## Returning the product of a list

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

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

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)
```

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

Program for product of array, C program to find product of array. // elements. #include <stdio.h>. int product( int ar[], int n). {. int result = 1;. for ( int i = 0; i < n; i++). result = result * ar[i];. return  Here is how to write a method that gives the product of the elements of a list: import functools import operator def product(seq): """Product of a sequence.""" return functools.reduce(operator.mul, seq, 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`.

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

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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.

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