How to perform element-wise multiplication of two lists?

I want to perform an element wise multiplication, to multiply two lists together by value in Python, like we can do it in Matlab.

This is how I would do it in Matlab.

a = [1,2,3,4]
b = [2,3,4,5]
a .* b = [2, 6, 12, 20]

A list comprehension would give 16 list entries, for every combination x * y of x from a and y from b. Unsure of how to map this.

If anyone is interested why, I have a dataset, and want to multiply it by Numpy.linspace(1.0, 0.5, num=len(dataset)) =).

Use a list comprehension mixed with zip():.

[a*b for a,b in zip(lista,listb)]

Since you're already using numpy, it makes sense to store your data in a numpy array rather than a list. Once you do this, you get things like element-wise products for free:

In [1]: import numpy as np

In [2]: a = np.array([1,2,3,4])

In [3]: b = np.array([2,3,4,5])

In [4]: a * b
Out[4]: array([ 2,  6, 12, 20])

Use np.multiply(a,b):

import numpy as np
a = [1,2,3,4]
b = [2,3,4,5]
np.multiply(a,b)

You can try multiplying each element in a loop. The short hand for doing that is

ab = [a[i]*b[i] for i in range(len(a))]

Yet another answer:

-1 ... requires import
+1 ... is very readable

import operator
a = [1,2,3,4]
b = [10,11,12,13]

list(map(operator.mul, a, b))

outputs [10, 22, 36, 52]

Fairly intuitive way of doing this:

a = [1,2,3,4]
b = [2,3,4,5]
ab = []                        #Create empty list
for i in range(0, len(a)):
     ab.append(a[i]*b[i])      #Adds each element to the list

you can multiplication using lambda

foo=[1,2,3,4]
bar=[1,2,5,55]
l=map(lambda x,y:x*y,foo,bar)

For large lists, we can do it the iter-way:

product_iter_object = itertools.imap(operator.mul, [1,2,3,4], [2,3,4,5])

product_iter_object.next() gives each of the element in the output list.

The output would be the length of the shorter of the two input lists.

create an array of ones; multiply each list times the array; convert array to a list

import numpy as np

a = [1,2,3,4]
b = [2,3,4,5]

c = (np.ones(len(a))*a*b).tolist()

[2.0, 6.0, 12.0, 20.0]

The map function can be very useful here. Using map we can apply any function to each element of an iterable.

Python 3.x

>>> def my_mul(x,y):
...     return x*y
...
>>> a = [1,2,3,4]
>>> b = [2,3,4,5]
>>>
>>> list(map(my_mul,a,b))
[2, 6, 12, 20]
>>>

Of course:

map(f, iterable)

is equivalent to

[f(x) for x in iterable]

So we can get our solution via:

>>> [my_mul(x,y) for x, y in zip(a,b)]
[2, 6, 12, 20]
>>>

In Python 2.x map() means: apply a function to each element of an iterable and construct a new list. In Python 3.x, map construct iterators instead of lists.

Instead of my_mul we could use mul operator

Python 2.7

>>>from operator import mul # import mul operator
>>>a = [1,2,3,4]
>>>b = [2,3,4,5]
>>>map(mul,a,b)
[2, 6, 12, 20]
>>>

Python 3.5+

>>> from operator import mul
>>> a = [1,2,3,4]
>>> b = [2,3,4,5]
>>> [*map(mul,a,b)]
[2, 6, 12, 20]
>>>

Please note that since map() constructs an iterator we use * iterable unpacking operator to get a list. The unpacking approach is a bit faster then the list constructor:

>>> list(map(mul,a,b))
[2, 6, 12, 20]
>>>

Can use enumerate.

a = [1, 2, 3, 4]
b = [2, 3, 4, 5]

ab = [val * b[i] for i, val in enumerate(a)]

gahooa's answer is correct for the question as phrased in the heading, but if the lists are already numpy format or larger than ten it will be MUCH faster (3 orders of magnitude) as well as more readable, to do simple numpy multiplication as suggested by NPE. I get these timings:

0.0049ms -> N = 4, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)]
0.0075ms -> N = 4, a = [i for i in range(N)], c = a * b
0.0167ms -> N = 4, a = np.arange(N), c = [a*b for a,b in zip(a, b)]
0.0013ms -> N = 4, a = np.arange(N), c = a * b
0.0171ms -> N = 40, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)]
0.0095ms -> N = 40, a = [i for i in range(N)], c = a * b
0.1077ms -> N = 40, a = np.arange(N), c = [a*b for a,b in zip(a, b)]
0.0013ms -> N = 40, a = np.arange(N), c = a * b
0.1485ms -> N = 400, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)]
0.0397ms -> N = 400, a = [i for i in range(N)], c = a * b
1.0348ms -> N = 400, a = np.arange(N), c = [a*b for a,b in zip(a, b)]
0.0020ms -> N = 400, a = np.arange(N), c = a * b

i.e. from the following test program.

import timeit

init = ['''
import numpy as np
N = {}
a = {}
b = np.linspace(0.0, 0.5, len(a))
'''.format(i, j) for i in [4, 40, 400] 
                  for j in ['[i for i in range(N)]', 'np.arange(N)']]

func = ['''c = [a*b for a,b in zip(a, b)]''',
'''c = a * b''']

for i in init:
  for f in func:
    lines = i.split('\n')
    print('{:6.4f}ms -> {}, {}, {}'.format(
           timeit.timeit(f, setup=i, number=1000), lines[2], lines[3], f))

To maintain the list type, and do it in one line (after importing numpy as np, of course):

list(np.array([1,2,3,4]) * np.array([2,3,4,5]))

or

list(np.array(a) * np.array(b))

you can use this for lists of the same length

def lstsum(a, b):
    c=0
    pos = 0
for element in a:
   c+= element*b[pos]
   pos+=1
return c

import ast,sys
input_str = sys.stdin.read()

input_list = ast.literal_eval(input_str)

list_1 = input_list[0]

list_2 = input_list[1]

import numpy as np

array_1 = np.array(list_1)

array_2 = np.array(list_2)

array_3 = array_1*array_2


print(list(array_3))