Just a word about efficiency: if you think to the tuple as an array, you
have to shift N elements one position left. With an array this requires
O(N) time, which is the complexity of my proposed solution.
Cheers,
raph
On Sat, Oct 4, 2008 at 11:26 PM, Chris Rathman <[EMAIL PROTECTED]>wrote:
> Thanks. That did the trick. Not sure it is an efficient solution, but if
> I was worried about efficiency, then I should be using lists anyway.
>
> Thanks,
> Chris
>
> Raphael Collet wrote:
>
>> You have to rebuild a brand new tuple, since the input field number 2 is
>> the output field number 1, and so on. The simplest, in my opinion, is to
>> get the fields in a list, and rebuild a tuple from the tail of that list.
>>
>> %% return tuple T without its first field, and its remaining fields
>> %% moved one position "to the left"
>> fun {Foo T}
>> {List.toTuple {Label T} {Record.toList T}.2}
>> end
>>
>> Cheers,
>> raph
>>
>> On Fri, Oct 3, 2008 at 6:44 AM, Chris Rathman <[EMAIL PROTECTED]<mailto:
>> [EMAIL PROTECTED]>> wrote:
>>
>> What I want is take a record that has implicit indexes such as:
>>
>> a(10 20 30 40)
>>
>> and return a record with one less arity, stripping off the first
>> value:
>>
>> a(20 30 40)
>>
>> I can get close with:
>>
>> {Record.filterInd a(10 20 30 40) fun {$ I X} I > 1 end}
>>
>> But that returns the value of
>>
>> a(2:20 3:30 4:40)
>>
>> which is not quite the same thing.
>>
>> Is there a way to do what I am trying to do?
>>
>> Thanks,
>> Chris
>>
>>
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