Hello,
I have an array of events captured at different pixels at different times. Each row has x, y and time values.
Here a simple example (x, y, time) with a short list of events:
1, 3, 2
1, 3, 2.3
1, 3, 2
1, 3, 2.1
1, 3, 2.4
4, 5, 1
4, 5, 1.4
4, 5 , 1
4, 5, 1.2
3 , 4, 1
3, 4, 1.4
3, 4 ,2
As can be seen, some times are duplicates for the same x,y.
I would like to create a 3D matrix made of x,y slices arranged according to time.
Duplicate times for the same pixel should be summed.
I would appreciate any suggestion.

 Respuesta aceptada

Try this —
xyt = [1, 3, 2
1, 3, 2.3
1, 3, 2
1, 3, 2.1
1, 3, 2.4
4, 5, 1
4, 5, 1.4
4, 5 , 1
4, 5, 1.2
3 , 4, 1
3, 4, 1.4
3, 4 ,2];
[Uxytr,~,ix] = unique(xyt(:,3));
Out = accumarray(ix,(1:size(xyt,1))',[],@(x){[sum(xyt(x,[1 2]),1) unique(xyt(x,3))]})
Out = 7×1 cell array
{[ 11 14 1]} {[4 5 1.2000]} {[7 9 1.4000]} {[ 5 10 2]} {[1 3 2.1000]} {[1 3 2.3000]} {[1 3 2.4000]}
There are 7 unique times, and the (x,y) values for those times are summed, with the times themselves remaining unchanged.
.

4 comentarios

Ilan
Ilan el 16 de Mayo de 2022
I realize that my question was not accurate enough:
I would like to create a 3D matrix in which each pixel (x,y) will contain the number of time occurences (say f) for that pixel. For example, the outcome from the example above should be for pix 1,3 time=2:
(x, y, f, t) 1, 3, 2, 2
for pix 4,5, t=1: 4, 5, 2, 1
for pix 4,5, t=1.2: 4, 5, 1, 1.2
I hope it is clear now.
Thanks.
Star Strider
Star Strider el 16 de Mayo de 2022
Editada: Star Strider el 16 de Mayo de 2022
I hope it is clear now.
Not really. I have no idea what result you want.
I was going by: ‘Duplicate times for the same pixel should be summed.
My code does exactly that.
EDIT — (16 May 2022 at 21:36)
To collect them together in cells is straightforward —
xyt = [1, 3, 2
1, 3, 2.3
1, 3, 2
1, 3, 2.1
1, 3, 2.4
4, 5, 1
4, 5, 1.4
4, 5 , 1
4, 5, 1.2
3 , 4, 1
3, 4, 1.4
3, 4 ,2];
[Uxytr,~,ix] = unique(xyt(:,3));
Out = accumarray(ix,(1:size(xyt,1))',[],@(x){xyt(x,:)})
Out = 7×1 cell array
{3×3 double } {[4 5 1.2000]} {2×3 double } {3×3 double } {[1 3 2.1000]} {[1 3 2.3000]} {[1 3 2.4000]}
Out{1}
ans = 3×3
4 5 1 4 5 1 3 4 1
Out{3}
ans = 2×3
4.0000 5.0000 1.4000 3.0000 4.0000 1.4000
Out{4}
ans = 3×3
1 3 2 1 3 2 3 4 2
.
Ilan
Ilan el 17 de Mayo de 2022
Thanks, this is what I wanted.
Star Strider
Star Strider el 17 de Mayo de 2022
As always, my pleasure!

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