Calculation of primes and adding to the list

Question

I recently started trying to solve problems with a Euler project using python and hit this road bump trying to calculate primes and add them to the list. I wrote the following code, but I'm confused why it doesn't output anything when I run it.

import math

primes = []

def isPrime(i):
    if number<=1:
        return False
    if number==2:
        return True
    if number%2==0:
        return False
    for i in range(3,int(sqrt(number))+1):
        if number%i==0:
            return False
    return True

for i in range (1, 9999999):
    if isPrime(i) == True:
        primes.append(i)
    else:
        continue
print(primes)

Answer 1

Try:

import math

primes = []

def isPrime(number):
    if number<=1:
        return False
    if number==2:
        return True
    if number%2==0:
        return False
    for i in range(3,int(math.sqrt(number))+1):
        if number%i==0:
            return False
    return True

for i in range (1, 9999999):
    if isPrime(i) == True:
        primes.append(i)

print(primes)

Answer 2

The easiest way to do this is to use something called a sieve. Here's how to get all the prime numbers up to a million.

def mark_sieve(sieve, x):
    for i in range(x+x, len(sieve), x):
        sieve[i] = False

sieve = [True]*(10**7+1)
for x in range(2, int(len(sieve)**0.5 + 1)):
    if sieve[x]: mark_sieve(sieve, x)

, sieve True, , 1 () . False sieve. , 1 . ? , . , , .

, , , sieve[some_number]. False, . , [x for x in range(len(sieve)) if sieve[x]]

-

import timeit
import math

def isPrime(number):
    if number<=1:
        return False
    if number==2:
        return True
    if number%2==0:
        return False
    for ind in range(3,int(math.sqrt(number))+1):
        if number%ind==0:
            return False
    return True

def mark_sieve(sieve, x):
    for i in range(x+x, len(sieve), x):
        sieve[i] = False


# Other approaches
time_0 = timeit.default_timer()
primes = []
for i in range (1, 10**6+1):
    if isPrime(i) == True:
        primes.append(i)
    else:
        continue

# Sieve Approach
time_1 = timeit.default_timer()
sieve = [True]*(10**6+1)
sieve[0] = False #because we wont consider zero and one as primes
sieve[1] = False
for x in range(2, int(len(sieve)**0.5 + 1)):
    if sieve[x]: mark_sieve(sieve, x)

primes_2 = [x for x in range(len(sieve)) if sieve[x]]

time_2 = timeit.default_timer()
time_1-time_0 # 12.423080921173096 seconds
time_2-time_1 # 0.9901950359344482 seconds

100 , , 12 . 90. , . , .

Answer 3

, .

, :

for i in range(3, int(math.sqrt(number)) + 1):

1009 ~ 30 , 10 1009, . .

:

primes = [2]

for number in range(3, 9999999 + 1, 2):  # only test odd numbers

    for prime in primes:
        if prime * prime > number:  # we're past sqrt, a prime!
            primes.append(number)
            break

        if number % prime == 0:  # a composite
            break

print(*primes[:10], '...', *primes[-10:])

, @ClockSlave , , , .

Answer 4

:

import math

primes = []

def isPrime(number):
    if number<=1:
        return False
    if number==2:
        return True
    if number%2==0:
        return False
    for ind in range(3,int(math.sqrt(number))+1):
        if number%ind==0:
            return False
    return True

for i in range (1, 100):
    if isPrime(i) == True:
        primes.append(i)
    else:
        continue

print(primes)

, 100:

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

Answer 5

:

primes = [
    i for i in range(1, 9999999) 
    if i == 2 
    or i > 2  # Anything less than 2 is not prime.
    and i % 2  # No evens (except for 2 above)
    and all(i % n for n in range(3, int(i ** 0.5) + 1))]

>>> primes[:10]
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

>>> primes[-10:]
[9999889,
 9999901,
 9999907,
 9999929,
 9999931,
 9999937,
 9999943,
 9999971,
 9999973,
 9999991]

Answer 6

, 999

import math

primes = []


def isPrime(number):
    if number <= 1:
        return False
    for i in range(2, int(math.sqrt(number)) + 1):
        if number % i == 0:
            return False
    return True

for i in range(1, 999):
    if isPrime(i):
        primes.append(i)

print(primes)

[]:

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]

Answer 7

Your algorithm for getting all primes in [0,9999999] is not very efficient. To do this, you need to spend a lot of time so that you can not see the result when it is executed. Just because it was not done. For a faster algorithm, you can check this out