I have a M x N x O matrix and I would like to reduce it to a MxN matrix in MATLAB using a vector b of size M that contains the index of the element in the third dimension that is to be kept.
What it does then is build a 2d array with its entries selected from various pages of the original 3d array.
I have this loop but I am interested in a loopless solution.
for i = 1:M
for j = 1:N
tmp(i, j) = P(i, j, b(i));
end
end
The easiest way may just be to remove the j loop in your code:
for ii = 1:M
tmp(ii, :) = P(ii, :, b(ii));
end
But for the sake of comparison, here's a solution without a loop.
Given a 3d array P:
M = 7;
N = 5;
O = 6;
P = ones(M, N, O) .* permute(1:O, [3 1 2]);
(in this case I've used a 3d array where each element is equal to its O index)
and b, of size Mx1 with values from 1..O:
b = randi(O, M, 1)
you can construct the subscripts of all of the elements of P(:,:,1) and use b to select which plane to use:
[rr, cc] = ndgrid(1:M, 1:N);
inds = sub2ind(size(P), rr(:), cc(:), b(rr(:)));
tmp = reshape(P(inds), M, N)
For:
b.' = 5 4 1 5 3 1 3
we get:
tmp =
5 5 5 5 5
4 4 4 4 4
1 1 1 1 1
5 5 5 5 5
3 3 3 3 3
1 1 1 1 1
3 3 3 3 3
The elements of each row corresponds to the element in b as expected.
Related
I am trying to sort an array based on another array. I tried the sort method with index return, but it is somehow behaving strangely.
y = [1 2 3; 2 3 4]
x = [1 3 4; 2 2 3]
[yy, ii] = sort(y,'descend');
yy =
2 3 4
1 2 3
ii =
2 2 2
1 1 1
But my x(ii) is not the matrix sorted based on y.
x(ii) =
2 2 2
1 1 1
I am expecting the matrix to be:
x(ii) =
2 2 3
1 3 4
How can I sort the matrix x according to another matrix y?
ii are row subscripts but are being inputted by you as linear indices.
You need to convert them to relevant linear indices before proceeding i.e.
>> szx = size(x);
>> x(sub2ind(szx, ii, repmat(1:szx(2),szx(1),1)))
ans =
2 2 3
1 3 4
I have two vectors, R and C, which have the number of rows and columns, respectively, of submatrices that I need to assemble in a ones matrix I (40x20). There's 12 submatrices total.
R = [4 2 4 4 2 4];
C = [4 16 16 4];
Moreover, all the elements of each submatrix have its value stored in vector k:
k = [3 2 3 3 2 3 2 1 2 2 1 2 2 1 2 2 1 2 3 2 3 3 2 3 ]; % 24 elements
Thus for instance, submatrix M(1:4,1:4) has 4 rows, and 4 columns and value equal to k(1) = 1.
QUESTION: How can I assemble matrix M with all submatrices?
Any ideas?
Thanks!
EDIT:
The matrix M should look like this:
and the submatrices:
and the values of k:
Here is a vectorized solution:
R1 = repelem(1:numel(R), R);
C1 = repelem(1:numel(C), C);
[CC RR] = meshgrid(C1, R1);
idx = sub2ind([numel(R), numel(C)], RR, CC);
result = k(idx);
Instead you can use cell array, fill it with sub matrices and then convert the cell array to a matrix.
carr = cell(numel(R), numel(C));
k1 = reshape(k,numel(R),numel(C));
for ii = 1:numel(R)
for jj = 1:numel(C)
carr(ii,jj)=repmat(K1(ii,jj), R(ii), C(jj));
end
end
result = cell2mat(carr)
A simple example to illustrate all elements of a matrix multiplying each element of a vector to generate a 3D array.
M = reshape(1:12,4,3);
V = 1:2;
n = length(V);
A = nan([size(M),n]);
for ii = 1 : n
A(:,:,ii) = M * V(ii);
end
then
A(:,:,1) =
1 5 9
2 6 10
3 7 11
4 8 12
A(:,:,2) =
2 10 18
4 12 20
6 14 22
8 16 24
Or by repmat both M and V to the size of [4,3,2],
A = repmat(M,1,1,n) * reshape(V(ones(size(M)),:),[size(M),n])
It creates two 3D array by repmat besides the resulting 3d array A.
How to make it efficiently WITHOUT for loop and save the memory use?
According to the answer by #Lincoln,
A = bsxfun(#times, repmat(M,1,1,n), reshape(1:n, 1, 1, n));
repmat the vector V to 3d is not necessary.
