I simplify my problem, let says I have three matrices.
I want to extract the red-boxed sub-matrices. I define
S = [1 4;
2 5]
that are the linear indices of the above matrices. So, A(S), B(S) and C(S) can extract the entries of the three matrices.
I pack them into vector by V = [ A(S)(:); B(S)(:); C(S)(:) ]. Let says after some manipulations, I obtain a new vector
V_new = [12 9 8 12 21 8 7 5 3 12 11 10]'
Here comes to my problem:
E.g for matrix A, I want to obtain
2->12, 5->9, 4->8 and 6->12
which are the first four entries of my V_new.
Since I have around 200 matrices, I have no idea to swap along the 200 matrices and the updated vector, V_new at the same time. Is writing a for-loop best way to do this purpose?
Thanks in advance.
Assuming that your A, B and C matrices have the same dimensions, rather work with a 3D matrix.
e.g. assuming your example matrices
M = cat(3,A,B,C)
No to extract those 4 upper left elements:
M_subset = M(1:2,1:2,:)
And then to reshape them into the vector you had:
V = M_subset(:)
then manipulate it to get V_new and finally put it back in the original:
M(1:2,1:2,:) = reshape(V_new,2,2,[])
Related
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).';
What I intend to do is very simple but yet I haven't found a proper way to do it. I have a function handle which depends on two variables, for example:
f = #(i,j) i+j
(mine is quite more complicated, though)
What I'd like to do is to create a matrix M such that
M(i,j) = f(i,j)
Of course I could use a nested loop but I'm trying to avoid those. I've already managed to do this in Maple in a quite simple way:
f:=(i,j)->i+j;
M:=Matrix(N,f);
(Where N is the dimension of the matrix) But I need to use MATLAB for this. For now I'm sticking to the nested loops but I'd really appreciate your help!
Use bsxfun:
>> [ii jj] = ndgrid(1:4 ,1:5); %// change i and j limits as needed
>> M = bsxfun(f, ii, jj)
M =
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
If your function f satisfies the following condition:
C = fun(A,B) accepts arrays A and B of arbitrary, but equal size and returns output of the same size. Each element in the output array C is the result of an operation on the corresponding elements of A and B only. fun must also support scalar expansion, such that if A or B is a scalar, C is the result of applying the scalar to every element in the other input array.
you can dispose of ndgrid. Just add a transpose (.') to the first (i) vector:
>> M = bsxfun(f, (1:4).', 1:5)
Function handles can accept matrices as inputs. Simply pass a square matrix of size N where the values corresponds to the row number for i, and a square matrix of size N where the values correspond to the column number for j.
N = 5;
f = #(i,j) i+j;
M = f(meshgrid(1:N+1), meshgrid(1:N+1)')
I'm new to Matlab. I need help with matrix comparison.
I have three matrices: R, S and T (size: 95956 x 1) and I need to compare all of their elements and see which elements are equal. Then we multiply the same values by a fourth matrix.
In a Matlab file, I read the coordinates and separate them into three matrices. Thus obtaining matrices R, S and T saved in "data.mat". Then I read in another file "data.mat" and used a For loop:
for t=1:1:length(CoordinateIndex)
index = R == S;
ts=ts+1;
end
Is this correct and how I will multiply only those values which are equal to a fourth matrix?
Using some simple example matrices and logical indexing, we can find the values that are common among all three via:
R = [1:4 1:4 1:4 1:4];
S = [1:2 1:2 1:4 1:6 1:2];
T = [1:6 1:2 1:8];
R_common = R(R==S&R==T);
This produces:
R_common =
1 2 1 2 3 4
If you want the indices that are common among all three matrices, use:
I = find(R==S&R==T);
which gives
I =
1 2 9 10 11 12
EDIT: As #Adiel suggested, you may be interested in the values that are common among the matrices regardless of their location in the matrices. This can be accomplished simply via the intersect function (also see this):
RST_intersection = intersect(intersect(R,S),T)
For my example matrices this gives
RST_intersection =
1 2 3 4
Note that the arguments to intersect can be different lengths, which may be helpful.
In matlab, after meeting a specific criterion, I used to return back the pixel itself and store it in the vector pixels as follows:
pixels(index) = y(i,j);
Now, I would like to return the location of those pixels. Should I do the following?
pixels(index) = i,j;
EDIT
If I want to then set those indexes to the value 1, I do the following, right?
for i=1:m
for j=1:n
y(i,j)=1
end
end
Thanks.
