Combine matrixes with like same values

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Conner Carriere
Conner Carriere am 11 Nov. 2022
Kommentiert: Conner Carriere am 21 Nov. 2022
I know this has been asked before, but im just not sure how to approach it.
I have a cell that looks
bbox
[0,241,637,168]
[204,181,382,286]
[56,314,185,243]
[0,59,574,506]
[8,58,230,546]
and another "key" cell that looks like this
1
1
2
2
3
I also have a cell that called "class"
5
11
12
9
10
I need an output like this
row bbox class
1 [0,241,637,168; 204,181,382,286] [5;11]
2 [56,314,185,243; 0,59,574,506] [12;9]
3 [8,58,230,546] [10]
Any info would help, I am sure that it uses either the unique() or accumarray() function

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Jan
Jan am 11 Nov. 2022
Bearbeitet: Jan am 11 Nov. 2022
bbox = {[0,241,637,168]; ...
[204,181,382,286]; ...
[56,314,185,243]; ...
[0,59,574,506]; ...
[8,58,230,546]};
key = [1, 1, 2, 2, 3];
result = splitapply(@(c) {cat(1, c{:})}, bbox, key(:))
result = 3×1 cell array
{2×4 double } {2×4 double } {[8 58 230 546]}
Another option is a simple loop:
ukey = unique(key);
result = cell(numel(ukey), 1);
for k = 1:numel(ukey)
result{k} = cat(1, bbox{key == ukey(k)});
end
result
result = 1×3 cell array
{2×4 double} {2×4 double} {[8 58 230 546]}
  5 Kommentare
Conner Carriere
Conner Carriere am 21 Nov. 2022
What would with that simple loop if I wanted to add another set of numbers to it. This one would be tied to the bounding box. Wherever the bounding box moves, the value would be "tied" to it and relocate to where it does. I changed the question to match
Conner Carriere
Conner Carriere am 21 Nov. 2022
realized I can run the function twice and combine them... Its late

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