On Mon, Dec 16, 2013 at 09:49:23AM +0100, Rafael Knuth wrote:
> That's actually a good example. Let me explain what I feel confused about.
> Print actually runs through the entire loop.
No it doesn't. Print has nothing to do with loops. It just happens that
the body of the loop -- the thing that
On 12/16/2013 09:49 AM, Rafael Knuth wrote:
Hej there,
number = 9
for element in range(2,9):
3 % 2 != 0:
My assumption is that the program should end the loop after the first
iteration again and it then should return True.
No. If it did that, it wouldn't be a *loop* at all, would it? The whol
On 12/16/2013 11:34 AM, spir wrote:
On 12/15/2013 05:54 PM, Rafael Knuth wrote:
PS: using "print" as Mark proposes is indeed you best friend to unsertand loops
(and also recursion).
Denis
___
Tutor maillist - Tutor@python.org
To unsubscribe or ch
On 12/15/2013 05:54 PM, Rafael Knuth wrote:
Hej,
I stumbled upon this program here (Python 3.3.0) and I don't quite
understand how the for loop plays with the return True statement:
def is_prime(number):
for element in range(2, number):
if number % element == 0:
retur
On 16.12.2013 09:49, Rafael Knuth wrote:
That's the tiny little detail I am confused about: What does return
exactly do? Does it return only the first value within a loop or does
it iterate through all values within a loop? (unless a given condition
is met)
The return statement has nothing to d
Hej there,
>> number = 9
>> for element in range(2,9):
>> 3 % 2 != 0:
>> My assumption is that the program should end the loop after the first
>> iteration again and it then should return True.
>
> No. If it did that, it wouldn't be a *loop* at all, would it? The whole
> reason loops (for and whil
On 15/12/13 16:54, Rafael Knuth wrote:
I stumbled upon this program here (Python 3.3.0) and I don't quite
understand how the for loop plays with the return True statement:
It doesn't.
Remember that indentation is all important in Python.
The return true statement is outside the loop so only
g
On 15/12/2013 16:54, Rafael Knuth wrote:
Hej,
I stumbled upon this program here (Python 3.3.0) and I don't quite
understand how the for loop plays with the return True statement:
def is_prime(number):
for element in range(2, number):
if number % element == 0:
return F
On Sun, Dec 15, 2013 at 05:54:10PM +0100, Rafael Knuth wrote:
> Hej,
>
> I stumbled upon this program here (Python 3.3.0) and I don't quite
> understand how the for loop plays with the return True statement:
>
> def is_prime(number):
> for element in range(2, number):
> if number % el
On Sun, Dec 15, 2013 at 11:54 AM, Rafael Knuth wrote:
> Hej,
>
> I stumbled upon this program here (Python 3.3.0) and I don't quite
> understand how the for loop plays with the return True statement:
>
> def is_prime(number):
> for element in range(2, number):
> if number % element ==
On Thu, 7 Oct 2010 12:41:07 pm Ryan Bridges wrote:
> Hi,
> I'm using Python 2.6.2 and I need to write a code so that when I use
> an input, it will say if the number is prime or not. How do I do
> this?
Using the keyboard is probably the best way, although at a pinch you
could copy and paste ind
On Thu, Oct 7, 2010 at 07:11, Ryan Bridges wrote:
> Hi,
> I'm using Python 2.6.2 and I need to write a code so that when I use an
> input, it will say if the number is prime or not. How do I do this?
>
>
Following links would be useful:
http://en.wikipedia.org/wiki/Prime_number
http://en.wikiped
On 03/28/2010 09:57 PM, yd wrote:
> It's not homework i just want to be able to convert my algorithm into
> good code, and the only way to do that is by actually writing it. I'm
> just writing it to learn how it's done.
In most cases, when:
1) the code is effective (i.e. it always gives correct an
>
>
> > What 'problem' are you trying to solve?
> In general, anytime you can use a premade solution, you are at an
> advantage,
> not cheating.
> That's one of the marks of a truly good programmer - being able to reuse as
> much code as possible.
> Unless it's a homework problem and he said "don't
On Sat, Mar 27, 2010 at 5:08 PM, yd wrote:
>
> Having a problem finding the first 1000 prime numbers, here is my code:-
>
> print(2)
> n =3
> counter =1
> while counter <=1000:
> for x in range(3,int((n**0.5)),2):
> if n%x != 0:
> print(n)
> n+=1
> counter+=1
> else:
>
On Sat, Mar 27, 2010 at 11:08 PM, yd wrote:
>
> Having a problem finding the first 1000 prime numbers, here is my code:-
>
> print(2)
> n =3
> counter =1
> while counter <=1000:
> for x in range(3,int((n**0.5)),2):
> if n%x != 0:
> print(n)
> n+=1
> counter+=1
> else:
>
>>> [x for x in range(3,int((n**0.5)),2)]
[]
your while loop is an infinite loop. Had you read the range documentations ?
>>> range(3,int((n**0.5)),2)
[]
>>> n**0.5
1.7320508075688772
>>> n = 3
>>> n ** 0.5
1.7320508075688772
>>> int ( n ** 0.5)
1
>>> range ( 3, 1, 2)
[]
On Sun, Mar 28, 2010 at
17 matches
Mail list logo
