How to index one array by assigned vector in certain dimension
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Warren Chuo
am 14 Jun. 2021
Kommentiert: Warren Chuo
am 21 Jun. 2021
Hi,
I'd like to form a new 2D array by indexing another 3D array, where the indexing is via one assigned vector.
Details like below:
Given arr_i,vec_i, want to find func_index as
arr_o = func_index(arr_i,vec_i);
%%%
arr_i1 = [111,112,113,114; ...
121,122,123,124];
arr_i2 = [211,212,213,214; ...
221,222,223,224];
arr_i3 = [311,312,313,314; ...
321,322,323,324];
arr_i = cat(3,arr_i1,arr_i2,arr_i3);
vec_i = [2,3,1,2];
arr_o = func_index(arr_i,vec_i);
%%%
With above operation, I expect arr_o will be
[211,312,113,214; ...
221,322,123,224];
The in/out relationship of func_index is:
arr_o(1:2,1) = arr_i(1:2,1,2), since vec_i(1) == 2;
arr_o(1:2,2) = arr_i(1:2,2,3), since vec_i(2) == 3;
arr_o(1:2,3) = arr_i(1:2,3,1), since vec_i(3) == 1;
arr_o(1:2,4) = arr_i(1:2,4,2), since vec_i(4) == 2;
Keypoint here is assigned vector actually index 3D array in certain one dimension (in this case is dim_3), and result in new 2D array.
Please propose matrix operation (avoiding loop, since the real array is quite big).
Thanks!!
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Akzeptierte Antwort
DGM
am 15 Jun. 2021
There are probably multiple ways of doing this. It could probably be done with a bunch of reshaping and such, but I did it this way. I'm not sure which would be faster or less confusing.
arr_i1 = [111,112,113,114; ...
121,122,123,124];
arr_i2 = [211,212,213,214; ...
221,222,223,224];
arr_i3 = [311,312,313,314; ...
321,322,323,324];
arr_i = cat(3,arr_i1,arr_i2,arr_i3);
% [211,312,113,214;
% 221,322,123,224];
% calculate all 8 subscripts for each axis
xvec = repelem(1:4,2);
yvec = repmat(1:2,[1 4]);
zvec = repelem([2 3 1 2],2);
% convert to linear indices
idx = sub2ind(size(arr_i),yvec,xvec,zvec);
reshape(arr_i(idx),2,[])
Otherwise, trying to do something directly like
arr_i(:,1:4,[2 3 1 2])
Would give the intersection of the subscript ranges. Such a result would be a much larger set than intended, and the desired result would be its block diagonal.
3 Kommentare
DGM
am 15 Jun. 2021
If you don't have repelem, you can probably just use kron()
x = 1:4
repelem(x,2)
kron(x,[1 1])
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