The infinite sequence of prime numbers

[Gribouillis]

The code is short but the performance is terrible. I modified your code to compare the two functions. Here are the results

Output:
primes(): 0.0711168740017456 seconds
infinite_primes(): 55.40710521499568 seconds
infinite_primes() is 779 times slower on 5000 primes.

The results rapidly get worse as the number of primes increase...

 
from heapq import heappush, heappop, heappushpop
import itertools as itt
 
 
def primes():
    """Return the infinite sequence of all prime numbers"""
    yield 2
    h = []
    n = 3
    x, i, seq = 4, 2, itt.count(6, 2)
    while True:
        if n < x:
            yield n
            n2 = n ** 2
            heappush(h, (n2, n, itt.count(n2 + n, n)))
        n = x + 1
        x, i, seq = heappushpop(h, (next(seq), i, seq))
 

def is_prime(num):
    mult = [i for i in range(1, num+1) if num % i == 0]
    if len(mult) == 2 and 1 in mult and num in mult:
        return True
    return False
 
def infinite_primes():
    counter = 1
    while True:
        if is_prime(counter):
            yield counter
        counter += 1

if __name__ == '__main__':
    from collections import deque
    from time import perf_counter
    exhaust = deque(maxlen=0).extend
    
    def perf(seq, n):
        start = perf_counter()
        exhaust(itt.islice(seq, 0, n))
        return perf_counter() - start

    N = 5000
    t1 = perf(primes(), N)
    print(f'primes(): {t1} seconds')
    t2 = perf(infinite_primes(), N)
    print(f'infinite_primes(): {t2} seconds')
    print(f'infinite_primes() is {int(t2/t1)} times slower on {N} primes.')