Develop Reference

Removing multiples of a number from a list in Python

I'm new to Python and i'm working on a program which finds the prime factors of a number. My code so far looks like this:
num = int(input('\nEnter a natural number greater than 1: '))
if num <= 1:
while num <= 1:
num = int(input('\nI said greater than 1: '))
if num == 2:
print('\n', num, 'is a prime number.')
else:
toggle = 0
flist = []
for i in range(2, num - 1):
if num % i == 0:
flist.append(i)
toggle = 1
if toggle == 0:
print('\n', num, 'is a prime number.')
if toggle == 1:
print('\n', num, 'is not a prime number.')
print('\nIt\'s prime factors are:', flist)
With an input of 30 for example i get this output:
Enter a natural number greater than 1: 30
30 is not a prime number.
It's prime factors are: [2, 3, 5, 6, 10, 15]
The prime factors in this case are 2, 3, 5. How can i remove their multiples ?
Thank you.

I would define an extra function to check if a number is prime
def is_prime(n):
if n>1:
for i in range(2,n):
if (n % i)==0:
return False
return True
Then I would use it inside you code to check if a number is prime before to add it to the final list of prime factors.
num = int(input('\nEnter a natural number greater than 1: '))
if num <= 1:
while num <= 1:
num = int(input('\nI said greater than 1: '))
if num == 2:
print('\n', num, 'is a prime number.')
else:
toggle = 0
flist = []
for i in range(2, num - 1):
if num % i == 0:
if is_prime(i): # append only prime numbers
flist.append(i)
toggle = 1
if toggle == 0:
print('\n', num, 'is a prime number.')
if toggle == 1:
print('\n', num, 'is not a prime number.')
print('\nIt\'s prime factors are:', flist)
This gives you the right output:
Enter a natural number greater than 1: 30
30 is not a prime number.
It's prime factors are: [2, 3, 5]

You are getting the multiples of your prime factors, such as 6 = 2 × 3, 10 = 2 × 5 and 15 = 3 × 5.
To avoid those, you should divide your initial number by the prime factors as you find them, so that it won't divide again by the multiples of those prime factors.
Also, you should do that repeatedly, since numbers often have an exponent of a prime factor (for instance, 12 = 2² × 3, so you should divide by 2 twice.)
for i in range(2, num):
while num % i == 0:
flist.append(i)
num = num // i
if num == 1:
break
When you get to 1, you know you can stop.
There are also other optimizations you can do, for instance when i² is larger than num, you know num must be a prime, since you already divided it by all factors smaller than i.
You can also skip even numbers after 2, since those will never be factors.
Using the simple code above, you'll know that num is a prime whenever flist is empty. (You don't need an extra toggle to track that.) That works, since you're never dividing by num itself, so if you get to the end and haven't divided once, that means it's a prime.
For a number such as 12, your flist will have [2, 2, 3]. If you want the unique prime factors, without repetitions, you can use set(flist) to get rid of the duplicates.

Here's a solution using the Sieve of Eratosthenes to first find all primes beneath N and then go through those to find all prime factors, this should be pretty efficient
import numpy as np
def findSieve(num):
allN = np.arange(1,num+1,1)
mask = allN == allN
maskN = np.ma.MaskedArray(allN)
p = 2
while maskN[p:].count() > 0:
mask = mask & ((allN%p != 0) | (allN <= p))
maskN = np.ma.MaskedArray(maskN, mask =~ mask)
p = maskN[p:].min()
return np.asarray(maskN[maskN.mask == False])
def findPrimeFactors(num):
sieve = findSieve(num)
flist = []
num_factored = num
for i in range(1,len(sieve)) :
while num_factored % sieve[i] == 0:
flist.append(sieve[i])
num_factored = num_factored // sieve[i]
if num_factored == 1:
break
prime = len(flist) == 1
if prime:
print('\n', num, 'is a prime number.')
else:
print('\n', num, 'is not a prime number.')
print('\nIt\'s prime factors are:', flist)
num = int(input('\nEnter a natural number greater than 1: '))
findPrimeFactors(num)

Checking for prime numbers: Any number is prime?

