[python] 목록의 제품 반품

다음을 수행하는 더 간결하고 효율적이거나 단순한 파이썬 방법이 있습니까?

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

편집하다:

실제로 이것은 operator.mul을 사용하는 것보다 약간 빠릅니다.

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__ import without_lambda,a")
print("without lambda:", t.timeit())
t = timeit.Timer("forloop(a)", "from __main__ import forloop,a")
print("for loop:", t.timeit())

t = timeit.Timer("with_lambda(b)", "from __main__ import with_lambda,b")
print("with lambda (no 0):", t.timeit())
t = timeit.Timer("without_lambda(b)", "from __main__ import without_lambda,b")
print("without lambda (no 0):", t.timeit())
t = timeit.Timer("forloop(b)", "from __main__ import forloop,b")
print("for loop (no 0):", t.timeit())

나에게 준다

('with lambda:', 17.755449056625366)
('without lambda:', 8.2084708213806152)
('for loop:', 7.4836349487304688)
('with lambda (no 0):', 22.570688009262085)
('without lambda (no 0):', 12.472226858139038)
('for loop (no 0):', 11.04065990447998)

답변

from operator import mul
reduce(mul, list, 1)

from operator import mul
import numpy as np
import numexpr as ne
# from functools import reduce # python3 compatibility

a = range(1, 101)
%timeit reduce(lambda x, y: x * y, a)   # (1)
%timeit reduce(mul, a)                  # (2)
%timeit np.prod(a)                      # (3)
%timeit ne.evaluate("prod(a)")          # (4)

a = range(1, 101)  # A
a = np.array(a)    # B
a = np.arange(1, 1e4, dtype=int) #C
a = np.arange(1, 1e5, dtype=float) #D

       | 1 | 2 | 3 | 4 |
------- + ----------- + ----------- ++ ---------- + ------ ----- +
 20.8 µs 13.3 µs 22.6 µs 39.6 µs
 B 106 µs 95.3 µs 5.92 µs 26.1 µs
 C 4.34ms 3.51ms 16.7µs 38.9µs
 D 46.6ms 38.5ms 180µs 216µs

       | 1 | 2 | 3 | 4 |
------- + ----------- + ----------- ++ ---------- + ------ ----- +
 23.6 µs 12.3 µs 68.6 µs 84.9 µs
 B 133µs 107µs 7.42µs 27.5µs
 C 4.79ms 3.74ms 18.6µs 40.9µs
 D 48.4ms 36.8ms 187µs 214µs

reduce(lambda x, y: x * y, list, 1)

from numpy import prod
prod(list)

eval('*'.join(str(item) for item in list))

import operator
reduce(operator.mul, list, 1)

import math

math.prod([2, 3, 4]) # 24