Assume I have a matrix called A.
The values of the matrix represent coordinates,
so row 2 and column 3 is the coordinate (2,3) in the 2D plan.
How can I map all the values of the matrix to different indices so that (0,0) would get the mapping value of 0 etc.?
(0,0) -> 0
(0,1) -> 1
(0,2) ->2
..
..
and so on.
Thanks.
Assuming that you are okay with the MATLAB indexing that starts with 1, this would work -
A1 = reshape([1:numel(A)],size(A,1),[])'
If you would like to start the mapping from 0, just subtract 1 -
A1 = reshape([1:numel(A)],size(A,1),[])' -1
"The sub2ind command determines the equivalent single index corresponding to a set of subscript values."
For example, if
i = sub2ind(size(A), 2, 3);
then
A(2,3) and A(i) refer to the same element in a matrix A.
In MATLAB you can index matrices linearly. Suppose you have the matrix:
a =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
Now, you can access the element in position (3,2) either using normal subscripts, or using the linear equivalent.
a(3,2)
ans = 7
a(7)
ans = 7
Assuming you have your indices as a list from 1 to numel(a) and don't really need a link between (3,2) and (7), this would be the simplest way to do it.
As you state, you want the element in position (0,0) to have index (0). Since MATLAB indexing starts at 1 you have two alternatives:
If you get a list (for instance from another program) where elements are listed from zero to (numel(a) - 1), such as ind = [0, 3, 6, 8], my suggestion is you simply do ind = ind + 1 (or ind_1 = ind + 1 if you don't want to overwrite the original vector.
Otherwise you can add one every time this way: x = a(ind + 1);.
However, if you really want a link between (3,2) and (7), I believe sub2ind is the way to go.
Related
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
I would like to create a column vector from the elements of a matrix A of size (3,3) that are not on the diagonal. Thus, I would have 6 elements in that output vector. How can I do this?
Use eye and logical negation, although this is no better than Divakar's original answer, and possibly significantly slower for very large matrices.
>> A = magic(4)
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
>> A(~eye(size(A)))
ans =
5
9
4
2
7
14
3
10
15
13
8
12
Use this to get such a column vector, assuming A is the input matrix -
column_vector = A(eye(size(A))==0)
If you don't care about the order of the elements in the output, you can also use a combination of setdiff and diag -
column_vector = setdiff(A,diag(A))
You can also use linear indexing to access the diagonal elements and null them. This will automatically reshape itself to a single vector:
A(1:size(A,1)+1:end) = [];
Bear in mind that this will mutate the original matrix A. If you don't want this to happen, make a copy of your matrix then perform the above operation on that copy. In other words:
Acopy = A;
Acopy(1:size(A,1)+1:end) = [];
Acopy will contain the final result. You need to create a vector starting from 1 and going to the end in increments of the rows of the matrix A added with 1 due to the fact that linear indices are column-major, so the linear indices used to access a matrix progress down each row first for a particular column. size(A,1) will allow us to offset by each column and we add 1 each time to ensure we get the diagonal coefficient for each column in the matrix.
Assuming that the matrix is square,
v = A(mod(0:numel(A)-1, size(A,1)+1) > 0).';
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
I would like to average every 3 values of an vector in Matlab, and then assign the average to the elements that produced it.
Examples:
x=[1:12];
y=%The averaging operation;
After the operation,
y=
[2 2 2 5 5 5 8 8 8 11 11 11]
Therefore the produced vector is the same size, and the jumping average every 3 values replaces the values that were used to produce the average (i.e. 1 2 3 are replaced by the average of the three values, 2 2 2). Is there a way of doing this without a loop?
I hope that makes sense.
Thanks.
I would go this way:
Reshape the vector so that it is a 3×x matrix:
x=[1:12];
xx=reshape(x,3,[]);
% xx is now [1 4 7 10; 2 5 8 11; 3 6 9 12]
after that
yy = sum(xx,1)./size(xx,1)
and now
y = reshape(repmat(yy, size(xx,1),1),1,[])
produces exactly your wanted result.
Your parameter 3, denoting the number of values, is only used at one place and can easily be modified if needed.
You may find the mean of each trio using:
x = 1:12;
m = mean(reshape(x, 3, []));
To duplicate the mean and reshape to match the original vector size, use:
y = m(ones(3,1), :) % duplicates row vector 3 times
y = y(:)'; % vector representation of array using linear indices
I have a matrix A with size (nr,nc), a vector of column indices B (so B has size (nr,1) and every element in B is an integer between 1 and nc), and I want to do something to every element in A that is of the form A(i,B(i)) for i between 1 and nr, efficiency being the key concern.
For concreteness, say C is a vector of size (nr,1), the goal is to do
for i=1:nr
A(i,B(i))=A(i,B(i))+C(i)
end
more efficiently. The context is usually that nr>>nc (because when nr is large vectorization is efficient for many operations). I have gotten a factor 3 speedup by using an indicator function approach:
for k=1:nc
A(:,k)=A(:,k)+(k==B).*C
end
Are there other ways (more efficient hopefully) to do this?
I guess this is similar to many questions on double-indexing, but it's concretely one I run into all the time.
Use linear indexing:
idx = sub2ind(size(A), 1:nr, B');
A(idx) = A(idx) + C';
or (edited version with one less transpose)
idx = sub2ind(size(A), (1:nr)', B);
A(idx) = A(idx) + C;
One way would be to use linear indexing of the matrix. You will need a vector v holding the offsets of the first element in each line, then index using A(v + B). For example:
>A=[1 2 3; 4 5 6; 7 8 9]
A =
1 2 3
4 5 6
7 8 9
>B = [1 2 3] % we want the 1st element of row 1, 2nd of row 2, 3rd of row 3
>ii = [0 3 6] + B
>a(ii)
1 5 9
Note: As groovingandi had shown, it is also possible (and more readable) to use sub2ind to generate the ii linear indices vector. The idea is essentially the same.