Hello dearest community,
I'm stuck on something very easy, I guess sometimes your just out of it. Anyhow, I wrote this to check for prime numbers:
num = int(input(""))
if num > 1:
for i in range(2, num-1):
if (num % i) == 0:
print(num, "is no Primenumber")
break
else:
print(i, "PRIME!!!")
(The original code doesn't print anything, but I had trouble with it, so I rewrote it for debugging purposes)
Anyway, it correctly identifies non prime numbers, but prints out this for prime numbers, lets say 7:
7
2 PRIME!!!
3 PRIME!!!
4 PRIME!!!
5 PRIME!!!

Use the below logic to find the number is prime or not. This is suited for python, as there is an else block for the for loop which gets exectued when the loop is not broken during execution which mean the number is prime.
if num > 1:
# Iterate from 2 to n / 2
for i in range(2, num//2):
# If num is divisible by any number between
# 2 and n / 2, it is not prime
if (num % i) == 0:
print(num, "is not a prime number")
break
else:
print(num, "is a prime number")
else:
print(num, "is not a prime number")

Not the most performance conscious, but will get the job done and is simple.
n = input("Enter a number to check if it is prime [returns True or False]: ")
n = int(n)
if n < 2:
quit()
if all(n % i for i in range(2, n)):
print(n, "PRIME!!!")

If you are concerned with performance for larger numbers it is better to use square root over number flooring division 2. Any number that is a factor greater than the square root will also have 1 that is less.
Here's a common solution:
def check_for_prime(p_in):
is_prime = True
for n in range(2, int(p_in ** 0.5) + 1):
if p_in % n == 0:
is_prime = False
break
return is_prime
num = int(input(""))
if check_for_prime(num):
print(f'{num} is prime')
else:
print(f'{num} is not prime')

Without third-party module, you can easily perform it this way:
def is_prime(number):
return all(number % n for n in range(2, number))
This way, you check:
all division between the specified number, and any number between 2 and the one before the number
the trick is to consider the fact that if any number % n gives 0, which means it is a multiple, so it can't be a prime number, and you will get a result immediately

How to find primes (Python)

I'm beginner in python programming and got stumpled on a Basic Question:
print the number of primes up to a given number.
For example the amount of primes before the number 100.
why my Code is not working? or what is wrong with my logic?
def count_prime(num):
newnumber = 0
for x in num:
if x%2 == 0:
newnumber = newnumber + 1
print(newnumber)
count_prime(100)

One very simple way (i.e. computationally slow) to test if a number is prime is to simply check if it can be divided by any other number.
def is_prime(n):
# If n is 0 or 1, it is not prime
if n <= 1:
return False
# Check every number between 2 and n-1
for i in range(2, n):
# If n is divisible by i, then the remainder will be zero
if n % i == 0:
return False
# Since n wasn't divisible by any other number, it must be prime
return True
To print all of the primes, you can then simply check each number.
for i in range(num):
if is_prime(i):
print(i)
This method will be very slow for large values of num. To see some much faster methods for checking the primality of a number you can have a look at this question.

num = int(input("Enter a number: "))
if num > 1:
for i in range(2,num):
if (num % i) == 0:
print(num,"is not a prime number")
print(i,"times",num//i,"is",num)
break
else:
print(num,"is a prime number")
else:
print(num,"is not a prime number")

First 100 prime numbers

I know there are a number of ways to find the first 100 prime numbers but please help me in my approach. I find the value of count to be increasing but for some reason the while loop condition doesn't apply:
count = 0
while(count <= 20):
for i in range(2, 20):
for j in range(2, i):
if i < j:
print("The number",i,"is prime")
elif i % j == 0:
break
else:
print("The number",i,"is prime")
count = count + 1
print(count)

You could use Sieve of Eratosthenes to find the first n prime numbers:
def primes_upto(limit):
prime = [True] * limit
for n in range(2, limit):
if prime[n]:
yield n # n is a prime
for c in range(n*n, limit, n):
prime[c] = False # mark composites
To get the first 100 primes:
>>> list(primes_upto(542))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, ... ,
499, 503, 509, 521, 523, 541]
To find the first n primes, you could estimate n-th prime (to pass the upper bound as the limit) or use an infinite prime number generator and get as many numbers as you need e.g., using list(itertools.islice(gen, 100)).

