Re: isPrime works but UnBoundLocalError when mapping on list

[norseman]

Mensanator wrote:

On Jul 15, 12:36 pm, defn noob <[EMAIL PROTECTED]> wrote:

On Jul 15, 7:28 pm, Mensanator <[EMAIL PROTECTED]> wrote:

On Jul 15, 11:26 am, defn noob <[EMAIL PROTECTED]> wrote:

isPrime works when just calling a nbr but not when iterating on a
list, why? adding x=1 makes it work though but why do I have to add
it?
Is there a cleaner way to do it?
def isPrime(nbr):
    for x in range(2, nbr + 1):
        if nbr % x == 0:
            break
    if x == nbr:
        return True
    else:
        return False

[isPrime(y) for y in range(11)]

Traceback (most recent call last):
  File "<pyshell#45>", line 1, in <module>
    [isPrime(y) for y in range(11)]
  File "C:\Python25\Progs\blandat\myMath.py", line 9, in isPrime
    if x == nbr:
UnboundLocalError: local variable 'x' referenced before assignment

map(isPrime, range(100))

Traceback (most recent call last):
  File "<pyshell#38>", line 1, in <module>
    map(isPrime, range(100))
  File "C:\Python25\Progs\blandat\myMath.py", line 9, in isPrime
    if x == nbr:
UnboundLocalError: local variable 'x' referenced before assignment>>> 
isPrime(10)
False

isPrime(11)

True
adding x=1 makes it work though:
def isPrime(nbr):
    x=1
    for x in range(2, nbr + 1):
        if nbr % x == 0:
            break
    if x == nbr:
        return True
    else:
        return False

[isPrime(y) for y in range(11)]

[False, True, True, True, False, True, False, True, False, False,
False]

No, it doesn't. You are falsely reporting that 1 is prime.
And instead of making the fake variable x, shouldn't you
instead test that nbr+1 is greater than 2? Or call it with
range(3,11) instead of range(11)? x isn't initialized
because if nbr+1 is <=2, the for loop has an invalid range
and doesn't even execute.

def isPrime(nbr):
    for x in range(2, nbr + 1):
        if nbr % x == 0:
            break
    if x == nbr:
        return True
    else:
        return False

this works for all primes, if i want to not include 1 i just do if
nbr<=1 return false

you are answering the wrong question.

No, I also mentioned the for loop having an invalid range,
which is why your original failed.

Pointing out that 1 isn't prime was a bonus.

anyway here is a clear one:
def isPrime(nbr):
    if nbr < 2:
        return False
    for x in range(2, nbr + 1):
        if nbr % x == 0:
            return nbr == x

I suppose you're not interested in knowing you don't
have to test anything higher than the square root of
the number.

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===========================
"don't...test...higher than the square root..."

I wondered when that was going to show up.
I too had a good math teacher.

Steve
[EMAIL PROTECTED]
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