Simple prime number generator in Python

Question

Could someone please tell me what I\'m doing wrong with this code? It is just printing \'count\' anyway. I just want a very simple prime generator (nothing fancy).

Answer 1

You can create a list of primes using list comprehensions in a fairly elegant manner. Taken from here:

>>> noprimes = [j for i in range(2, 8) for j in range(i*2, 50, i)]
>>> primes = [x for x in range(2, 50) if x not in noprimes]
>>> print primes
>>> [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

Answer 2

This is my implementation. Im sure there is a more efficient way, but seems to work. Basic flag use.

def genPrime():
    num = 1
    prime = False
    while True:
        # Loop through all numbers up to num
        for i in range(2, num+1):
            # Check if num has remainder after the modulo of any previous numbers
            if num % i == 0:
                prime = False
                # Num is only prime if no remainder and i is num
                if i == num:
                    prime = True
                break

        if prime:
            yield num
            num += 1
        else:
            num += 1

prime = genPrime()
for _ in range(100):
    print(next(prime))

Answer 3

How about this if you want to compute the prime directly:

def oprime(n):
counter = 0
b = 1
if n == 1:
    print 2
while counter < n-1:
    b = b + 2
    for a in range(2,b):
        if b % a == 0:
            break
    else:
        counter = counter + 1
        if counter == n-1:
            print b

Answer 4

def check_prime(x):
    if (x < 2): 
       return 0
    elif (x == 2): 
       return 1
    t = range(x)
    for i in t[2:]:
       if (x % i == 0):
            return 0
    return 1

Answer 5

Here is what I have:

def is_prime(num):
    if num < 2:         return False
    elif num < 4:       return True
    elif not num % 2:   return False
    elif num < 9:       return True
    elif not num % 3:   return False
    else:
        for n in range(5, int(math.sqrt(num) + 1), 6):
            if not num % n:
                return False
            elif not num % (n + 2):
                return False

    return True

It's pretty fast for large numbers, as it only checks against already prime numbers for divisors of a number.

Now if you want to generate a list of primes, you can do:

# primes up to 'max'
def primes_max(max):
    yield 2
    for n in range(3, max, 2):
        if is_prime(n):
            yield n

# the first 'count' primes
def primes_count(count):
    counter = 0
    num = 3

    yield 2

    while counter < count:
        if is_prime(num):
            yield num
            counter += 1
        num += 2

using generators here might be desired for efficiency.

And just for reference, instead of saying:

one = 1
while one == 1:
    # do stuff

you can simply say:

while 1:
    #do stuff

Answer 6

def primes(n): # simple sieve of multiples 
   odds = range(3, n+1, 2)
   sieve = set(sum([list(range(q*q, n+1, q+q)) for q in odds], []))
   return [2] + [p for p in odds if p not in sieve]

>>> primes(50)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

To test if a number is prime:

>>> 541 in primes(541)
True
>>> 543 in primes(543)
False