使用循环中的变量在循环中生成函数会导致变量的阴影效应

Question

我想在循环中生成从循环中访问变量的小函数。然后我想一次组合并应用所有功能。 MWE 看起来是这样说的：

from functools import reduce

def compose(*funcs):
    return lambda x: reduce(lambda y, f: f(y), reversed(funcs), x)


def replace_values_by_type(text, values, types) -> str:

    def peek(x):
        print('peek:', x)
        return x

    replacements = [lambda txt: peek(peek(txt).replace(peek(val), f'<{peek(typ)}>')) for val, typ in zip(values, types)]
    replace = compose(*replacements)

    text_with_replacements = replace(text)

    print(values)
    print(types)
    print(text)
    print(text_with_replacements)
    print()

    return text_with_replacements



replace_values_by_type('this is a test sentence', ['is', 'test'], ['A', 'B'])

当我运行我希望得到“this a sentence”。但是实际上只使用了循环中的最后一对 val 和 typ 。所以我猜会发生一些阴影或覆盖。你能解释一下吗？

-> % py mwe.py peek: this is a test sentence peek: test peek: B peek: this is a <B> sentence peek: this is a <B> sentence peek: test peek: B peek: this is a <B> sentence ['is', 'test'] ['A', 'B'] this is a test sentence this is a <B> sentence

顺便说一句。为了隔离问题，我还编写了这样的函数：

def replace_values_by_type(text, values, types) -> str: replacements = [] for val, typ in zip(values, types): def f(txt): return txt.replace(val, f'<{typ}>') replacements.append(f) text_with_replacements = text for f in replacements: text_with_replacements = f(text_with_replacements) return text_with_replacements print(replace_values_by_type('this is a test sentence', ['is', 'test'], ['A', 'B']))

问题依旧。

Answer 1

列表理解创建的所有闭包都在同一个变量范围内，循环中只有一个 val 和 typ 变量实例。当稍后调用闭包时，变量具有上次迭代的值。

您需要为每次迭代在唯一范围内生成闭包。一种方法是编写一个单独的函数来 returns 闭包，因为每个函数都会建立一个新的范围。

def replacer(val, typ):
    return lambda txt: peek(peek(txt).replace(peek(val), f'<{peek(typ)}>'))

replacements = [replacer(val, typ) for val, typ in zip(values, types)]

使用循环中的变量在循环中生成函数会导致变量的阴影效应

generating functions in a loop using variables from the loop cause shadowing effects of variables

python

variables

scope

composition