Just wanted to point out a little-known way to achieve the same with the
`kron` (Kronecker product) command:
julia> x=[1 2 3 4]
1x4 Array{Int64,2}:
1 2 3 4
julia> kron(x,int(ones(1,3)))
1x12 Array{Int64,2}:
1 1 1 2 2 2 3 3 3 4 4 4
-- mb
On Mon, Oct 13, 2014 at 6:05 PM, Rajn <[email protected]> wrote:
>
>
> Hi,
> I have two questions.
>
> First question:
>
> I have a fairly simple problem which can be solved in many ways.
>
> To begin I have a vector
> z1=Array(Float64,1,0)
>
> I have
> Z=1:0.01:100
>
> I would like to generate a vector in which each element is repeated
> length(Z) times i.e., [1 1 1 1....length(Z) times 1.01 1.01
> 1.01....length(Z) times and so on
>
> When I do
> for j=1:length(Z)
> z1=[z1 repmat([Z[j]],1,length(Z)]
> end
>
> and compare the time for this exact code with Matlab's/Octave, Julia takes
> 7 seconds in comparison to 1.5 seconds for Matlab/Octave.
>
> Of course I did not see any improvement using
> z1=[z1 ones(1,length(Z))*Z[j]]
>
> It took practically the same time.
>
> However, this reduced to fraction of a second...i.e., after I generated a
> vector of zeros and just replaced the zeros appropriately.
> z1=zeros(1,length(Z)^2)
> length(Z)=dd
> for j=1:dd
> v=1:dd:(dd-1)*dd
> z1[1,v[j]:(v[j+dd-1])]=Z[j];
> end
>
> Of course even Octave/Matlab became equally fast with this code.
>
> So what gives in repmat for Julia? Why is it slow? Have I gone wrong with
> my first code?
>
>
>
>
> Question 2:
> If I start with A=[]
> How do I vcat or hcat a row or a column vector to A in a for loop form a
> MATRIX.
> say my vector is A(10x1) and I want to make B=[A[1] A[2] A[3] A[4]] which
> is a 10x4 matrix.
> In Octave I would do
> B=[]
> B=[B A[j]] in a for loop.
>
> Thanks
>