This is a more simpler code. We have looped all the numbers from 0 to the number until we have printed 100 prime numbers.
n=0
i=0
while n<100:
i+=1
count=1
for j in range(2,i):
if i%j==0:
count=0
break
if count==1:
print(i,end=' ')
n+=1

If one is looking for a mix of efficiency and simple code, Numpy is worth a try. With some fancy indexing, the following code does the job of implementing the sieve of Eratosthenes.
import numpy as np
ns = np.array(range(2,N))
primes = []
last_prime=2
while last_prime:
primes.append(last_prime)
ns = ns[ns%last_prime != 0]
last_prime = ns[0] if len(ns) > 0 else None
print(primes[:100])
Then just adjust N until you do have 100 primes. A good first guess is something in the order of 100*log(100) ~ 460 (coming from the prime number theorem). This will give 88 primes. Increasing N to the value of 600, you will have enough primes.

prime_count=0
n=1
while(True):
count=0
i=2
n+=1
while(i<=n):
if(n==i):
print(n)
prime_count+=1
elif(n%i==0):
break
i+=1
if(prime_count==100):
break

Beginner looking for advice/help on this code (Python) [duplicate]

I was having issues in printing a series of prime numbers from one to hundred. I can't figure our what's wrong with my code.
Here's what I wrote; it prints all the odd numbers instead of primes:
for num in range(1, 101):
for i in range(2, num):
if num % i == 0:
break
else:
print(num)
break

You need to check all numbers from 2 to n-1 (to sqrt(n) actually, but ok, let it be n).
If n is divisible by any of the numbers, it is not prime. If a number is prime, print it.
for num in range(2,101):
prime = True
for i in range(2,num):
if (num%i==0):
prime = False
if prime:
print (num)
You can write the same much shorter and more pythonic:
for num in range(2,101):
if all(num%i!=0 for i in range(2,num)):
print (num)
As I've said already, it would be better to check divisors not from 2 to n-1, but from 2 to sqrt(n):
import math
for num in range(2,101):
if all(num%i!=0 for i in range(2,int(math.sqrt(num))+1)):
print (num)
For small numbers like 101 it doesn't matter, but for 10**8 the difference will be really big.
You can improve it a little more by incrementing the range you check by 2, and thereby only checking odd numbers. Like so:
import math
print 2
for num in range(3,101,2):
if all(num%i!=0 for i in range(2,int(math.sqrt(num))+1)):
print (num)
Edited:
As in the first loop odd numbers are selected, in the second loop no
need to check with even numbers, so 'i' value can be start with 3 and
skipped by 2.
import math
print 2
for num in range(3,101,2):
if all(num%i!=0 for i in range(3,int(math.sqrt(num))+1, 2)):
print (num)

