just change num to num ** 0.5 + 1, it's the same calculation but way faster for larger inputs.

my method is fast a half of time than your, but also slow

This one timed out. FAIL.

Are you saying it's better to perform O(sqrt(n)) times more operations to save up on one roughly O(1) operation?

True, but a power or square root operation is a very expensive operation.

Slow, you're iterating through way too many 'n's checking divisibility

this code will timeout if num is like 100000

range(2,int(num/2)) maybe quicker
the function any() ,its my first time see it~ learning~

this code should not range to num,just to the square root of num,when the num is big or there is not only a num,it's slow

... range(2,int(num ** 0.5) + 1) - it is faster

A prime number is a natural number greater than 1

nmadsen, that is not true: all will return as soon as it hits a false value. And this will happen at the same entry as the not any variant, since the expression is opposite.

From a logical standpoint, you are correct. However, not any() is faster than all() in this case for any non-prime number because any() will return true as soon as it hits a true value (i.e. it doesn't have to go through the whole range) whereas all() has to check the entire range before outputting its result.

If the number is negative, it is not a prime by definition. Using the absolute value would result in incorrect responses.

the question also prompted for negative inputs. Of course mathematically we define for positive integers.