返回列表的产品 [英] Returning the product of a list

问题描述

是否有更简洁、高效或简单的 Pythonic 方法来执行以下操作?

def product(lst):p = 1因为我在 lst:p *= i返回 p

我实际上发现这比使用 operator.mul 稍微快一点:

from 操作符 import mul# from functools import reduce # python3 兼容性def with_lambda(lst):减少(lambda x，y:x * y，lst)def without_lambda(lst):减少(多，lst)def forloop(lst):r = 1对于 lst 中的 x:r *= x返回导入时间a = 范围(50)b = range(1,50)#没有零t = timeit.Timer("with_lambda(a)", "from __main__ import with_lambda,a")打印(带有 lambda:"，t.timeit())t = timeit.Timer("without_lambda(a)", "from __main__ import without_lambda,a")打印(没有lambda:"，t.timeit())t = timeit.Timer("forloop(a)", "from __main__ import forloop,a")打印(for循环:"，t.timeit())t = timeit.Timer("with_lambda(b)", "from __main__ import with_lambda,b")打印(带有 lambda(无 0):"，t.timeit())t = timeit.Timer("without_lambda(b)", "from __main__ import without_lambda,b")打印(没有 lambda(没有 0):"，t.timeit())t = timeit.Timer("forloop(b)", "from __main__ import forloop,b")打印(for循环(无0):"，t.timeit())

给我

('with lambda:', 17.755449056625366)('没有 lambda:', 8.2084708213806152)('for 循环:', 7.4836349487304688)('带 lambda (无 0):', 22.570688009262085)('没有 lambda (no 0):', 12.472226858139038)('for 循环 (no 0):', 11.04065990447998)

解决方案

不使用 lambda:

from 操作符 import mul减少(多，列表，1)

它更好更快.使用 python 2.7.5

from 操作符 import mul将 numpy 导入为 np导入 numexpr 作为 ne# from functools import reduce # python3 兼容性a = 范围(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)

在以下配置中:

a = range(1, 101) # Aa = np.array(a) # Ba = np.arange(1, 1e4, dtype=int) #Ca = np.arange(1, 1e5, dtype=float) #D

python 2.7.5 的结果

<前>|1 |2 |3 |4 |-------+-----------+-----------+-----------+-----------+A 20.8 µs 13.3 µs 22.6 µs 39.6 µsB 106 微秒 95.3 微秒 5.92 微秒 26.1 微秒C 4.34 毫秒 3.51 毫秒 16.7 微秒 38.9 微秒D 46.6 毫秒 38.5 毫秒 180 微秒 216 微秒

结果:np.prod 是最快的，如果你使用 np.array 作为数据结构(小数组 18x，大数组 250x)

使用 python 3.3.2:

<前>|1 |2 |3 |4 |-------+-----------+-----------+-----------+-----------+A 23.6 µs 12.3 µs 68.6 µs 84.9 µsB 133 微秒 107 微秒 7.42 微秒 27.5 微秒C 4.79 毫秒 3.74 毫秒 18.6 微秒 40.9 微秒D 48.4 毫秒 36.8 毫秒 187 微秒 214 微秒

python 3 慢吗?

Is there a more concise, efficient or simply pythonic way to do the following?

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

EDIT:

I actually find that this is marginally faster than using operator.mul:

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

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

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

def forloop(lst):
    r = 1
    for x in lst:
        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())

gives me

('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)

解决方案

Without using lambda:

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

it is better and faster. With python 2.7.5

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)

In the following configuration:

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

Results with python 2.7.5

       |     1     |     2     |     3     |     4     |
-------+-----------+-----------+-----------+-----------+
 A       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       4.34 ms     3.51 ms     16.7 µs     38.9 µs
 D       46.6 ms     38.5 ms      180 µs      216 µs

Result: np.prod is the fastest one, if you use np.array as data structure (18x for small array, 250x for large array)

with python 3.3.2:

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

Is python 3 slower?

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