python script to give a list of prime no.

[Sathvik Babu Veligatla]

On Sunday, April 5, 2020 at 8:03:19 PM UTC+5:30, inhahe wrote:
> On Sun, Apr 5, 2020 at 8:26 AM Sathvik Babu Veligatla <
> sathvikveligatla at gmail.com> wrote:
> 
> > hi,
> > I am new to python, and i am trying to output the prime numbers beginning
> > from 3 and i cannot get the required output.
> > It stops after giving the output "7" and that's it.
> >
> > CODE:
> > a = 3
> > l = []
> > while True:
> >     for i in range(2,a):
> >         if a%i == 0:
> >             l.append(1)
> >         else:
> >             l.append(0)
> >
> >     if l.count(1) == 0:
> >         print(a)
> >         a = a + 2
> >     elif l.count(1) >> 0:
> >         a = a + 2
> >
> >
> >
> > Any help is appreciated,
> > Thank you.
> > --
> > https://mail.python.org/mailman/listinfo/python-list
> 
> 
> 
> "l = []" needs to go inside the "while True" loop. as it is, it keeps
> adding 1's and 0's to the list with each new "a" value, and then counts the
> 1's and 0's, without ever resetting the list back to empty, so once it gets
> to 9 the list has a 1 in it and l.count(1) will never be false again. 3, 5
> and 7 are prime so a 1 doesn't get added yet with those values, which is
> why 7 is the highest it gets to. so with "l = []" inside the while loop,
> "l" would be reset to empty with each new "a" value.
> 
> btw, this is a very inefficient way to compute prime numbers. for example,
> why add the 0's to the list? you never count the number of 0's so it's
> unnecessary. another example.. why use a list at all? all that matters is
> that it find *one* result where a%i==0, it doesn't matter how many results
> it found after that, so all you need is a boolean. and you might as well
> stop testing a%i for *all* i values up to i, because once it finds the
> first result that's 0, you know it's prime so you can just go to the next
> number. and you don't need i to go all the way to a, because a%i will
> always be > 0 for any i over a/2. and since you're not testing evens, it's
> actually any i over a/3. in actuality, though, computing any a%i where i >
> the square root of a is redundant because for each factor of a above the
> square root of a, you've already tested its cofactor, say a is 100. the
> square root of 100 is 10. if you're testing 100%20, that's unnecessary
> because 100/20 is 5 and you've already tested 100%5.

thank you buddy,
it worked...