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Python print non-prime numbers

I have a hackkerank coding challenge to print first n non prime numbers, i have the working code but the problem is that they have a locked code which prints numbers from 1 to n along with the output, in order to pass the test i need to print only the non prime numbers not 1...n numbers along with it. I cant comment the printing part of 1...n as it is blocked. please let me know the idea to print only 1st n non prime numbers:
here is my solution:
def manipulate_generator(generator, n):
if n>1:
ls=[1]
for elm in generator:
if elm>3 and len(ls)<n:
for k in range(2,elm):
if elm%k==0 and elm not in ls:
ls.append(elm)
print(elm)
if len(ls)==n:
return ls
That's the code I added but here is the code that's locked on which I have to write the code above to make it print the number one at a time
def positive_integers_generator():
n = 1
while True:
x = yield n
if x is not None:
n = x
else:
n += 1
k = int(input())
g = positive_integers_generator()
for _ in range(k):
n = next(g)
print(n)
manipulate_generator(g, n)
the point is the for _ in range(k): already prints out number which includes numbers I don't want printed out: This is the desired kind of output I want:for n=10 I want it to print out:
output:
1
4
6
8
9
10
12
14
15
16
I can't change this code but the one above is what I wrote and can be changed... Pleae help me out... Thanks in anticipation

Why not to throw away the numbers which we don't need? Look at this solution which I implemented...
def is_prime(n):
for i in range(2, n):
if n%i == 0:
return False
return True
def manipulate_generator(generator, n):
if is_prime(n+1):
next(generator)
manipulate_generator(generator, n+1)
Note: I understand that the logic can be improved to make it more efficient. But, its the idea of skipping unnecessary number printing which is important here !

You can print all the numbers from 1 up to the first prime number and then from that first number to the next one until you reach n.

I'm not sure of your hackerrank situation, but printing the first N non-prime numbers efficiently can be done this way.
def non_prime_numbers_till_n(n):
primes = set()
for num in range(2,number + 1):
if num > 1:
for i in range(2, math.sqrt(num)):
if (num % i) == 0:
break
else:
primes.add(num)
result = []
for i in range(1, n):
if i not in primes:
result.append(i)
return result
Depending on what your online editor expects you can either print them, or store them in a list and return the list.
Also bear in mind, you can only check upto the sqrt of the number to determine if its a prime or not.

I eventually came up with this answer which I believe should solve it but id there;s a better way to solve it please add your answers:
def manipulate_generator(generator, n):
for num in range(3,100000):
for q in range(2,num):
if num%q==0 and num>n:
generator.send(num-1)
return
this link python generator helped me to understand python generator

I just solved that right now. Like Swapnil Godse said you need to deal with all special cases to optimize computations. This link might be helpful: click here.
Here is the solution:
from math import sqrt
def is_prime(n):
if (n <= 1):
return False
if (n == 2):
return True
if (n % 2 == 0):
return False
i = 3
while i <= sqrt(n):
if n % i == 0:
return False
i = i + 2
return True
def manipulate_generator(g, n):
if is_prime(n+1):
next(g)
manipulate_generator(g, n+1)

prime = int(input('Please enter the range: '))
prime_number = []
for num in range(prime):
if num > 1:
for i in range(2,num):
if num % i == 0:
break
else:
prime_number.append(num)
print(f'Prime numbers in range {prime} is {prime_number}')
all_number = []
for i in range(2,prime+1):
all_number.append(i)
Non_prime_number = []
for element in all_number:
if element not in prime_number:
Non_prime_number.append(element)
print(f'Non Prime numbers in range {prime} is {Non_prime_number}')

Summation of primes : Using Generator vs w/o Generator

Project Euler 10th problem:
Find the sum of all the primes below two million.
Well I came up with two solutions :
USING GENERATOR FUNCTION
def gen_primes():
n = 2
primes = []
while True:
for p in primes:
if n % p == 0:
break
else:
primes.append(n)
yield n
n += 1
flag = True
sum=0
p=gen_primes()
while flag:
prime=p.__next__()
if prime > 2000000:
flag = False
else:
sum+=prime
print(sum)
W/O GENERATOR
def prime(number):
if number ==1:
return -1
else:
for a in (range(1,int(number**0.5)+1))[::2]:
if number%a==0 and a!=1:
return False
else:
return True
count=2
for i in (range(1,2000000))[::2]:
if prime(i)==True and i!=1:
count+=i
else:
continue
print(count)
Surprisingly, the latter (w/o g) takes 7.4 seconds whereas the former (using g) takes around 10 minutes !!!
Why is it so? Thought the generator would perform better because of fewer steps.

