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The infinite sequence of prime numbers - Gribouillis - Jan-28-2022
#!/usr/bin/env python
# SPDX-FileCopyrightText: 2022 Member Gribouillis of www.python-forum.io
#
# SPDX-License-Identifier: MIT
__version__ = '2022.01.28.2'
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))
if __name__ == '__main__':
s = primes()
# print the first prime numbers
for i in range(100):
p = next(s)
print(p)
RE: The infinite sequence of prime numbers - lucasbazan - Feb-11-2022
I made a function that verify if a number is prime, hope that is can help your code to be more tiny
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 Falsehttps://gist.github.com/lucasbazan/9a3c45c81ede877aa0bb3b3431c55532
RE: The infinite sequence of prime numbers - Gribouillis - Feb-11-2022
@lucasbazan How can it help?
RE: The infinite sequence of prime numbers - lucasbazan - Feb-11-2022
(Feb-11-2022, 06:27 PM)Gribouillis Wrote: @lucasbazan How can it help?If I understand correctly you can create an infinite loop with a counter using this function of mine, so I believe your code can be short. Some thing like this:
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):
print(counter)
counter += 1
if __name__ == '__main__':
infinite_primes()
RE: The infinite sequence of prime numbers - Gribouillis - Feb-11-2022
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.
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.')