I'm a proponent of not assuming the best solution and testing it. Below are some modifications I did to create simple classes of examples by both #igor-chubin and #user448810. First off let me say it's all great information, thank you guys. But I have to acknowledge #user448810 for his clever solution, which turns out to be the fastest by far (of those I tested). So kudos to you, sir! In all examples I use a values of 1 million (1,000,000) as n.
Please feel free to try the code out.
Good luck!
Method 1 as described by Igor Chubin:
def primes_method1(n):
out = list()
for num in range(1, n+1):
prime = True
for i in range(2, num):
if (num % i == 0):
prime = False
if prime:
out.append(num)
return out
Benchmark: Over 272+ seconds
Method 2 as described by Igor Chubin:
def primes_method2(n):
out = list()
for num in range(1, n+1):
if all(num % i != 0 for i in range(2, num)):
out.append(num)
return out
Benchmark: 73.3420000076 seconds
Method 3 as described by Igor Chubin:
def primes_method3(n):
out = list()
for num in range(1, n+1):
if all(num % i != 0 for i in range(2, int(num**.5 ) + 1)):
out.append(num)
return out
Benchmark: 11.3580000401 seconds
Method 4 as described by Igor Chubin:
def primes_method4(n):
out = list()
out.append(2)
for num in range(3, n+1, 2):
if all(num % i != 0 for i in range(2, int(num**.5 ) + 1)):
out.append(num)
return out
Benchmark: 8.7009999752 seconds
Method 5 as described by user448810 (which I thought was quite clever):
def primes_method5(n):
out = list()
sieve = [True] * (n+1)
for p in range(2, n+1):
if (sieve[p]):
out.append(p)
for i in range(p, n+1, p):
sieve[i] = False
return out
Benchmark: 1.12000012398 seconds
Notes: Solution 5 listed above (as proposed by user448810) turned out to be the fastest and honestly quiet creative and clever. I love it. Thanks guys!!
EDIT: Oh, and by the way, I didn't feel there was any need to import the math library for the square root of a value as the equivalent is just (n**.5). Otherwise I didn't edit much other then make the values get stored in and output array to be returned by the class. Also, it would probably be a bit more efficient to store the results to a file than verbose and could save a lot on memory if it was just one at a time but would cost a little bit more time due to disk writes. I think there is always room for improvement though. So hopefully the code makes sense guys.
2021 EDIT: I know it's been a really long time but I was going back through my Stackoverflow after linking it to my Codewars account and saw my recently accumulated points, which which was linked to this post. Something I read in the original poster caught my eye for #user448810, so I decided to do a slight modification mentioned in the original post by filtering out odd values before appending the output array. The results was much better performance for both the optimization as well as latest version of Python 3.8 with a result of 0.723 seconds (prior code) vs 0.504 seconds using 1,000,000 for n.
def primes_method5(n):
out = list()
sieve = [True] * (n+1)
for p in range(2, n+1):
if (sieve[p] and sieve[p]%2==1):
out.append(p)
for i in range(p, n+1, p):
sieve[i] = False
return out
Nearly five years later, I might know a bit more but I still just love Python, and it's kind of crazy to think it's been that long. The post honestly feels like it was made a short time ago and at the time I had only been using python about a year I think. And it still seems relevant. Crazy. Good times.

Instead of trial division, a better approach, invented by the Greek mathematician Eratosthenes over two thousand years ago, is to sieve by repeatedly casting out multiples of primes.
Begin by making a list of all numbers from 2 to the maximum desired prime n. Then repeatedly take the smallest uncrossed number and cross out all of its multiples; the numbers that remain uncrossed are prime.
For example, consider the numbers less than 30. Initially, 2 is identified as prime, then 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28 and 30 are crossed out. Next 3 is identified as prime, then 6, 9, 12, 15, 18, 21, 24, 27 and 30 are crossed out. The next prime is 5, so 10, 15, 20, 25 and 30 are crossed out. And so on. The numbers that remain are prime: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
def primes(n):
sieve = [True] * (n+1)
for p in range(2, n+1):
if (sieve[p]):
print p
for i in range(p, n+1, p):
sieve[i] = False
An optimized version of the sieve handles 2 separately and sieves only odd numbers. Also, since all composites less than the square of the current prime are crossed out by smaller primes, the inner loop can start at p^2 instead of p and the outer loop can stop at the square root of n. I'll leave the optimized version for you to work on.

break ends the loop that it is currently in. So, you are only ever checking if it divisible by 2, giving you all odd numbers.
for num in range(2,101):
for i in range(2,num):
if (num%i==0):
break
else:
print(num)
that being said, there are much better ways to find primes in python than this.
for num in range(2,101):
if is_prime(num):
print(num)
def is_prime(n):
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return False
return True

