To find first N prime numbers in python

Question

I am new to the programming world. I was just writing this code in python to generate N prime numbers. User should input the value for N which is the total number of prime n

Answer 1

Try this:

primeList = []
for num in range(2,10000):
    if all(num%i!=0 for i in range(2,num)):
        primeList.append(num)
x = int(raw_input("Enter n: "))
for i in range(x):
    print primeList[i]

Answer 2

Whilst playing with prime numbers in Python V3 I noticed that the smallest number by which a composite(non-prime) number is divisible is itself always a prime that is less than the square root of the number under test.

Below is my implementation of that finding to calculate the first N prime numbers.

first 1,000 primes in 0.028S | first 10,000 primes in 0.6S | first 100,000 primes in 14.3S

The snippet below also indicates how long the generation took and prints out the primes in a nice table format.

import time
import math

def first_n_Primes(n):
    number_under_test = 4
    primes = [2,3]
    while len(primes) < n:
        check = False
        for prime in primes:
            if prime > math.sqrt(number_under_test) : break
            if number_under_test % prime == 0:
                check = True
                break
        if not check:
            for counter in range(primes[len(primes)-1],number_under_test-1,2):
                if number_under_test % counter == 0:
                    check = True
                    break
        if not check:
            primes.append(number_under_test)
        number_under_test+=1
    return primes

start_time = time.time()
data = first_n_Primes(1000)
end_time = time.time()

i = 1
while i < len(data)+1:
    print('{0: <9}'.format(str(data[i-1])), end="")
    if i%10 == 0: print("")
    i+=1

print("\nFirst %d primes took %s seconds ---" % (len(data),end_time - start_time))

Answer 3

max = input("enter the maximum limit to check prime number");
if max>1 :
    for i in range (2,max):
        prime=0;
        for j in range (2,i):
            if(i%j==0):
                prime=1;
                break
        if(prime==0 and i!=0):
            print(i,"is prime number");
else:
    print("prime no start from 2");

Answer 4

Super quick sieve implementation by David Eppstein - takes 0.146s for the first 1000 primes on my PC:

def gen_primes():
    """ Generate an infinite sequence of prime numbers.
    """
    # Maps composites to primes witnessing their compositeness.
    # This is memory efficient, as the sieve is not "run forward"
    # indefinitely, but only as long as required by the current
    # number being tested.
    #
    D = {}  

    # The running integer that's checked for primeness
    q = 2  

    while True:
        if q not in D:
            # q is a new prime.
            # Yield it and mark its first multiple that isn't
            # already marked in previous iterations
            # 
            yield q        
            D[q * q] = [q]
        else:
            # q is composite. D[q] is the list of primes that
            # divide it. Since we've reached q, we no longer
            # need it in the map, but we'll mark the next 
            # multiples of its witnesses to prepare for larger
            # numbers
            # 
            for p in D[q]:
                D.setdefault(p + q, []).append(p)
            del D[q]

        q += 1

primes = gen_primes()


x = set()
y = 0
a = gen_primes()
while y < 10000:
  x |= set([a.next()])
  y+=1

print "x contains {:,d} primes".format(len(x))
print "largest is {:,d}".format(sorted(x)[-1])

Answer 5

You don't need to declare that many variables, see the below code is simple and easy to understand.

for num in range(1,50):     
   for i in range(2,num):    
      if num%i == 0:                  
         break 
   else:                 
      print(num,'is a prime')

First 50 prime numbers output

Answer 6

This might help:

import sys
from time import time
def prime(N):
    M=100
    l=[]
    while len(l) < N:
        for i in range(M-100,M):    
            num = filter(lambda y :i % y == 0,(y for y in range(2 ,(i/2)))) 
            if not num and i not in [0,1,4]:
                l.append(i)
        M +=100
    return l[:N]


def dotime(func, n):
    print func.__name__
    start = time()
    print sorted(list(func(n))),len(list(func(n)))
    print 'Time in seconds: ' + str(time() - start)


if __name__ == "__main__":
    dotime(prime, int(sys.argv[1]))