You need to splat (...) A: hcat(A...), in Gunnar's email, vs. hcat(A), in
your email. This will supply each element of A as a separate argument to
hcat().
Much like the prefix dot for elementwise computation, it's hard to see in a
proportional typeface.
On Sunday, August 24, 2014 10:36:29 AM UTC-5, Bradley Setzler wrote:
>
> Hi Gunnar,
>
> Correct me if I'm wrong, but hcat does not work for Any types. I'm looking
> for a version of hcat that works with Any.
>
> We have A:
>
> julia> A
> 3-element Array{Any,1}:
> [1,2]
> [3,4]
> [5,6]
>
> We want B:
>
> julia> B
> 2x3 Array{Int64,2}:
> 1 3 5
> 2 4 6
>
> So we try hcat(A) but it fails to return B:
>
> julia> hcat(A)
> 3x1 Array{Any,2}:
> [1,2]
> [3,4]
> [5,6]
>
>
> Thanks,
> Bradley
>
>
>
> On Sunday, August 24, 2014 4:01:25 AM UTC-5, Gunnar Farnebäck wrote:
>>
>> For maximum efficiency Diego's loop solution is likely best but for
>> succinctness, try hcat(A...).
>>
>> Den lördagen den 23:e augusti 2014 kl. 04:05:19 UTC+2 skrev Bradley
>> Setzler:
>>>
>>> Good evening,
>>>
>>> I often have Any types filled with vectors (especially as the return of
>>> a pmap), and need them as a matrix or DataFrame. For example, suppose I
>>> have,
>>>
>>> julia> A
>>> 3-element Array{Any,1}:
>>> [1,2]
>>> [3,4]
>>> [5,6]
>>>
>>> But I want,
>>>
>>> julia> B
>>> 2x3 Array{Int64,2}:
>>> 1 3 5
>>> 2 4 6
>>>
>>> I came up with the following, which successfully constructs B from A,
>>> but it's a bit inefficient/messy:
>>>
>>> B = A[1]
>>> for i=2:length(A)
>>> B = hcat(B,A[i])
>>> end
>>>
>>> Is there a better way to do this? Something like,
>>> julia> B = hcatForAny(A)
>>>
>>> Thanks,
>>> Bradley
>>>
>>>
