How can I multiply all items in a list together with Python?
PythonListMultiplicationPython Problem Overview
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.
Python Solutions
Solution 1 - Python
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
Solution 2 - Python
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+.
Solution 3 - Python
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))
Solution 4 - Python
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
Solution 5 - Python
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).
Solution 6 - Python
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.
Solution 7 - Python
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!)
Solution 8 - Python
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)
Solution 9 - Python
nums = str(tuple([1,2,3]))
mul_nums = nums.replace(',','*')
print(eval(mul_nums))
Solution 10 - Python
The simple way is:
import numpy as np
np.exp(np.log(your_array).sum())
Solution 11 - Python
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))
Solution 12 - Python
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
Solution 13 - Python
This is my code:
def product_list(list_of_numbers):
xxx = 1
for x in list_of_numbers:
xxx = xxx*x
return xxx
print(product_list([1,2,3,4]))
result : ('11234', 24)
Solution 14 - Python
How about using recursion?
def multiply(lst):
if len(lst) > 1:
return multiply(lst[:-1])* lst[-1]
else:
return lst[0]
Solution 15 - Python
My solution:
def multiply(numbers):
a = 1
for num in numbers:
a *= num
return a
Solution 16 - Python
'''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)
Solution 17 - Python
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