Returning the product of a list

Without using lambda:

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

from functools import reduce

a = [1, 2, 3]
reduce(lambda x, y: x * y, a, 1)

if you just have numbers in your list:

from numpy import prod
prod(list)

EDIT: as pointed out by @off99555 this does not work for large integer results in which case it returns a result of type numpy.int64 while Ian Clelland's solution based on operator.mul and reduce works for large integer results because it returns long.

Well if you really wanted to make it one line without importing anything you could do:

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

But don't.