这个python函数中的lambda表达式是怎么回事? [英] What's going on with the lambda expression in this python function?

问题描述

为什么这种尝试创建咖喱函数列表的方法不起作用?

Why does this attempt at creating a list of curried functions not work?

def p(x, num):
    print x, num

def test():
    a = []
    for i in range(10):
        a.append(lambda x: p (i, x))
    return a

>>> myList = test()
>>> test[0]('test')
9 test
>>> test[5]('test')
9 test
>>> test[9]('test')
9 test

这是怎么回事?

实际上可以实现上述功能的功能是:

A function that actually does what I expect the above function to do is:

import functools
def test2():
    a = []
    for i in range (10):
        a.append(functools.partial(p, i))
    return a


>>> a[0]('test')
0 test
>>> a[5]('test')
5 test
>>> a[9]('test')
9 test

推荐答案

在Python中，在循环和分支中创建的变量没有作用域.您使用lambda创建的所有函数均引用相同的i变量，该变量在循环的最后一次迭代时设置为9.

In Python, variables created in loops and branches aren't scoped. All of the functions you're creating with lambda have a reference to the same i variable, which is set to 9 on the last iteration of the loop.

解决方案是创建一个函数，该函数返回一个函数，从而确定迭代器变量的范围.这就是functools.partial()方法起作用的原因.例如:

The solution is to create a function which returns a function, thus scoping the iterator variable. This is why the functools.partial() approach works. For example:

def test():
    def makefunc(i):
        return lambda x: p(i, x)
    a = []
    for i in range(10):
        a.append(makefunc(i))
    return a

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