Your generator is doing a lot of unnecessary work. It checks all primes up to n when it only needs to check primes up to the square root of n. Here is a modified version of your generator function with the unnecessary work removed:
from time import time
t = time()
def gen_primes():
primes = []
yield 2
n = 3
while True:
is_prime = True
upper_limit = n ** .5
for p in primes:
if n % p == 0:
is_prime = False
break
elif p > upper_limit:
break
if is_prime:
primes.append(n)
yield n
n += 2 # only need to check divisibility by odd numbers
sum = 0
for prime in gen_primes():
if prime > 2_000_000:
break
else:
sum += prime
print("time:", time() - t)
print(sum)
This takes 2.3 seconds on my computer. The version without the generator takes 4.8 seconds on my computer.
If you're interested in a much more efficient algorithm for finding prime numbers, check out the Sieve of Eratosthenes.
https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

Here's a rework of your generator code that's almost twice as fast as the non-generator code. Similar to #DanielGiger's solution (+1) but attempts to be simpler and possibly a hair faster. It uses multiple squaring instead of a single square root (may or may not win out) and doesn't need a boolean flag:
def gen_primes():
yield 2
primes = [2]
number = 3
while True:
for prime in primes:
if prime * prime > number:
yield number
primes.append(number)
break
if number % prime == 0:
break
number += 2
total = 0
generator = gen_primes()
while True:
prime = next(generator)
if prime > 2_000_000:
break
total += prime
print(total)
Here's an equivalent rework of your non-generator code that also attempts to be simpler and just a hair faster. My rule of thumb is to deal with 2 and less as special cases and then write a clean odd divisor check:
def is_prime(number):
if number < 2:
return False
if number % 2 == 0:
return number == 2
for divisor in range(3, int(number**0.5) + 1, 2):
if number % divisor == 0:
return False
return True
total = 2
for i in range(3, 2_000_000, 2):
if is_prime(i):
total += i
print(total)
Also, your code generates an entire range of numbers and then throws away the even numbers -- we can use the third argument to range() to generate only the odd numbers in the first place.
In the end, the use of a generator has nothing to do with the general performance of your two solutions. You can rewrite your second solution as a generator:
def gen_primes():
yield 2
number = 3
while True:
for divisor in range(3, int(number**0.5) + 1, 2):
if number % divisor == 0:
break
else: # no break
yield number
number += 2
The difference between the two solutions is one uses only primes as trial divisors (at the cost of storage) and the other uses odd numbers (pseudoprimes) as trial divisors (with no storage cost). Time-wise, the one that does fewer trial divisions, wins.

Prime number calculator produces no output (python)

I am trying to find the 10001st prime using a basic logic:
1) Identify if a number is prime
2) Add it to a list if prime
3) Print the 10001st term in the list
Here is my code:
primelist=[]
import math
i=2
while True:
for x in range(2, int(math.sqrt(i))):
if i % x == 0:
continue
else:
primelist.append(i)
if len(primelist)== 10001:
break
print(primelist[-1])
Is the logic or code fundamentally wrong or inefficient?
What can I do to ameliorate/ make it work?
EDIT
I have incremented i (i+=1) and used all() to check if everything was indivisible, in response to comments
primelist=[2]
import math
i=3
while True:
for x in range(2, int(i**(1/2))):
if all(i%x!=0 for x in range(2, int(math.sqrt(i)))):
primelist.append(i)
i+=1
if len(primelist)== 10001:
break
print(primelist[-1])

Keep the algorithm/code structure the same (no optimization done), so we can easily share several language points, please see inline comments:
primelist=[]
import math
i=2
while True:
#changed to sqrt + 1, the second parameter of range is not inclusive,
#by adding 1, we make sure sqrt itself is included
for x in range(2, int(math.sqrt(i) + 1)):
if i % x == 0:
#you want break not continue, continue will try
#next possible factor, since you already figured out that this
#is not a prime, no need to keep trying
#continue
break
#a for/else block, many python users never encountered this python
#syntax. The else block is only triggered, if the for loop is naturally
#completed without running into the break statement
else:
#a little debugging, to visually confirm we generated all primes
print("adding {}".format(i))
primelist.append(i)
if len(primelist)== 11:
break
#advance to the next number, this is important,
#otherwise i will always be 2
i += 1
print(primelist[-1])
If you would like to optimize the algorithm, search online for "prime sieve".

