I give a small example: matrix A[3*3*2], only 9 elements of matrix A are '1',other elements of A are '0'. (the value of '1' and '0' do not matter, just indicate the number of voxels, and following 5 values belong to the 9 elements of matrix A.
matrix B = [ 12
34
61
81
42 ];
matrix C = [ 1 2 1
1 1 1
2 3 1
2 2 1
3 1 1];
The 5 elements belong to the 9 elements of matrix A.
How to assign the values of matrix B to matrix A based on the matrix C?
Here's how to do that using linear indexing via sub2ind:
A(sub2ind(size(A),C(:,1),C(:,2),C(:,3)))=B;
Related
I have some data in 10 matrices. Each matrix has a different number of rows, but the same number of columns.
I want to combine all 10 matrices to one matrix row-wise, interleaved, meaning the rows in that matrix will look like:
row 1 from matrix 0
...
row 1 from matrix 9
row 2 from matrix 0
...
row 2 from matrix 9
...
Example (with 3 matrices):
Matrix 1: [1 2 3 ; 4 5 6; 7 8 9]
Matrix 2: [3 2 1 ; 6 5 4]
Matrix 3: [1 1 1 ; 2 2 2 ; 3 3 3]
Combined matrix will be: [1 2 3 ; 3 2 1 ; 1 1 1 ; 4 5 6 ; 6 5 4 ; 2 2 2 ; 7 8 9 ; 3 3 3]
You can download the function interleave2 here https://au.mathworks.com/matlabcentral/fileexchange/45757-interleave-vectors-or-matrices
z = interleave2(a,b,c,'row')
you can see the way the function works in the source code of course
Here's a general solution that allows you to place however many matrices you want (with matching number of columns) into the starting cell array Result:
Result = {Matrix1, Matrix2, Matrix3};
index = cellfun(#(m) {1:size(m, 1)}, Result);
[~, index] = sort([index{:}]);
Result = vertcat(Result{:});
Result = Result(index, :);
This will generate an index vector 1:m for each matrix, where m is its number of rows. By concatenating these indices and sorting them, we can get a new index that can be used to sort the rows of the vertically-concatenated set of matrices so that they are interleaved.
I have been having problem with identifying two maximum values' position in 3D matrix (MATLAB). Say I have matrix A output as follows:
A(:,:,1) =
5 3 5
0 1 0
A(:,:,2) =
0 2 0
8 0 8
A(:,:,3) =
3 0 0
0 7 7
A(:,:,4) =
6 6 0
4 0 0
For the first A(:,:,1), I want to identify that the first row have the highest value (A=5). But I need the two index position, which in this case, 1 and 3. And this is the same as the other A(:,:,:).
I have searched through SO but since I am bad in MATLAB, I couldn't find way to work this through.
Please do help me on this. It would be better if I don't need to use for loop to get the desired output.
Shot #1 Finding the indices for maximum values across each 3D slice -
%// Reshape A into a 2D matrix
A_2d = reshape(A,[],size(A,3))
%// Find linear indices of maximum numbers for each 3D slice
idx = find(reshape(bsxfun(#eq,A_2d,max(A_2d,[],1)),size(A)))
%// Convert those linear indices to dim1, dim2,dim3 indices and
%// present the final output as a Nx3 array
[dim1_idx,dim2_idx,dim3_idx] = ind2sub(size(A),idx)
out_idx_triplet = [dim1_idx dim2_idx dim3_idx]
Sample run -
>> A
A(:,:,1) =
5 3 5
0 1 0
A(:,:,2) =
0 2 0
8 0 8
A(:,:,3) =
3 0 0
0 7 7
A(:,:,4) =
6 6 0
4 0 0
out_idx_triplet =
1 1 1
1 3 1
2 1 2
2 3 2
2 2 3
2 3 3
1 1 4
1 2 4
out_idx_triplet(:,2) is what you are looking for!
Shot #2 Finding the indices for highest two numbers across each 3D slice -
%// Get size of A
[m,n,r] = size(A)
%// Reshape A into a 2D matrix
A_2d = reshape(A,[],r)
%// Find linear indices of highest two numbers for each 3D slice
[~,sorted_idx] = sort(A_2d,1,'descend')
idx = bsxfun(#plus,sorted_idx(1:2,:),[0:r-1]*m*n)
%// Convert those linear indices to dim1, dim2,dim3 indices
[dim1_idx,dim2_idx,dim3_idx] = ind2sub(size(A),idx(:))
%// Present the final output as a Nx3 array
out_idx_triplet = [dim1_idx dim2_idx dim3_idx]
out_idx_triplet(:,2) is what you are looking for!
The following code gives you the column and row of the respective maximum.
The first step will obtain the maximum of each sub-matrix containing the first and second dimension. Since max works per default with the first dimension, the matrix is reshaped to combine the original first and second dimension.
max_vals = max(reshape(A,size(A,1)*size(A,2),size(A,3)));
max_vals =
5 8 7 6
In the second step, the index of elements equal to the respective max_vals of each sub-matrix is obtained using arrayfun over the third dimension. Since the output of arrayfun are cells, cell2mat is used to transform the output into a matrix. As a last step, the linear index from find is transformed into sub-indices by ind2sub.
[i,j] = ind2sub(size(A(:,:,1)),cell2mat(arrayfun(#(i)find(A(:,:,i)==max_vals(i)),1:size(A,3),'UniformOutput',false)))
i =
1 2 2 1
1 2 2 1
j =
1 1 2 1
3 3 3 2
Hence, the values in j are the ones you want to have.
N.B: This question is more complex than my previous question: Matlab: How I can make this transformation on the matrix A?
I have a matrix A 4x10000, I want to use it to find another matrix C, based on a predefined vector U.
