How do you find the product of n numbers in python?

I have just started with Python programming language. I tried to write a function which takes input either a list or multiple integers to find their product. I am trying to find the product of first million natural numbers but its displaying an MemoryError.

def product(*arg):
    answer=1
    if type(arg) == tuple:
        arg=str(arg)
        arg=arg.lstrip('[(')
        arg=arg.rstrip('],)')
        arg=arg.split(',')
        for i in arg:
            answer*=int(i)
        return answer
    else:
        for i in arg:
            answer*=int(i)
        return answer

j=range(1,1000000,1)
j=list(j)
print(product(j))

Steps:

• I convert the range object into list object if i am to pass a list as argument
• Now, within the function, i try to split the tuple by converting it string.
• I convert the resultant string into a list and then loop over the elements to find the product

Q1: How to avoid the memory error as i try to find the product of first Million natural numbers?

Q2 How to improve this code?

asked May 11, 2020 at 5:54

5

You can use a Generator in Python.

def generate_product():
    r = 1
    for i in range(1,1000000):
        r *= i + 1
        yield r

list(generate_product())[0]

It is more memory efficient and better in terms of performance.

Florian H

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answered May 11, 2020 at 6:30

HarshitHarshit

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To calculate the product of all numbers from 1 to 1 million use a simple loop:

r = 1
for l in range(1,1000000):
    r*=(i+1)
print(res)

But keep in mind that the result will be a pretty big number. That means that your calculation might take long and the resulting number will need a lot memory.

EDIT Then i missread your question a little. This is a function that multiplies the elements in a list:

def multiply_list_elements(_list):
    result = 1
    for element in _list:
        result*=element
    return result

multiply_list_elements([1,2,3,4])

>>> 24

The memory error probably came from the huge number as @ZabirAlNazi calculated so nicely.

answered May 11, 2020 at 6:18

Florian HFlorian H

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1

All of the solution is fine, but one point to make - your question is equivalent to find the factorial of 1 million.

The number of digits of n! = log10(1) + log10(2) + ... log10(n)

import math
num_dig = 1
for i in range(1,1000000):
  num_dig += math.log10(i)
print(num_dig)

So, the number of digits in your answer is 5565703 (approx.).

That's only the final n, if you also want the intermediate results it will require squared memory O(m^2).

import math
ans = 1
for i in range(2,1000001):
  ans *= i
print(ans)

N.B: You can approximate with logarithms and Stirling numbers with faster run-time.

answered May 11, 2020 at 6:38

Zabir Al NaziZabir Al Nazi

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A very simple solution would be:

    def prod_of():
        p=1
        for i in range(1,1000000):
            p* = i
        print(p)

answered May 11, 2020 at 8:48

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