This one should work if you want to stay with your algorithm:
import math
def is_prime(n):
for x in range(2, int(math.sqrt(n)) + 1):
if n % x == 0:
return False
return True
primelist = []
i = 2
while True:
if is_prime(i) is True:
primelist.append(i)
i += 1
if len(primelist) == 10001:
break
print(primelist[-1])

Here's a version of your code that's a bit faster, and which doesn't use much RAM. We really don't need to build a list of the primes we find. We don't use the numbers in that list to find more primes, and we're really only interested in its length. So instead of building a list we merely keep count of the primes we find.
# Include 2 in the count
count = 1
i = 3
while True:
if all(i % x for x in range(3, int(i ** 0.5) + 1, 2)):
count += 1
if count == 10001:
break
i += 2
print(i)
FWIW, here's a faster solution that uses sieving. We estimate the required size of the sieve using the prime number theorem. Wikipedia gives these bounds for p(n), the n'th prime number, which is valid for n >= 6:
log(n) + log(log(n)) - 1 < p(n) / n < log(n) + log(log(n))
We use the upper bound as the highest number in the sieve. For n < 6 we use a hard-coded list.
To save space, this sieve only holds odd numbers, so we treat the case of p(1) == 2 as a special case.
#!/usr/bin/env python3
''' Use a sieve of Eratosthenes to find nth prime
Written by PM 2Ring 2017.05.01
Sieve code derived from primes1 by Robert William Hanks
See http://stackoverflow.com/a/3035188/4014959
'''
from math import log
from sys import argv
def main():
num = int(argv[1]) if len(argv) > 1 else 10001
if num == 1:
# Handle 2 as a special case
print(1, 2)
return
elif num < 6:
# Use a pre-built table for (3, 5, 7, 11)
primes = [1, 1, 1, 1, 0, 1]
else:
# Compute upper bound from Prime number theorem
x = log(num)
hi = int(num * (x + log(x)))
print('upper bound', hi)
# Create a boolean list of odd primes in range(hi)
primes = [True] * (hi//2)
for i in range(3, 1 + int(hi**0.5), 2):
if primes[i//2]:
primes[i*i//2::i] = [False] * ((hi - i*i - 1) // (2*i) + 1)
# Count the primes until we get the nth one
k = 0
for i, b in enumerate(primes):
if b:
k += 1
if k == num:
break
print(num, 2*i+1)
if __name__ == "__main__":
main()
This code finds p(10001) = 104743 in under 0.15 seconds on my old single core 32 bit 2GHz machine running Python 3.6.0. It finds p(500000) = 7368787 in about 2.2 seconds.

How can I get this Python code to run more quickly? [Project Euler Problem #7]

I'm trying to complete this Project Euler challenge:
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can
see that the 6th prime is 13.
What is the 10 001st prime number?
My code seem to be right because it works with small numbers, e.g 6th prime is 13.
How can i improve it so that the code will run much more quickly for larger numbers such as 10 001.
Code is below:
#Checks if a number is a prime
def is_prime(n):
count = 0
for i in range(2, n):
if n%i == 0:
return False
break
else:
count += 1
if count == n-2:
return True
#Finds the value for the given nth term
def term(n):
x = 0
count = 0
while count != n:
x += 1
if is_prime(x) == True:
count += 1
print x
term(10001)
UPDATE:
Thanks for your responses. I should have been more clear, I am not looking to speed up the interpreter or finding a faster interpreter, because i know my code isn't great, so i was looking for ways of make my code more efficient.

A few questions to ponder:
Do you really need to check the division until n-1? How earlier can you stop?
Apart from 2, do you really need to check the division by all the multiples of two ?
What about the multiples of 3? 5? Is there a way to extend this idea to all the multiples of previously tested primes?

The purpose of Project Euler is not really to think learn programming, but to think about algorithms. On problem #10, your algorithm will need to be even faster than on #7, etc. etc. So you need to come up with a better way to find prime numbers, not a faster way to run Python code. People solve these problems under the time limit with far slower computers that you're using now by thinking about the math.
On that note, maybe ask about your prime number algorithm on https://math.stackexchange.com/ if you really need help thinking about the problem.

