配给生成器后，调用__next__()以生成值。任何帮助都是值得赞赏的。

Answer 1

您可以使用工厂函数。例如

import string

def make_inner_reducer_function(arr, required):

    red = ['0']

    alpha = string.printable[10:62]

    ss = ''

    it = len(arr) - required + 1

    for i in range(required):

        now = alpha[i]

        rd = '-'.join(red)

        ss += '\t' * (i + 1) + f'for {now} in range({it}-{rd}):\n'

        red.append(now)

    exec('def inner_reducer():\n' + ss + '\t' * (required + 1) + f'yield {red[-1]}')

    return locals()['inner_reducer']

f = make_inner_reducer_function([0, 1, 2, 3, 4], 4)

a = f()

print(a.__next__())

print(a.__next__())

print(a.__next__())

print(a.__next__())