The second one is broken. :( int(sqrt(len(myarray))) is not guaranteed to be the same as int(sqrt(i)), which is the last number that needs to be tested. Consider that for i = 49,

Thanks all for replies
I tested this one , no non-primes there , and x+1 is for the 49 problem :

myarray = [2,3]
from math import sqrt
def prime(x):
for i in xrange(3, x+1,2):
mylist = map(lambda x: i % x,myarray[0:int(sqrt(len(myarray)))])
if 0 not in mylist:
myarray.append(i)

It seems to me that this could not work in general, since the largest number to be tested has to be sqrt(n), and sqrt(len(primes(n)) < sqrt(n), since len(primes(n)) < n.

since the largest number to be tested has to be sqrt(n), and sqrt(len(primes(n)) < sqrt(n), since len(primes(n)) < n.

Note that it's not the sqrt(len(primes(n)) natural number, it's the sqrt(len(primes(n)) prime.

I'm curious as to how this works exactly, perhaps it's because the primes array grows faster than sqrt(n), I don't know.

EDIT: A basic debugging showed me some numbers were being tested by too many factors, which seems to corroborate with my theory (such as 99 being divided by 13).

As your intent seemed to be creating a golf, try to beat mine:

p = lambda x: [i for i in range(2,x+1) if all(map(lambda x: i % x,range(2,i**0.5+1)))]

EDIT2: Removing extraneous spaces:

p = lambda x:[i for i in range(2,x+1) if all(map(lambda x:i%x,range(2,i**0.5+1)))]
Filter is actually one char bigger, because you need a lambda...
[code=python]
p = lambda x:filter(lambda i:all(map(lambda x:i%x,range(2,i**0.5+1))),range(2,x+1))

Nice work ffao! Maybe you can post that in "Python Bitch", that would be a tough one to bitch about. I like the second code, since it also works with Python24.
From ffao ...

p=lambda x:[i for i in range(2,x+1)if sum(i%d<1 for d in range(2,i+1))<2]
print p(100)
"""
result -->
[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]
"""

I am currently practicing my java skills by doing this program called car rental management system. I am creating a login page where when the user login it ...