You can tell ones to construct a array on Ints by passing it a type
parameter:
ones(Int, 1, 3)
With a length of three it doesn't make much difference but with a bigger
array you save a lot of time avoiding the conversion.
On Tuesday, October 14, 2014 3:14:38 AM UTC+3, Miguel Bazdresch wrote:
>
> 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] <javascript:>>
> 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
>>
>
>
