Need insight on how to fix my Python prime number list generator-CodePudding

I'm new to programming, and to practice learning Python, I'm trying to write a program that will test which numbers in a range from (2, n-1) are prime, by testing if there are any factors num is divisible by between 2 and sqrt(n).

For the most part, the prime list is pretty accurate, but for some reason the squares of primes >= 11 make it through the loop. The for i in range part of the loop should be filtering them out. I have a feeling it has to do wih how I ordered the if-statements, or maybe how I indented everything, but I can't quite figure it out. Any help would be appreciated!

I also realize I could just keep adding num % 11 == 0 or num % 13 == 0 to line 7, but that kinda defeats the purpose of the program automating this for me.

Also, I know I could just find a million other Python prime checkers, but I'm trying to figure out where I'm going wrong here so I can improve. Thank you!!!

n = 362
prime = []
def prime_check(lst=range(2,n)):
  for num in lst:
    if num == 1 or num == 2 or num == 3 or num == 5 or num == 7:
      prime.append(num)
    if num % 2 == 0 or num % 3 == 0 or num % 5 == 0:
      continue
    for i in range(5, int(sqrt(n) 1), 2):
      if num % i == 0:
          continue
    else:
      prime.append(num)
  print(prime)

CodePudding user response：

import math
n = 362
prime = []
def prime_check(lst=range(2,n)):
    for num in lst:
        if num == 2 or num == 3 or num == 5 or num == 7:
            prime.append(num)
            continue
        if num % 2 == 0 or num % 3 == 0 or num % 5 == 0:
            continue
        flag = 0
        for i in range(5, int(math.sqrt(num) 1), 2):
            if num % i == 0:
                flag=1
                break
        if flag == 0:
            prime.append(num)
    return prime
print(prime_check())

You had one issue in the logic. The flag ensures a prime is added only once.

Note: As @Kelly Bundy mentioned wrongly, and I also got confused. The logic that you have wrote is wrong. That is exactly why the number 121 which is just 11x11 gets into the prime list. Note2: One portion I wrote wrong earlier regarding else part. (Corrected due to @Kelly Bundy)

def prime_check(lst=range(2,n)):
    for num in lst:
        flag = 0
        if num % 2 == 0:
             continue
        for i in range(3, int(math.sqrt(num) 1), 2):
            if num % i == 0:
                flag=1
                break
        if flag == 0:
            prime.append(num)
    return prime

Note: 1 is non-prime.

My earlier answer and even now the current answer, avoided for-else. Personally I, avoid it explicitly - using such syntactic constructs not only makes it less readable for me but the time to understand such things in large codebase is really difficult.

CodePudding user response：

You should use the correct root and use break and not treat 7 inconsistently.

Fixed code (Try it online!):

from math import sqrt

n = 362
prime = []
def prime_check(lst=range(2,n)):
  for num in lst:
    if num == 1 or num == 2 or num == 3 or num == 5:
      prime.append(num)
    if num % 2 == 0 or num % 3 == 0 or num % 5 == 0:
      continue
    for i in range(5, int(sqrt(num) 1), 2):
      if num % i == 0:
          break
    else:
      prime.append(num)
  print(prime)

prime_check()

Output:

[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, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359]