I made a matrix like this :
A = randi([-10, 10], 3, 3);
Which can have this result :
-1 1 -2
2 2 8
5 3 10
How can I transform it in a way that A(1) = -1, A(2) = 1 and A(3) = -2
(Accessing first line with terms 1,2,3)
Currently, A(1) = -1, A(2) = 2 and A(3) = 5 (columns)
Note : Not only the first line, but I want to access all elements sorted by lines
Thank you !
In Octave and MATLAB, data is stored in column-major order which means for your matrix, indices and elements are like this:
You need to take the transpose of the original matrix to access them the way you stated. In Octave, you can directly access them using A.'(1) , A.'(2), A.'(3) etc. In MATLAB, you cannot access them like Octave. So save them in a new matrix or replace the contents of the previous matrix. i.e. A = A.' and then you can use A(1), A(2), A(3) etc to access the desired elements.
A.' or transpose(A) for the given A actually gives:
So now as per the column major order, first, second and third elements are -1, 1 and -2 respectively and so on.
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 wish to append a row-vector (and later, also a column vector) to an existing x by y by z matrix. So basically "Add a new row (at the "bottom") for each z in the original 3d matrix. Consider the following short Matlab program
appendVector = [1 2 3 4 5]; % Small matrix for brevity. Actual matrices used are much larger.
origMatrix = ones(5,5,3);
appendMatrix = [origMatrix( ... ); appendVector];
My question is: How do I adress (using Matlab-style matrix adressing, not a "manual" C-like loop) origMatrix( ... ) in order to append the vector above? Feel free to also include a suggestion on how to do the same operation for a column-vector (I am thinking that the correct way to do the latter is to simply use the '-operator in Matlab).
A "row" in a 3D matrix is actually a multi-dimensional array.
size(origMatrix(1,:,:))
% 5 3
So to append a row, you would need to append a 5 x 3 array.
toAppend = rand(5, 3);
appendMatrix = cat(1, origMatrix, toAppend);
You could append just a 5 element vector and specify an index for the third dimension. In this case, the value for the "row" for all other indices in the third dimension would be filled with zeros.
appendVector = [1 2 3 4 5];
origMatrix = ones(5,5,3);
appendMatrix = origMatrix;
appendMatrix(end+1, :, 1) = appendVector;
If instead, you want to append the same vector along the third dimension, you could use repmat to turn your vector into a 1 x 5 x 3 array and then append that.
appendVector = repmat([1 2 3 4 5], 1, 1, size(origMatrix, 3));
appendMatrix = cat(1, origMatrix, appendVector);
For matrices with dimensions equal or less then 2 the command is:
For instance:
>> mat2str(ones(2,2))
ans =
[1 1;1 1]
However, as the help states, this does not work for higher dimensions:
>> mat2str(rand(2,2,2))
Error using mat2str (line 49)
Input matrix must be 2-D.
How to output matrices with higher dimensions than 2 with that is code compatible, without resorting to custom made for loops?
This isn't directly possible because there is no built-in character to represent concatenation in the third dimension (an analog to the comma and semicolon in 2D). One potential workaround for this would be to perform mat2str on all "slices" in the third dimension and wrap them in a call to cat which, when executed, would concatenate all of the 2D matrices in the third dimension to recreate your input matrix.
M = reshape(1:8, [2 2 2]);
arrays = arrayfun(#(k)mat2str(M(:,:,k)), 1:size(M, 3), 'uni', 0);
result = ['cat(3', sprintf(', %s', arrays{:}), ')'];
result =
'cat(3, [1 3;2 4], [5 7;6 8])'
isequal(eval(result), M)
1
UPDATE
After thinking about this some more, a more elegant solution is to flatten the input matrix, run mat2str on that, and then in the string used to recreate the data, we utilize reshape combined with the original dimensions to provide a command which will recreate the data. This will work for any dimension of data.
result = sprintf('reshape(%s, %s);', mat2str(M(:)), mat2str(size(M)));
So for the following 4D input
M = randi([0 9], 1, 2, 3, 4);
result = sprintf('reshape(%s, %s);', mat2str(M(:)), mat2str(size(M)));
'reshape([6;9;4;6;5;2;6;1;7;2;1;7;2;1;6;2;2;8;3;1;1;3;8;5], [1 2 3 4]);'
Now if we reconstruct the data using this generated string, we can ensure that we get the correct data back.
Mnew = eval(result);
size(Mnew)
1 2 3 4
isequal(Mnew, M)
1
By specifying both the class and precision inputs to mat2str, we can even better approximate the input data including floating point numbers.
M = rand(1,2,3,4,5);
result = sprintf('reshape(%s, %s);', mat2str(M(:),64,'class'), mat2str(size(M)));
isequal(eval(result), M)
1
I have written a for loop code and I want to write in more succinct way without using a for loop, but instead use matrix conditional.
I am teaching myself matlab and I would appreciate any feedback.
I want to create a new matrix, the first column is y, and the second column is filled with zero except for the y's whose indices are contained in the indices matrix. And in the latter case, add 1 instead of 0.
Thanks.
y=[1;2;3;4;5;6;7];
indices=[1;3;5];
[m,n]=size(y);
tem=zeros(m,1);
data=[y,tem];
[r,c]=size(indices);
for i=1:r
a=indices(i);
data(a,2 )=1;
end
Output:
data =
1 1
2 0
3 1
4 0
5 1
6 0
7 0
A shorter alternative:
data = [y(:), full(sparse(indices, 1, 1, numel(y), 1))];
The resulting matrix data is composed of two column vectors: y(:) and a sparse array, with "1"s at the positions corresponding to indices.
Using proper initialization and sparse matrices can be really useful in MATLAB.
How about
data = zeros( m, 2 );
data(:,1) = y;
data( indices, 2 ) = 1;
Suppose now I have two vectors of same length:
A = [1 2 2 1];
B = [2 1 2 2];
I would like to create a matrix C whose dim=m*n, m=max(A), n=max(B).
C = zeros(m,n);
for i = 1:length(A)
u = A(i);
v = B(i);
C(u,v)=C(u,v)+1;
end
and get
C =[0 2;
1 1]
More precisely, we treat the according indices in A and B as rows and columns in C, and C(u,v) is the number of elements in {k | A(i)=u and B(i)=v, i = 1,2,...,length(A)}
Is there a faster way to do that?
Yes. Use sparse. It assembles (i.e., sums up) the matrix values for repeating row-column pairs for you. You need an additional vector with the values that will be assembled into the matrix entries. If you use ones(size(A)), you will have exactly what you need - counting of repeated row-column pairs
spA=sparse(A, B, ones(size(A)));
full(spA)
ans =
0 2
1 1
The same can be obtained by simply passing scalar 1 to sparse function instead of a vector of values.
For matrices that have a large number of zero entries this is absolutely crucial that you use sparse storage. Another function you could use is accumarray. It can essentially do the same thing, but also works on dense matrix structure:
AA=accumarray([A;B]', 1);
AA =
0 2
1 1
You can pass size argument to accumarray if you want to create a matrix of specific size
AA=accumarray([A;B]', 1, [2 3]);
AA =
0 2 0
1 1 0
Note that you can actually also make it produce sparse matrices, and use a different operator in assembly (i.e., not necessarily a sum)
AA=accumarray([A;B]', 1, [2 3], #sum, 0, true)
will produce a sparse matrix (last parameter set to true) using sum for assembly and 0 as a fill value, i.e. a value which is used in cases a given row-column pair does not exist in A/B.