A faster interpreter won't cut it. Even an implementation written in C or assembly language won't be fast enough (to be in the "about one second" timeframe of project Euler). To put it bluntly, your algorithm is pathetic. Some research and thinking will help you write an algorithm that runs faster in a dog-slow interpreter than your current algorithm implemented in native code (I won't name any specifics, partly because that's your job and partly because I can't tell offhand how much optimization will be needed).

Many of the Euler problems (including this one) are designed to have a solution that computes in acceptable time on pretty much any given hardware and compiler (well, not INTERCAL on a PDP-11 maybe).
You algorithm works, but it has quadratic complexity. Using a faster interpreter will give you a linear performance boost, but the quadratic complexity will dwarf it long before you calculate 10,000 primes. There are algorithms with much lower complexity; find them (or google them, no shame in that and you'll still learn a lot) and implement them.

Without discussing your algorithm, the PyPy interpreter can be ridiculously faster than the normal CPython one for tight numerical computation like this. You might want to try it out.

to check the prime number you dont have to run till n-1 or n/2....
To run it more faster,you can check only until square root of n
And this is the fastest algorithm I know
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(sqrt(number))+1):
if number%i==0:
return False
return True

As most people have said, it's all about coming up with the correct algorithm. Have you considered looking at a
Sieve of Eratosthenes

import time
t=time.time()
def n_th_prime(n):
b=[]
b.append(2)
while len(b)<n :
for num in range(3,n*11,2):
if all(num%i!=0 for i in range(2,int((num)**0.5)+1)):
b.append(num)
print list(sorted(b))[n-1]
n_th_prime(10001)
print time.time()-t
prints
104743
0.569000005722 second

A pythonic Answer
import time
t=time.time()
def prime_bellow(n):
b=[]
num=2
j=0
b.append(2)
while len(b)-1<n:
if all(num%i!=0 for i in range(2,int((num)**0.5)+1)):
b.append(num)
num += 1
print b[n]
prime_bellow(10001)
print time.time()-t
Prints
104743
0.702000141144 second

import math
count = 0 <br/> def is_prime(n):
if n % 2 == 0 and n > 2:
return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
for i in range(2,2000000):
if is_prime(i):
count += 1
if count == 10001:
print i
break

I approached it a different way. We know that all multiples of 2 are not going to be prime (except 2) we also know that all non-prime numbers can be broken down to prime constituents.
i.e.
12 = 3 x 4 = 3 x 2 x 2
30 = 5 x 6 = 5 x 3 x 2
Therefore I iterated through a list of odd numbers, accumulating a list of primes, and only attempting to find the modulus of the odd numbers with primes in this list.
#First I create a helper method to determine if it's a prime that
#iterates through the list of primes I already have
def is_prime(number, list):
for prime in list:
if number % prime == 0:
return False
return True
EDIT: Originally I wrote this recursively, but I think the iterative case is much simpler
def find_10001st_iteratively():
number_of_primes = 0
current_number = 3
list_of_primes = [2]
while number_of_primes <= 10001:
if is_prime(current_number, list_of_primes):
list_of_primes.append(current_number)
number_of_primes += 1
current_number += 2
return current_number

A different quick Python solution:
import math
prime_number = 4 # Because 2 and 3 are already prime numbers
k = 3 # It is the 3rd try after 2 and 3 prime numbers
milestone = 10001
while k <= milestone:
divisible = 0
for i in range(2, int(math.sqrt(prime_number)) + 1):
remainder = prime_number % i
if remainder == 0: #Check if the number is evenly divisible (not prime) by i
divisible += 1
if divisible == 0:
k += 1
prime_number += 1
print(prime_number-1)

import time
t = time.time()
def is_prime(n): #check primes
prime = True
for i in range(2, int(n**0.5)+1):
if n % i == 0:
prime = False
break
return prime
def number_of_primes(n):
prime_list = []
counter = 0
num = 2
prime_list.append(2)
while counter != n:
if is_prime(num):
prime_list.append(num)
counter += 1
num += 1
return prime_list[n]
print(number_of_primes(10001))
print(time.time()-t)
104743
0.6159017086029053

