I have 3 vectors, v1, v2, v3. What I want to get is the difference between every possible pair of them, that is, v1-v2, v1-v3, v2-v3. How can I do this without looping in matlab?
Thank you.
Just use nchoosek to generate the combinations first and then use them to index into your array of row-vectors:
Test case:
numVectors = 3;
dim = 5;
Vs = rand(numVectors, dim);
Actual computation:
combs = nchoosek(1:size(Vs,1), 2);
differences = Vs(combs(:,1),:) - Vs(combs(:,2),:);
The above creates 3 random row vectors of dimension 5. So in your case, you may want to replace the creation of the random matrix with Vs = [v1; v2; v3]; if your vectors are row vectors; or transpose the vectors using Vs = [v1, v2, v3].'; if your data are column vectors.
Using bsxfun:
clear
clc
%// Sample vectors.
v1 = [1 2];
v2 = [10 20];
v3 = [0 0];
Out = bsxfun(#minus,[v1 v2 v3], [v1 v2 v3].')
Out =
0 1 9 19 -1 -1
-1 0 8 18 -2 -2
-9 -8 0 10 -10 -10
-19 -18 -10 0 -20 -20
1 2 10 20 0 0
1 2 10 20 0 0
Reasoning: Each difference is computed starting from the 1st element of the 1st vector until the 2nd element of the last vector.
The 1st column contains all the differences for the 1st element of the 1st vector, i.e. (1 -1), (1-2), (1-10), (1 - 20), (1 - 0), (1 - 0).
Then 2nd column, same thing but this time with the 2: (2 - 1), (2 - 2), (2 - 10), and so on.
Sorry if my explanations are unclear haha I don't know the right terms in english. Please ask for more details.
Code
%// Concatenate all vectors to form a 2D array
V = cat(2,v1(:),v2(:),v3(:),v4(:),v5(:))
N = size(V,2) %// number of vectors
%// Find all IDs of all combinations as x,y
[y,x] = find(bsxfun(#gt,[1:N]',[1:N])) %//'
%// OR [y,x] = find(tril(true(size(V,2)),-1))
%// Use matrix indxeing to collect vector data for all combinations with those
%// x-y IDs from V. Then, perform subtractions across them for final output
diff_array = V(:,x) - V(:,y)
Few points about the code
bsxfun with find gets us the IDs for forming pairwise combinations.
We use those IDs to index into the 2D concatenated array and perform subtractions between them to get the final output.
Bonus Stuff
If you look closely into the part where it finds the IDs of all combinations, that is basically nchoosek(1:..,2).
So, basically one can have alternatives to nchoosek(1:N,2) as:
[Y,X] = find(bsxfun(#gt,[1:N]',[1:N]))
[y,x] = find(tril(true(N),-1))
with [X Y] forming those pairwise combinations and might be interesting to benchmark them!
Related
I have a matrix index in Matlab with size GxN and a matrix A with size MxN.
Let me provide an example before presenting my question.
clear
N=3;
G=2;
M=5;
index=[1 2 3;
13 14 15]; %GxN
A=[1 2 3;
5 6 7;
21 22 23;
1 2 3;
13 14 15]; %MxN
I would like your help to construct a matrix Response with size GxM with Response(g,m)=1 if the row A(m,:) is equal to index(g,:) and zero otherwise.
Continuing the example above
Response= [1 0 0 1 0;
0 0 0 0 1]; %GxM
This code does what I want (taken from a previous question of mine - just to clarify: the current question is different)
Response=permute(any(all(bsxfun(#eq, reshape(index.', N, [], G), permute(A, [2 3 4 1])), 1), 2), [3 4 1 2]);
However, the command is extremely slow for my real matrix sizes (N=19, M=500, G=524288). I understand that I will not be able to get huge speed but anything that can improve on this is welcome.
Aproach 1: computing distances
If you have the Statistics Toolbox:
Response = ~(pdist2(index, A));
or:
Response = ~(pdist2(index, A, 'hamming'));
This works because pdist2 computes the distance between each pair of rows. Equal rows have distance 0. The logical negation ~ gives 1 for those pairs of rows, and 0 otherwise.
Approach 2: reducing rows to unique integer labels
This approach is faster on my machine:
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
It works by reducing rows to unique integer labels (using the third output of unique), and comparing the latter instead of the former.
