How can I transform this matlab code into Julia

General Usage
, ,
1 
>> val = zeros(3, 3);
>> row = zeros(3, 3);
>> A = [1 2 3; 6 5 1; 1 7 4];

[val(1, :), row(1, :)] = max(A)

val =

     6     7     4
     0     0     0
     0     0     0


row =

     2     3     3
     0     0     0
     0     0     0

How can I get the max value and row number and save them.

1 Like
2 
julia> vals, idx = findmax(A, dims=1)
([6 7 4], CartesianIndex{2}[CartesianIndex(2, 1) CartesianIndex(3, 2) CartesianIndex(3, 3)])

julia> vals
1×3 Matrix{Int64}:
 6  7  4

julia> idx
1×3 Matrix{CartesianIndex{2}}:
 CartesianIndex(2, 1)  CartesianIndex(3, 2)  CartesianIndex(3, 3)

If you need the row only and not the full index:

row = first.(Tuple.(idx))
1×3 Matrix{Int64}:
 2  3  3
4 Likes
3

yes, I only need the row only. Thanks, it solves my problem.

1 Like
4

Dear @skleinbo ,

how can I transform this matlab code into Julia code.

Here is the MATLAB code.

>> xx = zeros(2, 3)

xx =

     0     0     0
     0     0     0

>> yy = zeros(2, 3)

yy =

     0     0     0
     0     0     0

>> location = [1 2 3; 2 3 4]

location =

     1     2     3
     2     3     4

>> [xx(1,:), yy(1,:)] = ind2sub([2,2], location(1,:))

xx =

     1     2     1
     0     0     0


yy =

     1     1     2
     0     0     0

I want to transform this matlab code into Julia. and I try to rewrite the code,but get an error.

julia> xx = zeros(2, 3)
2×3 Matrix{Float64}:
 0.0  0.0  0.0
 0.0  0.0  0.0

julia> yy = zeros(2, 3)
2×3 Matrix{Float64}:
 0.0  0.0  0.0
 0.0  0.0  0.0


julia> location = [1 2 3; 2 3 4]
2×3 Matrix{Int64}:
 1  2  3
 2  3  4

julia> xx[1,:] .= CartesianIndices((2, 2))[location[1,:]][1]
ERROR: iteration is deliberately unsupported for CartesianIndex. Use `I` rather than `I...`, or use `Tuple(I)...`
Stacktrace:

Can you help me to correct my code to get the row and column index by the first row of location?

5

You can use LinearIndices and CartesianIndices to convert between linear and cartesian indices.

Edit: Ah, sorry, didn’t see you had discovered that already. You need to convert the cartesian indices to a tuple before indexing first, like the error suggests. Maybe

xx[1,:] .= first.(Tuple.(CartesianIndices((2, 2))[location[1,:]]))

As you can see, this is a bit unwieldy and inelegant and should prompt you to consider a less MATLABy way.

6

I try to use loop .

julia> for i in 1:2, j in 1:3
          xx[i, j] = CartesianIndices((2, 2))[location[i, j]][1]
       end

julia> for i in 1:2, j in 1:3
          yy[i, j] = CartesianIndices((2, 2))[location[i, j]][2]
       end

julia> xx
2×3 Matrix{Float64}:
 1.0  2.0  1.0
 2.0  1.0  2.0

julia> yy
2×3 Matrix{Float64}:
 1.0  1.0  2.0
 1.0  2.0  2.0

This is the result what I need. But I don’t think loop is the best way.

7

Thanks so much.This solves my problem.

8

Does Julia have the same function ind2sub or sub2ind? I think CartesianIndices or LinearIndices is worse than matlab function ind2sub.

9

Not for a long time PSA: replacement of ind2sub/sub2ind in Julia 0.7+

10

@skleinbo I have another problem. I have convert the results to Int, but why the type is still Float64?

 pol_ind_kp[iak_state, :] = Int.(first.(Tuple.(CartesianIndices((nk, na))[Int.(location[iak_state, :])])))
 
typeof(pol_ind_kp)
Matrix{Float64} (alias for Array{Float64, 2})
11

I have solved the problem by myself. I forget to set the zeros matrix Int when I define the pol_ind_kp.

12

Loops aren’t bad. They’re often the best way to solve a problem.

1 Like
13

Thanks for your reply. But I think loops may take a very long time.

14

They do in Matlab, but loops in Julia are fast.

2 Likes
15

Oh, It is really a good news to me. I use matlab and Stata very often. And loops always take a very long time. Thank you very much.

16

Matlab to Julia requires significant unlearning of bad habits. Another one for free: whenever you type rand(1) or rand(n,1), you almost certainly want rand() or rand(n).

17

Read carefully the Performance Tips · The Julia Language

Particularly the first one. There is where most commonly new Julia users fail to get a good performance.

18

Dear @skleinbo ,

how can I use LinearIndices to get the index. In matlab, we can use sub2ind.

>> A = [1 2 1; 2 2 1]

A =

     1     2     1
     2     2     1

>> inds = sub2ind([2,2], A(1,:), A(2,:))

inds =

     3     4     1
19

@ murphyk I try to use function sub2indv which @ murphyk wrote.

function subv2ind(shape, cindices)
    """Return linear indices given vector of cartesian indices.
    shape: d-tuple of dimensions.
    cindices: n-iterable of d-tuples, each containing cartesian indices.
    Returns: n-vector of linear indices.

    Based on:
    https://discourse.julialang.org/t/psa-replacement-of-ind2sub-sub2ind-in-julia-0-7/14666/8
    Similar to this matlab function:
    https://github.com/tminka/lightspeed/blob/master/subv2ind.m
    """
    lndx = LinearIndices(Dims(shape))
    n = length(cindices)
    out = Array{Int}(undef, n)
    for i = 1:n
        out[i] = lndx[cindices[i]...]
    end
    return out
end


julia> A = [1 2 1; 2 2 1]
2×3 Matrix{Int64}:
 1  2  1
 2  2  1

julia> subv2ind((2, 2), Tuple.(zip(A[1, :], A[2, :])))
3-element Vector{Int64}:
 3
 4
 1

Is there another simpler way to do this?

20 
julia> subv2ind((2, 2), [(i, j) for (i, j) in zip(A[1, :], A[2, :])])
3-element Vector{Int64}:
 3
 4
 1