based on the haskell code in the paper: The Genuine Sieve of Eratosthenes by Melissa E. O'Neill
from itertools import cycle, chain, tee, islice
wheel2357 = [2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10]
def spin(wheel, n):
for x in wheel:
yield n
n = n + x
import heapq
def insertprime(p,xs,t):
heapq.heappush(t,(p*p,(p*v for v in xs)))
def adjust(t,x):
while True:
n, ns = t[0]
if n <= x:
n, ns = heapq.heappop(t)
heapq.heappush(t, (ns.next(), ns))
else:
break
def sieve(it):
t = []
x = it.next()
yield x
xs0, xs1 = tee(it)
insertprime(x,xs1,t)
it = xs0
while True:
x = it.next()
if t[0][0] <= x:
adjust(t,x)
continue
yield x
xs0, xs1 = tee(it)
insertprime(x,xs1,t)
it = xs0
primes = chain([2,3,5,7], sieve(spin(cycle(wheel2357), 11)))
from time import time
s = time()
print list(islice(primes, 10000, 10001))
e = time()
print "%.8f seconds" % (e-s)
prints:
[104743]
0.18839407 seconds
from itertools import islice
from heapq import heappush, heappop
wheel2357 = [2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,
4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10]
class spin(object):
__slots__ = ('wheel','o','n','m')
def __init__(self, wheel, n, o=0, m=1):
self.wheel = wheel
self.o = o
self.n = n
self.m = m
def __iter__(self):
return self
def next(self):
v = self.m*self.n
self.n += self.wheel[self.o]
self.o = (self.o + 1) % len(self.wheel)
return v
def copy(self):
return spin(self.wheel, self.n, self.o, self.m)
def times(self, x):
return spin(self.wheel, self.n, self.o, self.m*x)
def adjust(t,x):
while t[0][0] <= x:
n, ns = heappop(t)
heappush(t, (ns.next(), ns))
def sieve_primes():
for p in [2,3,5,7]:
yield p
it = spin(wheel2357, 11)
t = []
p = it.next()
yield p
heappush(t, (p*p, it.times(p)))
while True:
p = it.next()
if t[0][0] <= p:
adjust(t,p)
continue
yield p
heappush(t, (p*p, it.times(p)))
from time import time
s = time()
print list(islice(sieve_primes(), 10000, 10001))[-1]
e = time()
print "%.8f seconds" % (e-s)
prints:
104743
0.22022200 seconds
import time
from math import sqrt
wheel2357 = [2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10]
list_prime = [2,3,5,7]
def isprime(num):
limit = sqrt(num)
for prime in list_prime:
if num % prime == 0: return 0
if prime > limit: break
return 1
def generate_primes(no_of_primes):
o = 0
n = 11
w = wheel2357
l = len(w)
while len(list_prime) < no_of_primes:
i = n
n = n + w[o]
o = (o + 1) % l
if isprime(i):
list_prime.append(i)
t0 = time.time()
generate_primes(10001)
print list_prime[-1] # 104743
t1 = time.time()
print t1-t0 # 0.18 seconds
prints:
104743
0.307313919067

Can this be made more pythonic?

I came across this (really) simple program a while ago. It just outputs the first x primes. I'm embarrassed to ask, is there any way to make it more "pythonic" ie condense it while making it (more) readable? Switching functions is fine; I'm only interested in readability.
Thanks
from math import sqrt
def isprime(n):
if n ==2:
return True
if n % 2 ==0 : # evens
return False
max = int(sqrt(n))+1 #only need to search up to sqrt n
i=3
while i <= max: # range starts with 3 and for odd i
if n % i == 0:
return False
i+=2
return True
reqprimes = int(input('how many primes: '))
primessofar = 0
currentnumber = 2
while primessofar < reqprimes:
result = isprime(currentnumber)
if result:
primessofar+=1
print currentnumber
#print '\n'
currentnumber += 1

Your algorithm itself may be implemented pythonically, but it's often useful to re-write algorithms in a functional way - You might end up with a completely different but more readable solution at all (which is even more pythonic).
def primes(upper):
n = 2; found = []
while n < upper:
# If a number is not divisble through all preceding primes, it's prime
if all(n % div != 0 for div in found):
yield n
found.append( n )
n += 1
Usage:
for pr in primes(1000):
print pr
Or, with Alasdair's comment taken into account, a more efficient version:
from math import sqrt
from itertools import takewhile
def primes(upper):
n = 2; foundPrimes = []
while n < upper:
sqrtN = int(sqrt(n))
# If a number n is not divisble through all preceding primes up to sqrt(n), it's prime
if all(n % div != 0 for div in takewhile(lambda div: div <= sqrtN, foundPrimes)):
yield n
foundPrimes.append(n)
n += 1

