How to "filter" an N-D array with matrix indexing?

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Antoine Purier
Antoine Purier el 15 de Nov. de 2017
Editada: Antoine Purier el 15 de Nov. de 2017
Hi Matlab users,
First post on Mathworks!
I am creating a generic function to calculate percentiles and confidence intervals for N-D input arrays. And that is where the problem is, because I want it to be able to deal with any array dimensions.
To explain the problem, I will take a simple 3D case. I want to avoid loops for this type of operation.
Basically I have an input 3D array like the following: a = randn(10,3,3)
a(:,:,1) =
1.3514 -0.8655 0.1837
-0.2248 -0.1765 -0.4762
-0.5890 0.7914 0.8620
-0.2938 -1.3320 -1.3617
-0.8479 -2.3299 0.4550
-1.1201 -1.4491 -0.8487
2.5260 0.3335 -0.3349
1.6555 0.3914 0.5528
0.3075 0.4517 1.0391
-1.2571 -0.1303 -1.1176
a(:,:,2) =
1.2607 0.4669 -0.9415
0.6601 -0.2097 -0.1623
-0.0679 0.6252 -0.1461
-0.1952 0.1832 -0.5320
-0.2176 -1.0298 1.6821
-0.3031 0.9492 -0.8757
0.0230 0.3071 -0.4838
0.0513 0.1352 -0.7120
0.8261 0.5152 -1.1742
1.5270 0.2614 -0.1922
a(:,:,3) =
-0.2741 0.2761 -3.0292
1.5301 -0.2612 -0.4570
-0.2490 0.4434 1.2424
-1.0642 0.3919 -1.0667
1.6035 -1.2507 0.9337
1.2347 -0.9480 0.3503
-0.2296 -0.7411 -0.0290
-1.5062 -0.5078 0.1825
-0.4446 -0.3206 -1.5651
-0.1559 0.0125 -0.0845
Everything is always calculated along the x dimensions (permutation step realized beforehand). I calculate previously in my script the index of the values I want to keep, my confidence intervals lower boundaries. ciLowerInd
ciLowerInd =
1
3
5
I want to keep the 1st, 3rd and 5th values of each columns in all the other dimensions than the first one, to end up with an output array of dimensions 3x3x3.
So the output would be :
b(:,:,1)
1.3514 -0.8655 0.1837
-0.5890 0.7914 0.8620
-0.8479 -2.3299 0.4550
b(:,:,2)
1.2607 0.4669 -0.9415
-0.0679 0.6252 -0.1461
-0.2176 -1.0298 1.6821
b(:,:,3)
-0.2741 0.2761 -3.0292
-0.2490 0.4434 1.2424
1.6035 -1.2507 0.9337
So I repmat these 3 index values to get a 3x3x3 dimensions array :
ciLowerInd(:,:,1) =
1 1 1
3 3 3
5 5 5
ciLowerInd(:,:,2) =
1 1 1
3 3 3
5 5 5
ciLowerInd(:,:,3) =
1 1 1
3 3 3
5 5 5
So my question is : Is there a way to kind of filter the input array ("a" in this case") to get an output array of same size as indexing matrix ciLowerInd and keeping only the 1st, 3rd and 5th values of each columns of a, to end up with something like array "b"? All this without looping to be able to deal with N dimensions.
I hope it was clear enough.
Thanks for your help
Antoine

Respuesta aceptada

Stephen23
Stephen23 el 15 de Nov. de 2017
Editada: Stephen23 el 15 de Nov. de 2017
One efficient way to achieve this for any ND array is to call subsref directly:
>> C = repmat({':'},1,ndims(a)); % colons for all trailing dimensions
>> C{1} = [1,3,5]; % indices for first dimension
>> b = subsref(a,substruct('()',C))
b(:,:,1) =
1.35140 -0.86550 0.18370
-0.58900 0.79140 0.86200
-0.84790 -2.32990 0.45500
b(:,:,2) =
1.260700 0.466900 -0.941500
-0.067900 0.625200 -0.146100
-0.217600 -1.029800 1.682100
b(:,:,3) =
-0.27410 0.27610 -3.02920
-0.24900 0.44340 1.24240
1.60350 -1.25070 0.93370
Or you can use C directly into a (thank you Andrei Bobrov):
>> b = a(C{:})
b(:,:,1) =
1.35140 -0.86550 0.18370
-0.58900 0.79140 0.86200
-0.84790 -2.32990 0.45500
b(:,:,2) =
1.260700 0.466900 -0.941500
-0.067900 0.625200 -0.146100
-0.217600 -1.029800 1.682100
b(:,:,3) =
-0.27410 0.27610 -3.02920
-0.24900 0.44340 1.24240
1.60350 -1.25070 0.93370
  4 comentarios
Stephen23
Stephen23 el 15 de Nov. de 2017
Editada: Stephen23 el 15 de Nov. de 2017
@Andrei Bobrov: true, I forgot that one can pass the char ':' as an index. Very neat.
Antoine Purier
Antoine Purier el 15 de Nov. de 2017
Editada: Antoine Purier el 15 de Nov. de 2017
Thanks Stephen and Andrei!
Now if one of my indices is below one (less than the first rank) or above the x-dimension size (11th rank on only 10 data), I would like to set the value of "b" to either -Inf or +Inf. And still keep the same output array size.
If I put a Nan or Inf in the C cell array it is not working obviously, so do I need to refilter after or is there a simpler way?

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Más respuestas (1)

José-Luis
José-Luis el 15 de Nov. de 2017
data = rand(10,3,3);
idx = (1:2:5)';
dims = size(data);
to_add = cumprod(dims);
for ii = 2:numel(dims)
shape = ones(1,ii);
shape(end) = dims(ii);
idx = bsxfun(@plus,idx,reshape((0:dims(ii)-1).*to_add(ii-1),shape));
end
result = data(idx);
  1 comentario
Antoine Purier
Antoine Purier el 15 de Nov. de 2017
Thanks for your post Jose-Luis! It seems that Stephens answer is the one that suits the best for what I want to achieve, in only 3 lines of code.

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