Is it possible to create NO 3d array if the final result wanted is 2d, the sum of A along the 3rd dim? By for loop, the code would be
M = reshape(1:12,4,3);
V = 1:2;
n = length(V);
A = 0;
for ii = 1 : n
A = A + M * V(ii);
end
Try replacing the for.. loop with:
M_rep = repmat(M,1,1,n) %//repeat M in the 3rd dimension n times
v = reshape(1:n, 1, 1, n) %//create a vector [1 2 .. n] pointing in the third dimension
A = bsxfun(#times, M_rep, v) %//vector multiply them in the third dimension
In your above example, n=2.
EDIT (to your added question): To sum without allocating A:
B = sum(bsxfun(#times, M_rep, v),3);
I'm Writing a function called large_elements that takes input an array named X that is a matrix or a vector. The function identifies those elements of X that are greater than the sum of their two indexes.
For example, if the element X(2,3) is 6, then that element would be identified because 6 > (2 + 3). The output of the function gives the indexes(row and column sub) of such elements found in row-major order. It is a matrix with exactly two columns. The first column contains the row indexes, while the second column contains the corresponding column indexes.
Here is an example, the statement
indexes = large_elements([1 4; 5 2; 6 0])
should give the output like this:
[1 2; 2 1; 3 1]
If no such element exists,
the function returns an
empty array.
I have came up with the following code
function indexes = large_elements(A)
[r c] = size(A);
ind = 1;
for ii = 1:r
for jj = 1:c
if A(ii,jj) > ii + jj
indexes(ind,:) = [ii jj];
ind = ind + 1;
else
indexes = [];
end
end
end
end
But the results are not as expected. Any help would be appreciated.
One vectorised approch using bsxfun, find and ind2sub
A = randi(8,5); %// Your matrix
%// finding sum of the indexes for all elements
indSum = bsxfun(#plus, (1:size(A,1)).', 1:size(A,2));
%// generating a mask of which elements satisfies the given condition (i.e A > indSum)
%// Transposing the mask and finding corresponding indexes
[c,r] = find(bsxfun(#gt, A, indSum).') ;
%// getting the matrix by appending row subs and col subs
out = [r,c]
Results:
Input A:
>> A
A =
4 4 7 2 2
1 3 4 8 3
8 8 2 8 7
8 3 4 5 1
4 1 1 1 1
Output in row-major order:
out =
1 1
1 2
1 3
2 4
3 1
3 2
3 4
4 1
Note: Getting subs in row-major order is tricky here
Also here is your correct loopy approach
[r, c] = size(A);
ind = 0;
indexes = [];
for ii = 1:r
for jj = 1:c
if A(ii,jj) > ii + jj
ind = ind + 1;
indexes(ind,:) = [ii jj];
end
end
end
That is because whenever you encounter an element which is smaller than the sum of its indices you are reinitializing the array to null. So the output is coming out to be null. You should not initialize it to null on the else condition.
Hi I want to reshape a matrix but the reshape command doesn't order the elements the way I want it.
I have matrix with elements:
A B
C D
E F
G H
I K
L M
and want to reshape it to:
A B E F I K
C D G H L M
So I know how many rows I want to have (in this case 2) and all "groups" of 2 rows should get appended horizontally. Can this be done without a for loop?
You can do it with two reshape and one permute. Let n denote the number of rows per group:
y = reshape(permute(reshape(x.',size(x,2),n,[]),[2 1 3]),n,[]);
Example with 3 columns, n=2:
>> x = [1 2 3; 4 5 6; 7 8 9; 10 11 12]
x =
1 2 3
4 5 6
7 8 9
10 11 12
>> y = reshape(permute(reshape(x.',size(x,2),n,[]),[2 1 3]),n,[])
y =
1 2 3 7 8 9
4 5 6 10 11 12
Cell array approach -
mat1 = rand(6,2) %// Input matrix
nrows = 3; %// Number of rows in the output
[m,n] = size(mat1);
%// Create a cell array each cell of which is a (nrows x n) block from the input
cell_array1 = mat2cell(mat1,nrows.*ones(1,m/nrows),n);
%// Horizontally concatenate the double arrays obtained from each cell
out = horzcat(cell_array1{:})
Output on code run -
mat1 =
0.5133 0.2916
0.6188 0.6829
0.5651 0.2413
0.2083 0.7860
0.8576 0.3032
0.1489 0.4494
out =
0.5133 0.2916 0.5651 0.2413 0.8576 0.3032
0.6188 0.6829 0.2083 0.7860 0.1489 0.4494