The given code is not very efficient. Alternative solution (just as inefficient):†
>>> from math import sqrt
>>> def is_prime(n):
... return all(n % d for d in range(2, int(sqrt(n)) + 1))
...
>>> def primes_up_to(n):
... return filter(is_prime, range(2, n))
...
>>> list(primes_up_to(20))
[2, 3, 5, 7, 11, 13, 17, 19]
This code uses all, range, int, math.sqrt, filter and list. It is not completely identical to your code, as it prints primes up to a certain number, not exactly n primes. For that, you can do:
>>> from itertools import count, islice
>>> def n_primes(n):
... return islice(filter(is_prime, count(2)), n)
...
>>> list(n_primes(10))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
That introduces another two functions, namely itertools.count and itertools.islice. (That last piece of code works only in Python 3.x; in Python 2.x, use itertools.ifilter instead of filter.)
†: A more efficient method is to use the Sieve of Eratosthenes.

A few minor things from the style guide.
Uses four spaces, not two. (Personally I prefer tabs, but that's not the Pythonic way.)
Fewer blank lines.
Consistent whitespace: n ==2: => n == 2:
Use underscores in your variables names: currentnumber => current_number

Firstly, you should not assign max to a variable as it is an inbuilt function used to find the maximum value from an iterable. Also, that entire section of code can instead be written as
for i in xrange(3, int(sqrt(n))+1, 2):
if n%i==0: return False
Also, instead of defining a new variable result and putting the value returned by isprime into it, you can just directly do
if isprime(currentnumber):

I recently found Project Euler solutions in functional python and it has some really nice examples of working with primes like this. Number 7 is pretty close to your problem:
def isprime(n):
"""Return True if n is a prime number"""
if n < 3:
return (n == 2)
elif n % 2 == 0:
return False
elif any(((n % x) == 0) for x in xrange(3, int(sqrt(n))+1, 2)):
return False
return True
def primes(start=2):
"""Generate prime numbers from 'start'"""
return ifilter(isprime, count(start))

Usually you don't use while loops for simple things like this. You rather create a range object and get the elements from there. So you could rewrite the first loop to this for example:
for i in range( 3, int( sqrt( n ) ) + 1, 2 ):
if n % i == 0:
return False
And it would be a lot better if you would cache your prime numbers and only check the previous prime numbers when checking a new number. You can save a lot time by that (and easily calculate larger prime numbers this way). Here is some code I wrote before to get all prime numbers up to n easily:
def primeNumbers ( end ):
primes = []
primes.append( 2 )
for i in range( 3, end, 2 ):
isPrime = True
for j in primes:
if i % j == 0:
isPrime = False
break
if isPrime:
primes.append( i )
return primes
print primeNumbers( 20 )

Translated from the brilliant guys at stacktrace.it (Daniele Varrazzo, specifically), this version takes advantage of a binary min-heap to solve this problem:
from heapq import heappush, heapreplace
def yield_primes():
"""Endless prime number generator."""
# Yield 2, so we don't have to handle the empty heap special case
yield 2
# Heap of (non-prime, prime factor) tuples.
todel = [ (4, 2) ]
n = 3
while True:
if todel[0][0] != n:
# This number is not on the head of the heap: prime!
yield n
heappush(todel, (n*n, n)) # add to heap
else:
# Not prime: add to heap
while todel[0][0] == n:
p = todel[0][1]
heapreplace(todel, (n+p, p))
# heapreplace pops the minimum value then pushes:
# heap size is unchanged
n += 1
This code isn't mine and I don't understand it fully (but the explaination is here :) ), so I'm marking this answer as community wiki.

You can make it more pythonic with sieve algorithm (all primes small than 100):
def primes(n):
sieved = set()
for i in range(2, n):
if not(i in sieved):
for j in range(i + i, n, i):
sieved.add(j)
return set(range(2, n)) - sieved
print primes(100)
A very small trick will turn it to your goal.