The best way to solve the above problem would be to use the "Miller Rabin Primality Test" algorithm. It uses a probabilistic approach to find if a number is prime or not. And it is by-far the most efficient algorithm I've come across for the same.
The python implementation of the same is demonstrated below:
def miller_rabin(n, k):
# Implementation uses the Miller-Rabin Primality Test
# The optimal number of rounds for this test is 40
# See http://stackoverflow.com/questions/6325576/how-many-iterations-of-rabin-miller-should-i-use-for-cryptographic-safe-primes
# for justification
# If number is even, it's a composite number
if n == 2:
return True
if n % 2 == 0:
return False
r, s = 0, n - 1
while s % 2 == 0:
r += 1
s //= 2
for _ in xrange(k):
a = random.randrange(2, n - 1)
x = pow(a, s, n)
if x == 1 or x == n - 1:
continue
for _ in xrange(r - 1):
x = pow(x, 2, n)
if x == n - 1:
break
else:
return False
return True

Igor Chubin's answer can be improved. When testing if X is prime, the algorithm doesn't have to check every number up to the square root of X, it only has to check the prime numbers up to the sqrt(X). Thus, it can be more efficient if it refers to the list of prime numbers as it is creating it. The function below outputs a list of all primes under b, which is convenient as a list for several reasons (e.g. when you want to know the number of primes < b). By only checking the primes, it saves time at higher numbers (compare at around 10,000; the difference is stark).
from math import sqrt
def lp(b)
primes = [2]
for c in range(3,b):
e = round(sqrt(c)) + 1
for d in primes:
if d <= e and c%d == 0:
break
else:
primes.extend([c])
return primes

My way of listing primes to an entry number without too much hassle is using the property that you can get any number that is not a prime with the summation of primes.
Therefore, if you divide the entry number with all primes below it, and it is not evenly divisible by any of them, you know that you have a prime.
Of course there are still faster ways of getting the primes, but this one already performs quite well, especially because you are not dividing the entry number by any number, but quite only the primes all the way to that number.
With this code I managed on my computer to list all primes up to 100 000 in less than 4 seconds.
import time as t
start = t.clock()
primes = [2,3,5,7]
for num in xrange(3,100000,2):
if all(num%x != 0 for x in primes):
primes.append(num)
print primes
print t.clock() - start
print sum(primes)

A Python Program function module that returns the 1'st N prime numbers:
def get_primes(count):
"""
Return the 1st count prime integers.
"""
result = []
x=2
while len(result) in range(count):
i=2
flag=0
for i in range(2,x):
if x%i == 0:
flag+=1
break
i=i+1
if flag == 0:
result.append(x)
x+=1
pass
return result

A simpler and more efficient way of solving this is storing all prime numbers found previously and checking if the next number is a multiple of any of the smaller primes.
n = 1000
primes = [2]
for i in range(3, n, 2):
if not any(i % prime == 0 for prime in primes):
primes.append(i)
print(primes)
Note that any is a short circuit function, in other words, it will break the loop as soon as a truthy value is found.

we can make a list of prime numbers using sympy library
import sympy
lower=int(input("lower value:")) #let it be 30
upper=int(input("upper value:")) #let it be 60
l=list(sympy.primerange(lower,upper+1)) #[31,37,41,43,47,53,59]
print(l)

