python 序列中前 n 项的乘积

Question

我正在尝试创建一个函数，它接受一个参数（一个数字）并返回该数字的阶乘。

例如 f(5) 将返回 1*2*3*4*5

到目前为止我所拥有的是

def product(n, term):
    """Return the product of the first n terms in a sequence.

    term -- a function that takes one argument
    """
    k, total = 1, 1
    while k <= n:
        k, total = k + 1, total * term(k, 1)
    return total


def factorial(n):
    """Return n factorial by calling product.

    >>> factorial(4)
    24
    """
    return product(n, mul)

但是，是否可以使该术语仅接受 1 个参数？

Answer 1

import math

def factorial(n):
    return math.factorial(n)

替代实现：

def factorial(n):
    return reduce(lambda x,y:x*y,range(1,n+1))

使用递归：

def factorial(n):
     if n == 0:
         return 1
     else:
         return n * factorial(n-1)

Answer 2

计算 n 的阶乘是递归函数的标准示例：

def fac(n):
    return n * fac(n-1) if n > 1 else 1

Answer 3

又怎样呢？

import operator

def product(nums):
    return reduce(operator.mul, nums, 1)

def factorial(num):
    return product(range(2, num+1))

Answer 4

如果您的意思是在 product(n, term) 中，term(n) 应该是一个从索引 n 串联到的函数该点的值；那么你的 factorial(n) 将被定义为 def Factorial(n): return Product(n, Identity) 其中，identity 是 defidentity(n): return n

换句话说：

def product(n, term):
    """Return the product of the first n terms in a sequence.

    term -- a function that takes one argument
    """
    k, total = 1, 1
    while k <= n:
        k, total = k + 1, total * term(k)
    return total


def identity(n):
    return n

def factorial(n):
    """Return n factorial by calling product.

    >>> factorial(4)
    24
    """
    return product(n, identity)