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Suppose I have a matrix of dimension [4x4], and a vector of [16x1], I need to multiply every 4 element in the vector in one element in the matrix, (instead of multiplying element in row by element in vector), how can I do that using loop ?
For example here below, the results of the first four elements in the resulted vector as shown in the below example, then the same thing for the second, third and fourth rows in the matrix. :
So the results in that example is supposed to be [16x1]
Thank you
Using kron you can use this one-liner:
%A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
%v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
sum(kron(A,ones(4,4)).'.*v).'/4
I use the kronecker tensor product to "replicate" 4x4 time the A matrice. After that it's pure algebra.
This is just matrix multiplication in disguise... If your tall vector was a matrix of the same size as the matrix shown, where each highlighted block is a row, it's matrix multiplication. We can set this up, then reshape back into a vector.
You can use indexing to turn this into simple matrix multiplication. A question I answered already today details how the below indexing works using bsxfun, then we just reshape at the end:
% Setup
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
% Matrix mutliplication
r = numel(v)/size(A,1);
b = A * v( bsxfun( #plus, (1:r:numel(v)).', 0:r-1 ) );
% Make result a column vector
b = reshape( b.', [], 1 );
See if this is what you want:
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
v = [2 2 2 2 0 0 0 0 1 1 1 1 3 3 3 3].';
r = reshape(sum(bsxfun(#times, permute(A, [3 2 1]), permute(reshape(v, 1, [], size(A,2)), [2 3 1])), 2), [], 1);
which gives
r =
17
17
17
17
41
41
41
41
65
65
65
65
89
89
89
89
There are details that I assumed, but this shoudl do the trick:
A=reshape(1:16,4,4).';
b=repelem([2,0,1,3],1,4).';
c=[];
for row=1:size(A,1)
c=[ c; sum(reshape(repelem(A(row,:),4).*b.',4,[]),2)];
end
I am assuming here that your demo for the vector is just a bad example and that you wont have repeated values, otherwise an easier version can be achieved by just not doing 3/4ths of the multiplications.
If you do not have access to repelem, have a look at alterative codes that do the same thing:Element-wise array replication in Matlab
Lets say I have a 4 dimensional matrix, from which I would like to retrieve the maximum values over the 2nd and 3rd dimension.
A = rand(4, 4, 4, 4);
[max_2, in_2] = max(A, [], 2);
[max_3, in_3] = max(max_2, [], 3);
How could I use ind_2 and ind_3 to obtain a logical 4 dimensional matrix, where a 1 entry means this entry is maximum in the 2nd and 3rd dimension?
I would use this approach:
A = rand(4, 4, 4, 4); % example data
B = permute(A, [1 4 2 3]); % permute dims 2 and 3 to the end
B = reshape(B, size(A,1), size(A,4), []); % collapse last two dims
C = bsxfun(#eq, B, max(B, [], 3)); % maximize over collapsed last dim
C = reshape(C, size(A,1), size(A,4), size(A,2), size(A,3)); % expand dims back
C = permute(C, [1 3 4 2]); % permute dims back. This is the final result
Here's an approach working with linear indices and uses argmax indices from max function, so it would only consider the first argmax in case of ties for the max value -
% Get size parameters
[m,n,p,q] = size(A);
% Reshape to merge second and third dims
[~, in_23] = max(reshape(A,m,[],q), [], 2);
% Get linear indices equivalent that could be mapped onto output array
idx1 = reshape(in_23,m,q);
idx2 = bsxfun(#plus,(1:m)', m*n*p*(0:q-1)) + (idx1-1)*m;
% Initialize output array an assign 1s at linear indices from idx2
out = false(m,n,p,q);
out(idx2) = 1;
Explanation with a sample
1) Input array :
>> A
A(:,:,1,1) =
9 8
9 1
A(:,:,2,1) =
2 9
8 1
A(:,:,1,2) =
1 7
8 1
A(:,:,2,2) =
8 5
9 7
2) Reshape array for a better visualization :
>> reshape(A,m,[],q)
ans(:,:,1) =
9 8 2 9
9 1 8 1
ans(:,:,2) =
1 7 8 5
8 1 9 7
3) The question is to take max value from each of the rows. For that, we had idx2 as the linear indices :
>> idx2
idx2 =
1 13
2 14
Looking back at the reshape version, thus we chose (bracketed ones) -
>> reshape(A,m,[],q)
ans(:,:,1) =
[9] 8 2 9
[9] 1 8 1
ans(:,:,2) =
1 7 [8] 5
8 1 [9] 7
So, looking closely, we see that for the first row, we had two 9s, but we are choosing the first one only.