Here's a simple and intuitive version of checking whether it's a prime in a RECURSIVE function! :) (I did it as a homework assignment for an MIT class)
In python it runs very fast until 1900. IF you try more than 1900, you'll get an interesting error :) (Would u like to check how many numbers your computer can manage?)
def is_prime(n, div=2):
if div> n/2.0: return True
if n% div == 0:
return False
else:
div+=1
return is_prime(n,div)
#The program:
until = 1000
for i in range(until):
if is_prime(i):
print i
Of course... if you like recursive functions, this small code can be upgraded with a dictionary to seriously increase its performance, and avoid that funny error.
Here's a simple Level 1 upgrade with a MEMORY integration:
import datetime
def is_prime(n, div=2):
global primelist
if div> n/2.0: return True
if div < primelist[0]:
div = primelist[0]
for x in primelist:
if x ==0 or x==1: continue
if n % x == 0:
return False
if n% div == 0:
return False
else:
div+=1
return is_prime(n,div)
now = datetime.datetime.now()
print 'time and date:',now
until = 100000
primelist=[]
for i in range(until):
if is_prime(i):
primelist.insert(0,i)
print "There are", len(primelist),"prime numbers, until", until
print primelist[0:100], "..."
finish = datetime.datetime.now()
print "It took your computer", finish - now , " to calculate it"
Here are the resuls, where I printed the last 100 prime numbers found.
time and date: 2013-10-15 13:32:11.674448
There are 9594 prime numbers, until 100000
[99991, 99989, 99971, 99961, 99929, 99923, 99907, 99901, 99881, 99877, 99871, 99859, 99839, 99833, 99829, 99823, 99817, 99809, 99793, 99787, 99767, 99761, 99733, 99721, 99719, 99713, 99709, 99707, 99689, 99679, 99667, 99661, 99643, 99623, 99611, 99607, 99581, 99577, 99571, 99563, 99559, 99551, 99529, 99527, 99523, 99497, 99487, 99469, 99439, 99431, 99409, 99401, 99397, 99391, 99377, 99371, 99367, 99349, 99347, 99317, 99289, 99277, 99259, 99257, 99251, 99241, 99233, 99223, 99191, 99181, 99173, 99149, 99139, 99137, 99133, 99131, 99119, 99109, 99103, 99089, 99083, 99079, 99053, 99041, 99023, 99017, 99013, 98999, 98993, 98981, 98963, 98953, 98947, 98939, 98929, 98927, 98911, 98909, 98899, 98897] ...
It took your computer 0:00:40.871083 to calculate it
So It took 40 seconds for my i7 laptop to calculate it. :)

# computes first n prime numbers
def primes(n=1):
from math import sqrt
count = 1
plist = [2]
c = 3
if n <= 0 :
return "Error : integer n not >= 0"
while (count <= n - 1): # n - 1 since 2 is already in plist
pivot = int(sqrt(c))
for i in plist:
if i > pivot : # check for primae factors 'till sqrt c
count+= 1
plist.append(c)
break
elif c % i == 0 :
break # not prime, no need to iterate anymore
else :
continue
c += 2 # skipping even numbers
return plist

You are terminating the loop too early. After you have tested all possibilities in the body of the for loop, and not breaking, then the number is prime. As one is not prime you have to start at 2:
for num in xrange(2, 101):
for i in range(2,num):
if not num % i:
break
else:
print num
In a faster solution you only try to divide by primes that are smaller or equal to the root of the number you are testing. This can be achieved by remembering all primes you have already found. Additionally, you only have to test odd numbers (except 2). You can put the resulting algorithm into a generator so you can use it for storing primes in a container or simply printing them out:
def primes(limit):
if limit > 1:
primes_found = [(2, 4)]
yield 2
for n in xrange(3, limit + 1, 2):
for p, ps in primes_found:
if ps > n:
primes_found.append((n, n * n))
yield n
break
else:
if not n % p:
break
for i in primes(101):
print i
As you can see there is no need to calculate the square root, it is faster to store the square for each prime number and compare each divisor with this number.

How about this? Reading all the suggestions I used this:
prime=[2]+[num for num in xrange(3,m+1,2) if all(num%i!=0 for i in range(2,int(math.sqrt(num))+1))]
Prime numbers up to 1000000
root#nfs:/pywork# time python prime.py
78498
real 0m6.600s
user 0m6.532s
sys 0m0.036s

Adding to the accepted answer, further optimization can be achieved by using a list to store primes and printing them after generation.
import math
Primes_Upto = 101
Primes = [2]
for num in range(3,Primes_Upto,2):
if all(num%i!=0 for i in Primes):
Primes.append(num)
for i in Primes:
print i

