CODEWITHSUNDEEP
python
Jack
Jack

Reputation: 461

How to check all the numbers in a range against a calculation?

So I'm trying to print all the prime numbers up to 100. Here's my code so far:

primes = []

for num in range(100):
    if num > 1:
        for i in range(2, num):
            if num % i != 0:
                primes += [num]
                break
            else:
                break
            
print(primes)

However when i run the code, it only does the calculation (if num % i != 0) for the first iteration in the range (for i in range(2, num)). This is a problem in the case of numbers like 15, where the first iteration in for i in range is 2. 15 divided by 2 does not give a whole number, and so it runs primes += [num]. It has taken the first iteration and has not run the rest of them. Is there a way to run all the iterations in the range?

Upvotes: 1

Views: 81

Answers (5)

Gabio
Gabio

Reputation: 9494

Try this:

primes = []

for num in range(100):
    if num > 1:
        for i in range(2, num):
            if num % i == 0:
                break
        else:
            primes.append(num)

print(primes)

The code will execute the else block only if the inner for loop is exhausted, i.e completes it's all iteration without breaking.

Upvotes: 1

Eeshaan
Eeshaan

Reputation: 1635

You got the logic somewhat wrong. Here's how you've to do it

primes = []

for num in range(2, 100): # 0 and 1 aren't primes
    for i in range(2, num):
        if num % i == 0:
            break
    else: # the else condition is for the for loop i.e. if it executes without break
        primes.append(num)
            
print(primes)

Upvotes: 1

HackLab
HackLab

Reputation: 539

tried to adapt to your approach although, I think there should be better ways to solve this problem

>>> primes = []
>>> for num in range(2,100):
...     p = True
...     for i in range(2, num):
...         if num % i == 0:
...             p = False
...     if p:
...         primes += [num]
...
>>> primes
[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]

Upvotes: 0

Pedro Rodrigues
Pedro Rodrigues

Reputation: 2658

Here is my take on it.

>>> def primes(limit=100):
...     for number in range(limit):
...         is_prime = all([number % divisor != 0 for divisor in range(2, number)])
...         if is_prime:
...             yield number
...
>>> list(primes(10))
[0, 1, 2, 3, 5, 7]
>>> list(primes(100))
[0, 1, 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]
>>>

Is not complete, 0 and 1 are not primes. But I think you get the idea.

Upvotes: 0

Koen Dercksen
Koen Dercksen

Reputation: 111

The all function may be what you're looking for!

Upvotes: 0

Related Questions