Python函数查找素数

Question

This is not about finding the prime numbers, it is about how do I transfer the code into a function.

so I have this codes to help me to print the prime numbers from 2-100:

pnumber = [x for x in range(2, 101) if all(x % i for i in range(2, x))]
print(pnumber)

If I def this as a function, to find the prime numbers in a range:

 def p_number(a, b):
     pnumber = [x for x in range(a, b+1) if all(x % i for i in range(2, b))]
     print(pnumber)

 p_number(2, 100)

You can see that I use a to replace 2, and b to replace 100, and change the codes accordingly. But somehow, this won't work, it will output an empty list.

I wonder why?

Answer 1

Mind that if you use the upperbound b in the check:

all(x % i for i in range(2, b))

this will include all prime numbers up to b . So 2 , 3 , 5 , etc. are also part of range(2, b) (given b is large enough). So that means that if we test for instance whether 3 is a prime, we will check for i = 3 , and 3 % 3 is 0 , so that will fail.

Furthermore it will have a bad impact on performance. The idea of a prime test is to check all numbers up to but excluding the number. So a quick fix is:

def p_number(a, b):
    pnumber = [x for x in range(a, b+1) if all(x % i for i in range(2, ))]
    print(pnumber)

We can easily boost it further by using int(sqrt(x))+1 instead of x :

def p_number(a, b):
    pnumber = [x for x in range(a, b+1) if all(x % i for i in range(2, x))]
    print(pnumber)

We can boost it further, for instance by only evaluating odd numbers (and add 2 to the result). But using sqrt will usually already result in a significant speedup.

Answer 2

Change your function to -

def p_number(a, b):
     pnumber = [x for x in range(a, b + 1) if all(x % i for i in range(a, x))]
     print(pnumber)

If you notice, you were iterating i from 2 to b, and not from 2 to x which is why you were getting an empty list.