I'll simplify my problem with a simple example:
from a matrix A
20 4 4 74 20 20 4
36 1 1 11 36 36 1
77 1 1 15 77 77 1
3 4 2 6 7 8 15
and
U=[2 3 4 6 7 8 2&4&15 7&8 4|6].
& : AND
| : OR
I want, first, to find an intermediate entity B:
2 3 4 6 7 8 2&4&15 7&8 4|6
[20 36 77] 0 1 0 0 1 1 0 1 0 4
[4 1 1] 1 0 1 0 0 0 1 0 1 4
[74 11 15] 0 0 0 1 0 0 0 0 1 2
we put 1 if the corresponding value of the first line and the vector on the left, made a column in the matrix A.
the last column of the entity B is the sum of 1 of each line.
at the end I want a matrix C, consisting of vectors which are left in the entity B, but only if the sum of 1 is greater than or equal to 3.
for my example:
20 4
C = 36 1
77 1
This was a complex one indeed and because of the many restrictions and labeling processes involved, it won't be as efficient as the solution to the previous problem. Here's the code to solve the posted problem -
find_labels1 = 2:8; %// Labels to be detected - main block
find_labels2 = {[2 4 15],[7 8],[4 6]}; %// ... side block
A1 = A(1:end-1,:); %// all of A except the last row
A2 = A(end,:); %// last row of A
%// Find unique columns and their labels for all of A execpt the last row
[unqmat_notinorder,row_ind,inv_labels] = unique(A1.','rows'); %//'
[tmp_sortedval,ordered_ind] = sort(row_ind);
unqcols = unqmat_notinorder(ordered_ind,:);
[tmp_matches,labels] = ismember(inv_labels,ordered_ind);
%// Assign labels to each group
ctl = numel(unique(labels));
labelgrp = arrayfun(#(x) find(labels==x),1:ctl,'un',0);
%// Work for the main comparisons
matches = bsxfun(#eq,A2,find_labels1'); %//'
maincols = zeros(ctl,numel(find_labels1));
for k = 1:ctl
maincols(k,:) = any(matches(:,labelgrp{k}),2);
end
%// Work for the extra comparisons added that made this problem extra-complex
lens = cellfun('length',find_labels2);
lens(end) = 1;
extcols = nan(ctl,numel(find_labels2));
for k = 1:numel(find_labels2)
idx = find(ismember(A2,find_labels2{k}));
extcols(:,k)=arrayfun(#(n) sum(ismember(labelgrp{n},idx))>=lens(k),1:ctl).'; %//'
end
C = unqcols(sum([maincols extcols],2)>=3,:).' %//'# Finally the output
I will give you a partial answer. I think you can take from here. Idea is to concatenate first 3 rows of A with each element of U replicated as last column. After you get the 3D matrix, replicate your original A and then just compare the rows. The rows which are equal, that is equivalent to putting one in your table.
B=(A(1:3,:).';
B1=repmat(B,[1 1 length(U)]);
C=permute(U,[3 1 2]);
D=repmat(C,[size(B1,1),1,1]);
E=[B1 D];
F=repmat(A',[1 1 size(E,3)]);
Now compare F and E, row-wise. If the rows are equal, then you put 1 in your table. For replicating & and |, you can form some kind of indicator vector.
Say,
indU=[1 2 3 4 5 6 7 7 7 8 8 -9 -9];
Same positive value indicates &, same negative value indicates |. Different value indicate different entries.
I hope you can take from here.
I have matrix a <500 x 500> and matrix b <500 x 2>.
Matrix b contains two types of values which are row and column coordinates for matrix a. I would like to use the values in matrix b to to copy all the values that fall on the row and column coordinates of matrix a.
see example below
matrix a matrix b output
1 2 3 4 5 1 5 1 2 3 4 5
6 7 8 9 10 2 5 7 8 9 10
11 12 13 14 15 1 3 11 12 13
Because every row will have a different length you'll need to save the values into a cell array.
Something like this should work:
output = cell( size(b,1),1);
for i = 1:size(a,1)
output{i} = a(i, b(i,1):b(i,2) )
end
I have data in two columns that looks as follows:
A B
1,265848208 3
-0,608043611 0
-0,285735893 0
0,006895134 7
0 7
-0,004526196 7
0,176326617 10
-0,159688071 2
0,22439945 2
-0,991045044 1
0,178022324 1
-0,270967397 4
0,285849994 4
1,881705539 23
1,057184204 10
NaN 10
For all unique values in B I want to extract the corresponding value in column A and move it to a new matrix. I'm looking to then compute the mean of all the corresponding values in A and use as a dependent variable (weighted by no of observations per value in B) in a regression with the common value of B being the independent variable to reduce noise. Any help would on how to do this in Matlab (except running the regression) would be great!
Thanks
Oscar
Here is an efficient solution:
X = [
1.265848208 3
-0.608043611 0
-0.285735893 0
0.006895134 7
0 7
-0.004526196 7
0.176326617 10
-0.159688071 2
0.22439945 2
-0.991045044 1
0.178022324 1
-0.270967397 4
0.285849994 4
1.881705539 23
1.057184204 10
NaN 10
];
%# unique values in B, and their indices
[valB,~,subs] = unique(X(:,2));
%# values of A for each unique number in B (cellarray)
valA = accumarray(subs, X(:,1), [], #(x) {x});
%# mean of each group
meanValA = cellfun(#nanmean, valA)
%# perform regression here...
The result:
%# B values, mean of corresponding values in A, number of A values
>> [valB meanValA cellfun(#numel,valA)]
ans =
0 -0.44689 2
1 -0.40651 2
2 0.032356 2
3 1.2658 1
4 0.0074413 2
7 0.00078965 3
10 0.61676 3
23 1.8817 1