4) Finally, we are assigning these into the output array initialized as logical zeros :
>> out
out(:,:,1,1) =
1 0
1 0
out(:,:,2,1) =
0 0
0 0
out(:,:,1,2) =
0 0
0 0
out(:,:,2,2) =
1 0
1 0
Is there a way to create matrix that is consisted of a certain numbers that are stored in an array? For example, I want to create a 10-by-1 matrix consisting only of numbers from an array a = [6,2,15,24], that are randomly stored in matrix elements. The final product should look something like this:
M = [15,24,2,15,2,6,24,15,2,15]
If you have the Statistics toolbox, you can use randsample with the third argument set to true to indicate that the data a is to be sampled with replacement:
a = [6 2 15 24];
M = randsample(a,10,true)
function b = resample( a, size )
indices = randi( numel( a ), size );
b = a( indices );
Example:
>> resample( [6,2,15,24], [4,5] )
ans =
2 6 15 2 6
2 2 15 15 6
24 6 6 2 6
2 6 24 6 2
Suppose I have a 3x4 matrix
Now how to calculate a matrix of the same size, which contains each element being a sum of itself and all elements, which are higher and lefter than it?
You must be looking for cumsum:
integ = cumsum( cumsum( A, 1 ), 2 );
For example:
A = [1 2 3;
4 5 6];
cumsum( cumsum( A, 1 ), 2 )
Results with:
1 3 6
5 12 21
PS,
This operation is sometimes refereed to as integral image.
I need help with my code. The code is used to find the minumin of a square-distance problem. I am providing my code through an example, I believe this will be the easiest way to explain what I need.
clear all
clc
x=10.8; % is a fixed value
y=34; % is a fixed value
z=12; % is a fixed value
A = [11 14 1; 5 8 18; 10 8 19; 13 20 16]; % a (4x3) matrix
B = [2 3 10; 6 15 16; 7 3 15; 14 14 19]; % a (4x3) matrix
I create a new matrix C which is composed in this following way:
C1 = bsxfun(#minus, A(:,1)',B(:,1));
C1=C1(:); % this is the first column of the new matrix C
C2 = bsxfun(#minus, A(:,2)',B(:,2));
C2=C2(:); % this is the second column of the new matrix C
C3 = bsxfun(#minus, A(:,3)',B(:,3));
C3=C3(:); % this is the third column of the new matrix C
C = [C1 C2 C3]; % the new matrix C of size (16x3)
C has to be formed in this way! And this is what I meant when I wrote in my title a composed-matrix
Then:
[d,p] = min((C(:,1)-x).^2 + (C(:,2)-y).^2 + (C(:,3)-z).^2);
d = sqrt(d);
outputs:
d = 18.0289;
p = 13;
Gives me the distance (d) and position (p) which satisfies this min problem.
MY PROBLEM:
I need to find which combinations of A and B has given my this p value, in other words I need the index from ´A,B´ which gives me this optimal C1,C2,C3:
C1 = bsxfun(#minus, A(?,1)',B(?,1));
C2 = bsxfun(#minus, A(?,2)',B(?,2));
C3 = bsxfun(#minus, A(?,3)',B(?,3));
The ? is the index position I need, in this case the index position of the matrix A and the index position of B.
Calculated by hand I have the following illustration:
I know that:
C = [9 11 -9
5 -1 -15
4 11 -14
-3 0 -18
3 5 8
-1 -7 2
-2 5 3
-9 -6 -1
8 5 9
4 -7 3
3 5 4
-4 -6 0
11 17 6
7 5 0
6 17 1
-1 6 -3]
And I know that my optimal index is given in the position 13th. This index positions goes back to:
[13-2 20-3 16-10]
Which is A(4,:) - B(1,:)
I need a code which can help me to find this indexes from A and B
Thanks in advance!
PS. I am using the code in parameter estimation problems of ODEs.
First case: vector-matrix case
subvals = bsxfun(#minus,A,[x y z])
[distance,index] = min(sqrt(sum(subvals.^2,2)))
Second case: Two matrices case
subvals = bsxfun(#minus,A,permute(B,[3 2 1]));
[distances,indices] = min(sqrt(sum(subvals.^2,2)),[],3)
Testing for second case:
%%// Get some random data into A and B
A = randi(20,8,3)
B = randi(20,4,3)
%%// Just to test out out code for correctness,
%%// let us make any one one row of B, say 3rd row equal to
%%// any one row of A, say the 6th row -
B(3,:) = A(6,:)
%%// Use the earlier code
subvals = bsxfun(#minus,A,permute(B,[3 2 1]));
[distances,indices] = min(sqrt(sum(subvals.^2,2)),[],3)
%%// Get the minimum row index for A and B
[~,min_rowA] = min(distances)
min_rowB = indices(min_rowA)
Verification
min_rowA =
6
min_rowB =
3
Edit 1 [Response to simple example posted in question]:
The title says you are interested in finding the difference of two matrices and then find the shortest distance between it to a vector [x y z]. So I am hoping this is what you need -
x=10.8;
y=34;
z=12;
A = [11 14 1; 5 8 18; 10 8 19; 13 20 16];
B = [2 3 10; 6 15 16; 7 3 15; 14 14 19];
C = A -B; %%// Distance of two vectors as posted in title
subvals = bsxfun(#minus,C,[x y z])
[distance,index] = min(sqrt(sum(subvals.^2,2)))
Output
distance =
31.0780
index =
3
Edit 2: After you have done this -
[d,p] = min((C(:,1)-x).^2 + (C(:,2)-y).^2 + (C(:,3)-z).^2);
If you are looking to find the corresponding indices of A and B , you may do this -
[minindex_alongB,minindex_alongA] = ind2sub(size(A),p)