How to catch StopIteration?

Chris cwitts at gmail.com
Tue Jun 17 08:49:08 CEST 2008


On Jun 17, 8:43 am, Chris <cwi... at gmail.com> wrote:
> On Jun 17, 5:50 am, ccy56... at gmail.com wrote:
>
>
>
> > I'm writing to see calcuration process.
> > And so, I can't catch StopIteration...
>
> > What is mistake?
>
> > def collatz(n):
> >   r=[]
> >   while n>1:
> >     r.append(n)
> >     n = 3*n+1 if n%2 else n/2
> >     yield r
>
> > for i, x in enumerate(collatz(13)):
> >   try:
> >     last = x[:i+1]
> >     print x[:i+1]
> >   except StopIteration:
> >     print last.appnd(1)
>
> > Output:
> > [13]
> > [13, 40]
> > [13, 40, 20]
> > [13, 40, 20, 10]
> > [13, 40, 20, 10, 5]
> > [13, 40, 20, 10, 5, 16]
> > [13, 40, 20, 10, 5, 16, 8]
> > [13, 40, 20, 10, 5, 16, 8, 4]
> > [13, 40, 20, 10, 5, 16, 8, 4, 2]
> > last.appnd(1) <= [13, 40, 20, 10, 5, 16, 8, 4, 2, 1]  # i want this
> > list
>
> def collatz(n):
>   r=[]
>   while n>1:
>     r.append(n)
>     n = 3*n+1 if n%2 else n/2
>     yield r
>
> i = 1
> while 1:
>     try:
>         last = x[:i]
>         print x[:i]
>         i += 1
>     except StopIteration
>         last.append(1)
>         break
>
> You will have to control the for loop yourself otherwise the
> StopIteration is handled for you.

forgot to put
x = collatz(13)
before the while loop starts and a x.next() inside the while, but
hopefully you get the point :p



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