Ytdyklly

Here's my implementation to factorize a positive integer:

def is_prime(n):

    if n == 1:

        return False

    if n % 2 == 0:

        return False

    i = 3

    while i * i <= n:

        if n % i == 0:

            return False

        i += 2

    return True





# @timeit

def prime_factors(n):

    prime_factor_list = 

    while not n % 2:

        prime_factor_list.append(2)

        n //= 2

    while not n % 3:

        prime_factor_list.append(3)

        n //= 3

    i = 5

    while n != 1:

        if is_prime(i):

            while not n % i:

                prime_factor_list.append(i)

                n //= i

        i += 2



    return prime_factor_list

How can I improve this code in speed and make it pythonic?

prime_factors()

The single greatest improvement to performance that you could make would be to remove the call to is_prime() altogether. If all factors smaller than i have been removed from n already, then i could not possibly be composite.

The next improvement you could make would be to turn prime_factors() into a generator. The function would no longer need to resize and recopy prime_factor_list whenever the list needs to grow beyond its estimated size. Furthermore, the caller could start making use of initial results before the entire factorization is finished.

Using three lines (i = 5, while n != 1, and i += 2) to structure the loop — something that would be succinctly written in C as a for (i = 5; n != 1; i += 2) header — is un-Pythonic. Such counting loops are usually better written using some kind of range(), xrange(), enumerate(), or something from itertools. Here, itertools.chain() and itertools.count() would be appropriate.

Without changing the spirit of your algorithm, I would write:

import itertools



def prime_factors(n):

    for i in itertools.chain([2], itertools.count(3, 2)):

        if n <= 1:

            break

        while n % i == 0:

            n //= i

            yield i

is_prime()

As mentioned above, you don't need is_prime() at all, if your only goal is to implement prime_factors().

However, if you wanted to implement is_prime() anyway, you could tighten it up a bit. My main recommendation would be to avoid performing the i * i multiplication with each iteration. You would probably be better off computing the $sqrt{n}$ limit just once.

Then, as with the other function, you should express the loop more Pythonically using range() or xrange(). Actually, you wouldn't need an explicit loop; you could use any() instead.

So, if we stick to your algorithm in spirit, we could tighten up the implementation like this:

def is_prime(n):

    if n < 3 or n % 2 == 0:

        return n == 2

    else:

        return not any(n % i == 0 for i in range(3, int(n**0.5 + 2), 2))

Your code is written fairly well, yet in is_prime you forgot to check whether the input integer is 2; just add this:

if n == 2:

    return True

Also, for some negative integers is_prime returns True. I suggest you add the following statement to prune them away:

if n <= 1:

    return False

prime_factors never calls is_prime with a number that's divisible by 2 or 3. is_prime could be faster if it used this knowledge. However, if is_prime will use this knowledge, then it may be a bit misleading in global scope. So you could move the definition of is_prime inside of prime_factors.

Although the code is Pythonic alright, it would be more idiomatic to use a prime generator to get the next prime factor, instead of iterating from 5 upwards. Given a prime generator gen, the final loop of the code would become:

while n != 1:

    prime = next(gen)

    while not n % prime:

        prime_factor_list.append(prime)

        n //= prime

Few additional thoughts to Fast Number Factorization in Python answer.

is_prime()

In case if you have multiple consequent calls you should use something like Sieve_of_Eratosthenes. If you will, time to generate sieve will depend on maximum value of number to factorize but total time will be reduces.

prime_factors()

There is one thing you miss in your code. Lets take prime number, let is be
$ 10007 $ and multiply it by $ 2 $, we will receive $ 20014 $. Its factorization will be $ 20014 = 10007 times 2 $.

Now lets analyze your prime_factors. You will find that $ 2 $ is prime divisor of $ 20014 $ and will continue to iterate through all prime numbers up to $ 10007 $ but actually you have to iterate up to square root of initial number, because if you cross this line and reminder still not equal $ 1 $ than it is prime.

This trick can reduce time for such cases.

As the result, you can assume that you need to generate sieve up to square root of highest number.

def print_factors(x):enter code here
print("The factors of",x,"are:")enter code here
for i in range(1, num+1):enter code here
print()enter code here
print(i,"#--->",end=' ')enter code here
for j in range(1,i+1):enter code here
if i % j == 0:enter code here
print(j,end=', ')enter code here
num = int(input("Enter a limit of the natural sequenced numbers: "))enter code here
print_factors(num)enter code here