Here is the simplest logic for beginners to get prime numbers:
p=[]
for n in range(2,50):
for k in range(2,50):
if n%k ==0 and n !=k:
break
else:
for t in p:
if n%t ==0:
break
else:
p.append(n)
print p

n = int(input())
is_prime = lambda n: all( n%i != 0 for i in range(2, int(n**.5)+1) )
def Prime_series(n):
for i in range(2,n):
if is_prime(i) == True:
print(i,end = " ")
else:
pass
Prime_series(n)
Here is a simplified answer using lambda function.

def function(number):
for j in range(2, number+1):
if all(j % i != 0 for i in range(2, j)):
print(j)
function(13)

for i in range(1, 100):
for j in range(2, i):
if i % j == 0:
break
else:
print(i)

Print n prime numbers using python:
num = input('get the value:')
for i in range(2,num+1):
count = 0
for j in range(2,i):
if i%j != 0:
count += 1
if count == i-2:
print i,

def prime_number(a):
yes=[]
for i in range (2,100):
if (i==2 or i==3 or i==5 or i==7) or (i%2!=0 and i%3!=0 and i%5!=0 and i%7!=0 and i%(i**(float(0.5)))!=0):
yes=yes+[i]
print (yes)

min=int(input("min:"))
max=int(input("max:"))
for num in range(min,max):
for x in range(2,num):
if(num%x==0 and num!=1):
break
else:
print(num,"is prime")
break

This is a sample program I wrote to check if a number is prime or not.
def is_prime(x):
y=0
if x<=1:
return False
elif x == 2:
return True
elif x%2==0:
return False
else:
root = int(x**.5)+2
for i in xrange (2,root):
if x%i==0:
return False
y=1
if y==0:
return True

n = int(raw_input('Enter the integer range to find prime no :'))
p = 2
while p<n:
i = p
cnt = 0
while i>1:
if p%i == 0:
cnt+=1
i-=1
if cnt == 1:
print "%s is Prime Number"%p
else:
print "%s is Not Prime Number"%p
p+=1

Using filter function.
l=range(1,101)
for i in range(2,10): # for i in range(x,y), here y should be around or <= sqrt(101)
l = filter(lambda x: x==i or x%i, l)
print l

for num in range(1,101):
prime = True
for i in range(2,num/2):
if (num%i==0):
prime = False
if prime:
print num

f=0
sum=0
for i in range(1,101):
for j in range(1,i+1):
if(i%j==0):
f=f+1
if(f==2):
sum=sum+i
print i
f=0
print sum

The fastest & best implementation of omitting primes:
def PrimeRanges2(a, b):
arr = range(a, b+1)
up = int(math.sqrt(b)) + 1
for d in range(2, up):
arr = omit_multi(arr, d)

Here is a different approach that trades space for faster search time. This may be fastest so.
import math
def primes(n):
if n < 2:
return []
numbers = [0]*(n+1)
primes = [2]
# Mark all odd numbers as maybe prime, leave evens marked composite.
for i in xrange(3, n+1, 2):
numbers[i] = 1
sqn = int(math.sqrt(n))
# Starting with 3, look at each odd number.
for i in xrange(3, len(numbers), 2):
# Skip if composite.
if numbers[i] == 0:
continue
# Number is prime. Would have been marked as composite if there were
# any smaller prime factors already examined.
primes.append(i)
if i > sqn:
# All remaining odd numbers not marked composite must be prime.
primes.extend([i for i in xrange(i+2, len(numbers), 2)
if numbers[i]])
break
# Mark all multiples of the prime as composite. Check odd multiples.
for r in xrange(i*i, len(numbers), i*2):
numbers[r] = 0
return primes
n = 1000000
p = primes(n)
print "Found", len(p), "primes <=", n

Adding my own version, just to show some itertools tricks v2.7:
import itertools
def Primes():
primes = []
a = 2
while True:
if all(itertools.imap(lambda p : a % p, primes)):
yield a
primes.append(a)
a += 1
# Print the first 100 primes
for _, p in itertools.izip(xrange(100), Primes()):
print p