CODEWITHSUNDEEP
python
Shay Kaha
Shay Kaha

Reputation: 1

Find multiple prime factors Python

I have to write code that finds prime factors. Also, I have to keep in mind the factors that appear multiple times. For 12, I know how to write the code that returns 3 and 2.

def prime_factors(n):
for possible_prime in range(2,int(math.sqrt(n)+1)):
    redundant=n%possible_prime
    if redundant==0:
        for last_check in range(2,int(math.sqrt(possible_prime)+1)):
            redundant2=possible_prime%last_check
            if redundant2!=0:
                print(possible_prime)

But what I need to get is 2, 2, 3. Can anyone help? I am supposed to use loops and lists.

Thanks in advance.

Shai

Upvotes: 0

Views: 367

Answers (4)

Vlad Bezden
Vlad Bezden

Reputation: 89507

Here is an example of how to find prime factors using recursion and walrus operator

def prime_factors(n):
    for i in range(2, int(n ** 0.5) + 1):
        if (q_r := divmod(n, i))[1] == 0:
            return [i] + factor_list(q_r[0])
    return [n]

>>> prime_factors(12)
[2, 2, 3]

Upvotes: 0

TomR8
TomR8

Reputation: 102

It is often better to keep things simple and stupid (KISS principle). Although there are more efficient algorithms to do prime factorization, this one is straight forward:

import math

def prime_factors(n):
    res = []
    # handle factors 2 first
    while n%2 == 0: 
        res.append(2)
        n = n//2
    fac = 3
    # handle all odd factors until limit reached
    while fac < int(math.sqrt(n))+1:
        while n%fac == 0:
            res.append(fac)
            n = n//fac
        fac += 2
    # append remaining prime factor
    if n > 2:
        res.append(n)
    return res

print (prime_factors(2222222222))

Note use of // for integer division in Python 3.

Upvotes: 1

Sentice
Sentice

Reputation: 23

I am new to Python but will give it a try (please correct me if the synthax is not perfect)

def prime_factors(n):
    prime_list = []
    job_done = False
    first_prime = 2
    while not job_done:
        for possible_prime in range(first_prime,n+1):
            if n%possible_prime == 0:
                prime_list.append(possible_prime)
                first_prime = possible_prime
                n = int(n/possible_prime)
                if n == 1:
                    job_done = True
                break
    return prime_list

Upvotes: 0

Adirio
Adirio

Reputation: 5286

First of all you should remove all even numbers from the range() to optimize the loop, I would also suggest adding one to the integer form of the sqrt to keep types consistent. The second loop is not optimized either, as every time the first loop results in a redundant == 0 the possible prime is either prime or a combination of already found primer factors, thus reducing the amount of numbers you need to check against

def prime_factors(n):
    factors = []
    for possible_prime in [2] + range(3, int(math.sqrt(n))+1, 2):
        if not (n % possible_prime):
            limit = int(math.sqrt(possible_prime)
            is_prime = True
            for factor in factors:
                if factor > limit:
                    break
                if not (possible_prime % factor):
                    is_prime = False
                    break
            if is_prime:
                factors.append(possible_prime)
    return factors

As you said, this gives you all prime factors, now we just want to duplicate (or further) if apropiate:

def prime_factors(n):
    factors = []
    for possible_prime in [2] + range(3, int(math.sqrt(n))+1, 2):
        if not (n % possible_prime):
            limit = int(math.sqrt(possible_prime)
            is_prime = True
            for factor in factors:
                if factor > limit:
                    break
                if not (possible_prime % factor):
                    is_prime = False
                    break
            if is_prime:
                factors.append(possible_prime)
    factors_copy = factors[:]
    for factor in factors_copy:
        i = 0
        j = n
        while not(j % factor):
            i += 1
            j /= factor
        for k in range(i-1):
            factors.append(factor)
    return sorted(factors)

Upvotes: 0

Related Questions