Coding exclusions in a for loop (python)

Question

i'm trying to figure out how to exclude certain numbers (numbers that aren't prime) in my program. For example, I don't want 15 to be printed so I used the modulo to exclude numbers divisible by 5 but that eliminates 5, which is a prime number. Any help on how to code this would be huge help! Thank you...

start = int(input("Start number: "))
end = int(input("End number: "))

while start < 0 or end < 0:
    print ("Start and end number must be positive. Try again.")
    start = int(input("Start number: "))
    end = int(input("End number: "))

while start > end:
    print ("End number must be greater than start number. Try again.")


for x in range (start, end):
    is_prime = True

for x in range (start, end):
    trial = x % 2
    trial1 = x % 5
    trial2 = x % 3
    if trial != 0 and trial1 != 0 and trial2 != 0:
        print (x, "is a prime number")
        x = x + 1
    else:
        is_prime = False

http://stackoverflow.com/questions/18833759/python-prime-number-checker — user5219763, Mar 23 '16 at 15:21

score 0 · Answer 1 · answered Mar 23 '16 at 15:24

0

You can create a list of numbers you want to use in your loop let's call it loop_list and then simply use for x in loop_list, that avoid you setting up exclusions

answered Mar 23 '16 at 15:24

Alessandro

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score 0 · Answer 2 · answered Mar 23 '16 at 15:24

0

If you want to be sure that N is a prime number, you have to check for all number to squareroot(N) if N%number != 0.

Are you looking for optimisation? If no, you just need that.

answered Mar 23 '16 at 15:24

Whitefret

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score 0 · Answer 3 · answered Mar 23 '16 at 16:07

0

You can try this prime number logic :-

def check_prime(num):
    notPrimeFlag = 'N'
    for i in range(2, num):
        if num % i == 0:
            notPrimeFlag = 'Y'
            break
    if notPrimeFlag == 'N':
        print("Prime number")

answered Mar 23 '16 at 16:07

Sourav

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Coding exclusions in a for loop (python)

3 Answers3