For your size values this takes approximately 1 second on my computer:
clear
N = 19; M = 500; G = 524288;
index = randi(5,G,N); A = randi(5,M,N);
tic
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
toc
gives
Elapsed time is 1.081043 seconds.
MATLAB has a multitude of functions for working with sets, including setdiff, intersect, union etc. In this case, you can use the ismember function:
[~, Loc] = ismember(A,index,'rows');
Which gives:
Loc =
1
0
0
1
2
And Response would be constructed as follows:
Response = (1:size(index,1) == Loc).';
Response =
2×5 logical array
1 0 0 1 0
0 0 0 0 1
You could reshape the matrices so that each row instead lies along the 3rd dimension. Then we can use implicit expansion (see bsxfun for R2016b or earlier) for equality of all elements, and all to aggregate on the rows (i.e. false if not all equal for a given row).
Response = all( reshape( index, [], 1, size(index,2) ) == reshape( A, 1, [], size(A,2) ), 3 );
You might even be able to avoid some reshaping by using all in another dimension, but it's easier for me to visualise it this way.
I have a matrix suppX in Matlab with size GxN and a matrix A with size MxN. I would like your help to construct a matrix Xresponse with size GxM with Xresponse(g,m)=1 if the row A(m,:) is equal to the row suppX(g,:) and zero otherwise.
Let me explain better with an example.
suppX=[1 2 3 4;
5 6 7 8;
9 10 11 12]; %GxN
A=[1 2 3 4;
1 2 3 4;
9 10 11 12;
1 2 3 4]; %MxN
Xresponse=[1 1 0 1;
0 0 0 0;
0 0 1 0]; %GxM
I have written a code that does what I want.
Xresponsemy=zeros(size(suppX,1), size(A,1));
for x=1:size(suppX,1)
Xresponsemy(x,:)=ismember(A, suppX(x,:), 'rows').';
end
My code uses a loop. I would like to avoid this because in my real case this piece of code is part of another big loop. Do you have suggestions without looping?
One way to do this would be to treat each matrix as vectors in N dimensional space and you can find the L2 norm (or the Euclidean distance) of each vector. After, check if the distance is 0. If it is, then you have a match. Specifically, you can create a matrix such that element (i,j) in this matrix calculates the distance between row i in one matrix to row j in the other matrix.
You can treat your problem by modifying the distance matrix that results from this problem such that 1 means the two vectors completely similar and 0 otherwise.
This post should be of interest: Efficiently compute pairwise squared Euclidean distance in Matlab.
I would specifically look at the answer by Shai Bagon that uses matrix multiplication and broadcasting. You would then modify it so that you find distances that would be equal to 0:
nA = sum(A.^2, 2); % norm of A's elements
nB = sum(suppX.^2, 2); % norm of B's elements
Xresponse = bsxfun(#plus, nB, nA.') - 2 * suppX * A.';
Xresponse = Xresponse == 0;
We get:
Xresponse =
3×4 logical array
1 1 0 1
0 0 0 0
0 0 1 0
Note on floating-point efficiency
Because you are using ismember in your implementation, it's implicit to me that you expect all values to be integer. In this case, you can very much compare directly with the zero distance without loss of accuracy. If you intend to move to floating-point, you should always compare with some small threshold instead of 0, like Xresponse = Xresponse <= 1e-10; or something to that effect. I don't believe that is needed for your scenario.
Here's an alternative to #rayryeng's answer: reduce each row of the two matrices to a unique identifier using the third output of unique with the 'rows' input flag, and then compare the identifiers with singleton expansion (broadcast) using bsxfun:
[~, ~, w] = unique([A; suppX], 'rows');
Xresponse = bsxfun(#eq, w(1:size(A,1)).', w(size(A,1)+1:end));
I have a situation analogous to the following
z = magic(3) % Data matrix
y = [1 2 2]' % Column indices
So,
z =
8 1 6
3 5 7
4 9 2
y represents the column index I want for each row. It's saying I should take row 1 column 1, row 2 column 2, and row 3 column 2. The correct output is therefore 8 5 9.
I worked out I can get the correct output with the following
x = 1:3;
for i = 1:3
result(i) = z(x(i),y(i));
end
However, is it possible to do this without looping?