It is extremely inefficient to do so in a nested loop in Matlab.
Using sub2ind can help you do so much faster:
y( sub2ind( size(y), i, j ) ) = 1;
EDIT - sub2ind
What sub2ind does?
Suppose you have a matrix M of size [4 6]:
M = [ 1 5 9 13 17 21
2 6 10 14 18 22
3 7 11 15 19 23
4 8 12 16 20 24 ];
You wish to access two elements: the one at the first row and second column, and another at the fourth row and the fifth column.
In that case you have the rows you wish to access r = [ 1 4 ] and the columns you wish to access c = [ 2 5 ]. However, if you try and access
>> M( r, c )
This is a 2x2 matrix
ans =
5 17
8 20
And not the two elements you were looking for (which are 5 and 20).
What sub2ind does is convert the row/column indices you have into linear indices
>> sub2ind( size(M), r, c )
ans =
5 20
which happens to be the linear indices of the requested entries.
You can think of linear indices as the single index required to access an element in a matrix in the case that the matrix was converted to a vector stacking its columns one after the other.
A few comments:
Matlab has a few ways of indexing matrices: by row / column indices (like i and j in your question). By linear indices (like index in your question). However, the more efficient way is to use logical indexing: that is, using a matrix of the same size as y with true for the entries you wish to set / get.
So, in your example, if you could get such a logical matrix instead of index or i and j it would have been better.
Matlab has many advantages over other programing languages. One of them is its ability to perform vector/matrix operations extremely efficient. Resorting to loops, or worse, nested loops, is something that should be avoided in Matlab.
It is not a good practice to use i and j as variables in Matlab.
If you want to find the occurrence of a value y(i,j) simply evaluate
idx = (pixels == y(i,j));
Depending on your variables you can then probably do
index(idx) = 1;
In MATLAB I have a very large matrix (matrix A). Now I would like to find the row-index of the row which contain certain values in the second column. These values - which I'm looking for in Matrix A - are stored in anonther Matrix (Matrix B) with consists out of a row (800 numbers).
Besides I would like to redo this calculation for the same matrix A, but for ten different matrices, with different sizes (which contain the values I'm looking for in different columns of matrix A).
Because of the sizes of the matrix I think i need a loop to extract the row in matrix A which contain te value of Matrix B. How can I do this?
regards,
V
Thanks for the quick response! Indeed the problem is maybe a bit complex to answer without an example, and indeed duplicate entries cause some problems. Therefore hereby an example
For example I have a -simplified- matrix A:
1 2 3 4
9 9 9 9
4 3 2 1
And a -simplified- matrix (row) B: [9 3]
And a -simplified- matrix (row) C: [9 2]
Then I would like to get matrix D and matrix E.These matrices should contain in the first column the numbers from the original matrix D(or E) and in the second column the corresponding row-location of this value in matrix A.
So, matrix D =
9 2
3 3
matrix E =
9 2
2 3
As represented in this example matrix B and matrix C can contain data which is present in several column of matrix A (like the nine). However, martix B should "search" in column 2 of matrix A. Likewise, should matrix C "search" in column 3 of Matrix A, resulting in matrix D and E as given in the example.
As mentionned by Shai in his comment, your question is quite vague and a lot of special case could arise (duplicate entries, relative size of A and B, etc.). But in all generality I tried a small piece of code that seems to do what you want. There are certainly quicker ways of doing it, and certainly more information on your problem could help optimize this.
colA=2;
% Example
nmax=10;
nA=5;
A=randi(nmax,[nA nA]);
nB=3;
B=randi(nmax,[1 nB]);
% Find rows
rows=cell(size(B));
for i=1:numel(B)
rows(i)={find(A(:,colA)==B(i))};
end
The input / output was:
A =
3 7 8 5 4
9 7 3 7 5
8 2 9 9 8
9 5 9 7 9
3 3 4 6 8
B =
1 7 5
rows =
[0x1 double] [1;2] [4]
Assuming you have two vectors, largeDataIndex (the second column of your matrix) and interestingIndex (your b) and you want the following:
For each value of interestingIndex , find the position in largeDataIndex
Then an easy method would be this:
result = zeros(size(interestingIndex))
for i = 1:length(result)
result(i) = find(interestingIndex(i) == largeDataIndex)
end
Note that this assumes there is always just one entry that matches, otherwise you should define result as a cell array rather than a vector.