Positions in an array

General Usage
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1

Hello!

I have a 15360x3 array and the last column may only assume three different values: -1, 1, or 2. I’d like to obtain arrays with the positions in this array where the third column assumes each of these possible values (for example, what are the lines in the array where the last column has the value of -1?).

I tried to use the indexin function but didn’t succeed.

2

Hi Mariana,

have you looked at findall?
Maybe this example helps (EDIT: directly use the array shape you asked for):

julia> a = rand(-1:2, (15360, 3));

julia> idx = findall(==(-1), a[:, 3])
3817-element Vector{Int64}:
     8
    12
    19
    20
    21
    25
    30
    37
    38
    40
     ⋮
 15321
 15328
 15339
 15340
 15341
 15345
 15348
 15349
 15350

julia> all(a[idx, 3] .== -1)
true
1 Like
3

I’m not sure I understand what you need, but this might help:

julia> A = hcat(rand(4,2),[1, -1, 2, 2])
4×3 Matrix{Float64}:
 0.118909  0.147115   1.0
 0.33352   0.229109  -1.0
 0.152295  0.923474   2.0
 0.111448  0.916297   2.0

julia> [A[row,3]==1 for row=axes(A,1), col=axes(A,2)]
4×3 Matrix{Bool}:
 1  1  1
 0  0  0
 0  0  0
 0  0  0

julia> [A[row,3]==2 for row=axes(A,1), col=axes(A,2)]
4×3 Matrix{Bool}:
 0  0  0
 0  0  0
 1  1  1
 1  1  1
1 Like
4

Thank you very much for the answers, they solved my problem!

1 Like
5

You could also consider spitting out all desired indices from matrix m in one go:

[v .== m[:,3] for v in [-1,1,2]]

or better:

collect(eachcol(m[:,3] .== [-1 1 2]))
6

Another simple way to find indices for each value in one go, without manually listing possible values:

using SplitApplyCombine
idx = groupfind(a[:, 3])

# indices of -1: idx[-1]
# indices of 1: idx[1]
# ...

looks like every other data-related question here can be solved by a SplitApplyCombine function

2 Likes
7

@aplavin, that is an extremelly nice solution!