Two other possible ways I can suggest is to use sub2ind to find the linear indices that you can use to sample the matrix directly:
z = magic(3);
y = [1 2 2];
ind = sub2ind(size(z), 1:size(z,1), y);
result = z(ind);
We get:
>> result
result =
8 5 9
Another way is to use sparse to create a sparse matrix which you can turn into a logical matrix and then sample from the matrix with this logical matrix.
s = sparse(1:size(z,1), y, 1, size(z,1), size(z,2)) == 1; % Turn into logical
result = z(s);
We also get:
>> result
result =
8
5
9
Be advised that this only works provided that each row index linearly increases from 1 up to the end of the rows. This conveniently allows you to read the elements in the right order taking advantage of the column-major readout that MATLAB is based on. Also note that the output is also a column vector as opposed to a row vector.
The link posted by Adriaan is a great read for the next steps in accessing elements in a vectorized way: Linear indexing, logical indexing, and all that.
there are many ways to do this, one interesting way is to directly work out the indexes you want:
v = 0:size(y,2)-1; %generates a number from 0 to the size of your y vector -1
ind = y+v*size(z,2); %generates the indices you are looking for in each row
zinv = z';
zinv(ind)
>> ans =
8 5 9
Suppose I have 2 input vectors x and reset of the same size
x = [1 2 3 4 5 6]
reset = [0 0 0 1 0 0]
and an output y which is the cumulative sum of the elements in x. Whenever the value of resets corresponds to 1, the cumulative sum for the elements reset and start all over again just like below
y = [1 3 6 4 9 15]
How would I implement this in Matlab?
One approach with diff and cumsum -
%// Setup few arrays:
cx = cumsum(x) %// Continuous Cumsumed version
reset_mask = reset==1 %// We want to create a logical array version of
%// reset for use as logical indexing next up
%// Setup ID array of same size as input array and with differences between
%// cumsumed values of each group placed at places where reset==1, 0s elsewhere
%// The groups are the islands of 0s and bordered at 1s in reset array.
id = zeros(size(reset))
diff_values = x(reset_mask) - cx(reset_mask)
id(reset_mask) = diff([0 diff_values])
%// "Under-compensate" the continuous cumsumed version cx with the
%// "grouped diffed cumsum version" to get the desired output
y = cx + cumsum(id)
Here's a way:
result = accumarray(1+cumsum(reset(:)), x(:), [], #(t) {cumsum(t).'});
result = [result{:}];
This works because if the first input to accumarray is sorted, the order within each group of the second input is preserved (more about this here).
I have a <206x193> matrix A. It contains the values of a parameter at 206 different locations at 193 time steps. I am interested in the maximum value at each location over all times as well as the corresponding indices. I have another matrix B with the same dimensions of A and I'm interested in values for each location at the time that A's value at that location was maximal.
I've tried [max_val pos] = max(A,[],2), which gives the right maximum values, but A(pos) does not equal max_val.
How exactly does this function work?
I tried a smaller example as well. Still I don't understand the meaning of the indices....
>> H
H(:,:,1) =
1 2
3 4
H(:,:,2) =
5 6
7 8
>> [val pos] = max(H,[],2)
val(:,:,1) =
2
4
val(:,:,2) =
6
8
pos(:,:,1) =
2
2
pos(:,:,2) =
2
2
The indices in idx represent the index of the max value in the corresponding row. You can use sub2ind to create a linear index if you want to test if A(pos)=max_val
A=rand(206, 193);
[max_val, idx]=max(A, [], 2);
A_max=A(sub2ind(size(A), (1:size(A,1))', idx));
Similarly, you can access the values of B with:
B_Amax=B(sub2ind(size(A), (1:size(A,1))', idx));
From your example:
H(:,:,2) =
5 6
7 8
[val pos] = max(H,[],2)
val(:,:,2) =
6
8
pos(:,:,2) =
2
2
The reason why pos(:,:,2) is [2; 2] is because the maximum is at position 2 for both rows.
max is a primarily intended for use with vectors. In normal mode, even the multi-dimensional arrays are treated as a series of vectors along which the max function is applied.
So, to get the values in B at each location at the time where A is maximum, you should
// find the maximum values and positions in A
[c,i] = max(A, [], 2);
// iterate along the first dimension, to retrieve the corresponding values in B
C = [];
for k=1:size(A,1)
C(k) = B(k,i(k));
end
You can refer to #Jigg's answer for a more concise way of creating matrix C