Sathvik Babu Veligatla <sathvikveliga...@gmail.com> writes:
> 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.
As others have already noticed, the l.count(1) >> 0: is faulty. You probably
meant
l.count(1) > 0:, but else would be simpler.
But then, once there is a 1 in l that never disappears, so after that it will
never print any more numbers. You should put the l = [] AFTER the while True,
so that you start with a new empty list every cycle.
And then, putting 0 an 1's in a list and then counting how many 1's there are
is quite inefficient. Just counting from the beginning would be more efficient.
Like:
a = 3
while True:
count = 0
for i in range(2,a):
if a%i == 0:
count += 1
if count == 0:
print(a)
a = a + 2
else:
a = a + 2
Which can be simplified to:
a = 3
while True:
count = 0
for i in range(2,a):
if a%i == 0:
count += 1
if count == 0:
print(a)
a = a + 2
And you effectively use the counter as a boolean, so replace is by a boolean.
a = 3
while True:
is_prime = True
for i in range(2,a):
if a%i == 0:
is_prime = False
# once we found a divisor, it is no use to continue
break
if is_prime:
print(a)
a = a + 2
Further optimizations are possible, for example use range(2,a/2) or even range
(2, sqrt(a)).
--
Pieter van Oostrum
www: http://pieter.vanoostrum.org/
PGP key: [8DAE142BE17999C4]
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