python - Returning the product of a list

Question

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)

Answer 1

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?