How can I multiply all items in a list together with Python?

Question

I need to write a function that takes a list of numbers and multiplies them together. Example: [1,2,3,4,5,6] will give me 1*2*3*4*5*6. I could really use your help.

Answer 1

Python 3: use functools.reduce:

>>> from functools import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720

Python 2: use reduce:

>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720

For compatible with 2 and 3 use pip install six, then:

>>> from six.moves import reduce
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720

Answer 2

You can use:

import operator
import functools
functools.reduce(operator.mul, [1,2,3,4,5,6], 1)

See reduce and operator.mul documentations for an explanation.

You need the import functools line in Python 3+.

Answer 3

I would use the numpy.prod to perform the task. See below.

import numpy as np
mylist = [1, 2, 3, 4, 5, 6] 
result = np.prod(np.array(mylist))  

Answer 4

If you want to avoid importing anything and avoid more complex areas of Python, you can use a simple for loop

product = 1  # Don't use 0 here, otherwise, you'll get zero 
             # because anything times zero will be zero.
list = [1, 2, 3]
for x in list:
    product *= x

Answer 5

Starting Python 3.8, a .prod function has been included to the math module in the standard library:

math.prod(iterable, *, start=1)

The method returns the product of a start value (default: 1) times an iterable of numbers:

import math
math.prod([1, 2, 3, 4, 5, 6])

>>> 720

If the iterable is empty, this will produce 1 (or the start value, if provided).

Answer 6

Here's some performance measurements from my machine. Relevant in case this is performed for small inputs in a long-running loop:

import functools, operator, timeit
import numpy as np

def multiply_numpy(iterable):
    return np.prod(np.array(iterable))

def multiply_functools(iterable):
    return functools.reduce(operator.mul, iterable)

def multiply_manual(iterable):
    prod = 1
    for x in iterable:
        prod *= x

    return prod

sizesToTest = [5, 10, 100, 1000, 10000, 100000]

for size in sizesToTest:
    data = [1] * size

    timerNumpy = timeit.Timer(lambda: multiply_numpy(data))
    timerFunctools = timeit.Timer(lambda: multiply_functools(data))
    timerManual = timeit.Timer(lambda: multiply_manual(data))

    repeats = int(5e6 / size)
    resultNumpy = timerNumpy.timeit(repeats)
    resultFunctools = timerFunctools.timeit(repeats)
    resultManual = timerManual.timeit(repeats)
    print(f'Input size: {size:>7d} Repeats: {repeats:>8d}    Numpy: {resultNumpy:.3f}, Functools: {resultFunctools:.3f}, Manual: {resultManual:.3f}')

Results:

Input size:       5 Repeats:  1000000    Numpy: 4.670, Functools: 0.586, Manual: 0.459
Input size:      10 Repeats:   500000    Numpy: 2.443, Functools: 0.401, Manual: 0.321
Input size:     100 Repeats:    50000    Numpy: 0.505, Functools: 0.220, Manual: 0.197
Input size:    1000 Repeats:     5000    Numpy: 0.303, Functools: 0.207, Manual: 0.185
Input size:   10000 Repeats:      500    Numpy: 0.265, Functools: 0.194, Manual: 0.187
Input size:  100000 Repeats:       50    Numpy: 0.266, Functools: 0.198, Manual: 0.185

You can see that Numpy is quite a bit slower on smaller inputs, since it allocates an array before multiplication is performed. Also, watch out for the overflow in Numpy.

Answer 7

I personally like this for a function that multiplies all elements of a generic list together:

def multiply(n):
    total = 1
    for i in range(0, len(n)):
        total *= n[i]
    print total

It's compact, uses simple things (a variable and a for loop), and feels intuitive to me (it looks like how I'd think of the problem, just take one, multiply it, then multiply by the next, and so on!)

Answer 8

The simple way is:

import numpy as np
np.exp(np.log(your_array).sum())

Answer 9

Numpy has the prod() function that returns the product of a list, or in this case since it's numpy, it's the product of an array over a given axis:

import numpy
a = [1,2,3,4,5,6]
b = numpy.prod(a)

...or else you can just import numpy.prod():

from numpy import prod
a = [1,2,3,4,5,6]
b = prod(a)

Answer 10

nums = str(tuple([1,2,3]))
mul_nums = nums.replace(',','*')
print(eval(mul_nums))

Answer 11

Found this question today but I noticed that it does not have the case where there are None's in the list. So, the complete solution would be:

from functools import reduce

a = [None, 1, 2, 3, None, 4]
print(reduce(lambda x, y: (x if x else 1) * (y if y else 1), a))

In the case of addition, we have:

print(reduce(lambda x, y: (x if x else 0) + (y if y else 0), a))

Answer 12

I would like this in following way:

    def product_list(p):
          total =1 #critical step works for all list
          for i in p:
             total=total*i # this will ensure that each elements are multiplied by itself
          return total
   print product_list([2,3,4,2]) #should print 48

Answer 13

How about using recursion?

def multiply(lst):
    if len(lst) > 1:
        return multiply(lst[:-1])* lst[-1]
    else:
        return lst[0]

Answer 14

My solution:

def multiply(numbers):
    a = 1
    for num in numbers:
        a *= num
    return a

Answer 15

'''the only simple method to understand the logic use for loop'''

Lap=[2,5,7,7,9] x=1 for i in Lap: x=i*x print(x)

Answer 16

It is very simple do not import anything. This is my code. This will define a function that multiplies all the items in a list and returns their product.

def myfunc(lst):
    multi=1
    for product in lst:
        multi*=product
    return product