[Tutor] Handling Generator exceptions in Python 2.5

[Joe Python]

Thanks everyone for the responses.

- Joe

On Fri, Jun 19, 2009 at 11:09 AM, vince spicer <vinces1979 at gmail.com> wrote:

> Well*
>
> *result = [(ListA[i] - ListB[i-1])/ListA[i] for i in range(len(ListA))*if
> not ListA[i] == 0*]
>
> will exclude any results where listA[i] is 0, if you still want these in
> the result
> you may want to use good'ol for statement instead of list comprehension
>
>
> results = []
> for x in range(len(listA)):
>     y = ListA[i] - ListB[i-1]
>     if not ListA[i] == 0:
>         y = y / ListA[i]
>     results.append(y)
>
> print results
>
> Vince
>
>
> On Fri, Jun 19, 2009 at 8:55 AM, Joe Python <jopython at gmail.com> wrote:
>
>> I have a generator as follows to do list calculations.
>>
>> *result = [(ListA[i] - ListB[i-1])/ListA[i] for i in range(len(ListA))]*
>>
>> The above generator, throws  '*ZeroDivisionError*' exception if ListA[i]
>> = 0.
>> Is there a way to say 'Don't divide by ListA[i] if its equal to 0 (within
>> that statement)'.
>>
>> Sorry if this question sounds too stupid.
>>
>> TIA
>> Joe
>>
>> _______________________________________________
>> Tutor maillist  -  Tutor at python.org
>> http://mail.python.org/mailman/listinfo/tutor
>>
>>
>
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