Why doesn't this work (Python)?

Question

So I'm trying to learn Python using some tutorials, and I decided to create my own exercise. I wanted to create a script that figures out how many prime numbers there are between 1 and 1000, as well as printing the prime numbers. This is what I have so far:

numberlist = []
a = 1
prime = True
while a < 1000:
    a = a + 1
    for divisor in range(2,a):
        if a/divisor==int(a/divisor):
            prime=False
        if prime == True:
            print a
            numberlist.append(a)

print "Number of prime numbers between 1 and 1000:", len(numberlist)

However, when I start the application, it returns

"Number of prime numbers between 1 and 1000: 0"

I don't know what I've done wrong. Can someone please clarify what I screwed up on?

Thank you for your help.

Edit: So now the code looks like this, but the same problem is occurring:

numberlist = []

a = 1

for a in xrange(1, 1000):
    for divisor in range(2,a):
        if a % divisor == 0:
           prime=False
        else:
           prime=True
if prime == True:
    print a
    numberlist.append(a)

print "Number of prime numbers between 1 and 1000:", len(numberlist)

Just to let you know, you can significantly improve this algorithm by doing for divisor in numberlist (if you make a start at 2, or do for divisor in numberlist[1:] and just start with numberlist=[1]). — Brendan Long
– Brendan Long, Commented Jul 26, 2011 at 21:10

stefan · Accepted Answer · 2011-07-26 21:54:25Z

There are a number of problems with this code. First, a/divisor == int(a/divisor) is always True, because division in Python is integral (you will always get an integer result). Use a % divisor == 0 instead.

Another problem is that prime = True is outside the loop, meaning that as soon as one value is declared not prime, no more values can be prime! (prime will never get set back to True anywhere.)

A third issue, more of style, is that it is preferred to use a for ... in loop in python, eg

for a in xrange(1, 1000):
    for divisor in xrange(2, a): ...

Edit: Regarding your modified code: your if statement at the very end is not indented, and therefore not part of the for loop. It is only being executed once, after the loop ends, meaning prime will be the last value set (where a is 999, not a prime). You want to indent the entire if statement to be in your inner for loop. You could condense this, though:

for a in xrange(1, 1000):
    for divisor in xrange(2, a):
        if a % divisor == 0:
            break
    else:
        print a
        numberlist.append(a)

Note that this uses the for ... else clause. Putting else at the end of a for loop will cause that else block to run only if the loop is not broken by break.

As a further comment, starting the whole loop at 1 will have 1 in the list, but 1 is not a prime.

Patrick87 · Accepted Answer · 2011-07-26 21:07:41Z

3

a/divisor evaluates to an int since both a and divisor are ints, so a/divisor == int(a/divisor) is always true. Try checking the remainder after division... the modulus... I think that in Python, maybe

  if a % divisor == 0:
     ...

This basically says "if the remainder after dividing a by divisor is zero, then".

answered Jul 26, 2011 at 21:07

Patrick87

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1 Comment

Agustin Cautin Over a year ago

This is changed on the version 3.0. "An expression like 1/2 returns a float. Use 1//2 to get the truncating behavior. (The latter syntax has existed for years, at least since Python 2.2.)" http://www.python.org/dev/peps/pep-0238/

John Percival Hackworth · Accepted Answer · 2011-07-26 21:07:51Z

3

You never reset prime to True.

answered Jul 26, 2011 at 21:07

John Percival Hackworth

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Comments

Alex Reynolds · Accepted Answer · 2011-07-26 21:09:17Z

1

I think you forgot to reset prime's value to True on each iteration.

answered Jul 26, 2011 at 21:09

Alex Reynolds

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Comments

user448810 · Accepted Answer · 2011-07-27 16:50:31Z

0

It may help you when you get further along to know that there are 168 primes less than 1000.

answered Jul 27, 2011 at 16:50

user448810

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