相关疑难解决方法(0)

是否有更简洁,高效或简单的pythonic方式来执行以下操作？

def product(list):
    p = 1
    for i in list:
        p *= i
    return p

编辑:

我实际上发现这比使用operator.mul要快一些:

from operator import mul
# from functools import reduce # python3 compatibility

def with_lambda(list):
    reduce(lambda x, y: x * y, list)

def without_lambda(list):
    reduce(mul, list)

def forloop(list):
    r = 1
    for x in list:
        r *= x
    return r

import timeit

a = range(50)
b = range(1,50)#no zero
t = timeit.Timer("with_lambda(a)", "from __main__ import with_lambda,a")
print("with lambda:", t.timeit())
t = timeit.Timer("without_lambda(a)", "from __main__ …