reduce 函数是如何工作的? [英] How does reduce function work?

问题描述

据我所知，reduce 函数需要一个列表 l 和一个函数 f.然后，它在列表的前两个元素上调用函数 f，然后用下一个列表元素和前一个结果重复调用函数 f.

As far as I understand, the reduce function takes a list l and a function f. Then, it calls the function f on first two elements of the list and then repeatedly calls the function f with the next list element and the previous result.

所以，我定义了以下函数:

So, I define the following functions:

以下函数计算阶乘.

def fact(n):
    if n == 0 or n == 1:
        return 1
    return fact(n-1) * n


def reduce_func(x,y):
    return fact(x) * fact(y)

lst = [1, 3, 1]
print reduce(reduce_func, lst)

现在，这不应该给我 ((1! * 3!) * 1!) = 6 吗?但是，它给出了 720.为什么是 720?它似乎也需要 6 的阶乘.但是，我需要了解原因.

Now, shouldn't this give me ((1! * 3!) * 1!) = 6? But, instead it gives 720. Why 720? It seems to take the factorial of 6 too. But, I need to understand why.

有人可以解释为什么会发生这种情况以及解决方法吗?

Can someone explains why this happens and a work-around?

我基本上想计算列表中所有条目的阶乘乘积.备份计划是运行一个循环并计算它.但是，我更喜欢使用 reduce.

I basically want to compute the product of factorials of all the entries in the list. The backup plan is to run a loop and compute it. But, I would prefer using reduce.

推荐答案

好的，知道了:

我需要先将数字映射到它们的阶乘，然后使用乘法运算符调用 reduce.

I need to map the numbers to their factorials first and then call reduce with multiply operator.

所以，这行得通:

lst_fact = map(fact, lst)
reduce(operator.mul, lst_fact)

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