Further to Bill’s point Nollaig, your approach using rank fails for a rank 0
item …
$ ,:`]@.(<:@$^:2 fee) fee=:'ab'
1 2
$ ,:`]@.(<:@$^:2 fee) fee=:,'a'
1 1
$ ,:`]@.(<:@$^:2 fee) fee=:'a'
For this solution maybe better to just use $ rather than ,: as shown (easier to
follow and efficient):
$(_2{. 1 1,$ fee)$fee=:'ab'
1 2
so …
mat=:((_2{. 1 1,$) $ ]) NB: Note in the left tine of the fork (_2{. 1
1,$) the argument is supplied by definition of the fork
$mat 'ab'
1 2
$mat ,'a'
1 1
$mat 'a'
1 1
Rob
> On 3 Oct 2015, at 11:28 am, bill lam <[email protected]> wrote:
>
> $$ looks like a scalar but is a rank 1 array. use #$ instead.
> ,:`]@.(<:@#@$) fee=:'ab'
>
> the verb
> ,:`]@.(<:@$^:2 fee) is reduced to ] and then run as ] fee
> On Oct 3, 2015 8:27 AM, "Nollaig MacKenzie" <[email protected]>
> wrote:
>
>>
>>
>> I have a variable, fee, which might have shape
>> N,2 or just 2. I want a function that converts
>> fee when it has shape 2 to shape 1 2.
>>
>> It seems straightforward: if the rank is 2,
>> make no change, if 1, apply ,:
>>
>> So something like ,:`]@.(<:@$^:2) should
>> do the trick. Indeed, this works:
>>
>> $ ,:`]@.(<:@$^:2 fee) fee=:'ab'
>> 1 3
>>
>> But when I try to define something to do
>> what I want, I get lost.
>>
>> ,:`]@.(<:@$^:2) fee=:'ab'
>> |rank error
>> | ,:`]@.(<:@$^:2)fee=:'ab'
>> ,:`]@.(<:@$^:2 ]) fee=:'ab'
>> |domain error
>> | ,:`]@.(<:@$^:2]) fee=:'ab'
>> ,:`]@.((<:@$^:2)@: ]) fee=:'ab'
>> |rank error
>> | ,:`]@.((<:@$^:2)@:]) fee=:'ab'
>>
>>
>>
>> (I'll spare you more)
>>
>> I must be missing something obvious.
>>
>> --
>> Nollaig MacKenzie
>> http://www.yorku.ca/nollaig
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>>
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