Print series of prime numbers in python

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)

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)

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

I was inspired by Igor and made a code block that creates a list:

def prime_number():

for num in range(2, 101):
    prime = True
    for i in range(2, num):
        if (num % i == 0):
            prime = False
    if prime and num not in num_list:
        num_list.append(num)
    else:
        pass
return num_list


num_list = []
prime_number()
print(num_list)

a=int(input('enter the lower no.'))
b=int(input('enter the higher no.'))
print("Prime numbers between",a,"and",b,"are:")
for num in range(a,b):

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

First we find factor of that number

def fac(n):
  res = []
  for i in range(1,n+1):
    if n%i == 0:
res.append(i)

Script to check prime or not

def prime(n):
return(fac(n) == [1,n])

Script to print all prime number upto n

def prime_list(n):
  pri_list = []
  for i in range(1,n+1):
    if prime(i)
      pri_list.append(i)
return(pri_list)

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

as one line comprehension list solution:

>>> [num for num in range(2, 101) if all(num % i != 0 for i in range(2, num))]
[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]

which 101 represent the maximum range value between 2 and 100 of list

1 can only be divided by one number, 1 itself, so with this definition 1 is not a prime number. The primary number must have only two positive factors 1 and itself.

num= int(input("Enter the numbner"))
isDivisible= False
int=2

while i<num:
    if num%i==0
    isDivisible True
    print("The number {} is divisible by {}.".format(num,i))
    i +=1

if isDivisible:
    print("The number {} is not prime.".format(num))
else:
    print("The number {} is a prime number.".format(num))

min = int(input("Enter lower range: ")) max = int(input("Enter upper range: "))

print("The Prime numbes between",min,"and",max,"are:"

for num in range(min,max + 1): if num > 1: for i in range(2,num): if (num % i) == 0: break else: print(num)