I have a txt file with 4 column of real numbers like this
1 0 2 5
0 1 -6 2.5
-1 2 7 9
3 5 9 -2
ecc
Each column need to be a 200x200 matrix. What I should to is to read each single element of a the first column and put it into a matrix 200x200 matrix. As it is the file, it should not matter if the element in the second row (0 in the example) becomes the (1,2) element of the matrix or (2,1), since this matrix should be symmetric.
Can you help me?
You can use the following approach, which in this example extracts the first matrix:
data = load(<path to txt file>)
N = size(data,1);
mat1 = zeros(sqrt(N),sqrt(N));
mat1(:) = data(:,1);
It is also possible to keep the 4 matrices in a cell, by using the following:
data = load(<path to file>)
N = size(data,1);
matCell = cell(size(data,2),1);
for ii=1:length(matCell)
matCell{ii} = zeros(sqrt(N),sqrt(N));
matCell{ii}(:) = data(:,ii);
end
In this case you can use matCell{1} to access the first 200x200 matrix.
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 wonder how to do this in MATLAB.
I have a={1;2;3} and would like to create a cell array
{{1,1};{1,2};{1,3};{2,1};{2,2};{2,3};{3,1};{3,2};{3,3}}.
How can I do this without a for loop?
You can use allcomb from MATLAB File-exchange to help you with this -
mat2cell(allcomb(a,a),ones(1,numel(a)^2),2)
Just for fun, using kron and repmat:
a = {1;2;3}
b = mat2cell([kron(cell2mat(a),ones(numel(a),1)) repmat(cell2mat(a),numel(a),1)])
Here square brackets [] are used to perform a concatenation of both column vectors, where each is defined either by kron or repmat.
This can be easily generalized, but I doubt this is the most efficient/fast solution.
Using repmat and mat2cell
A = {1;2;3};
T1 = repmat(A',[length(A) 1]);
T2 = repmat(A,[1 length(A)]);
C = mat2cell(cell2mat([T1(:),T2(:)]),ones(length(T1(:)),1),2);
You can use meshgrid to help create unique permutations of pairs of values in a by unrolling both matrix outputs of meshgrid such that they fit into a N x 2 matrix. Once you do this, you can determine the final result using mat2cell to create your 2D cell array. In other words:
a = {1;2;3};
[x,y] = meshgrid([a{:}], [a{:}]);
b = mat2cell([x(:) y(:)], ones(numel(a)*numel(a),1), 2);
b will contain your 2D cell array. To see what's going on at each step, this is what the output of the second line looks like. x and y are actually 2D matrices, but I'm going to unroll them and display what they both are in a matrix where I've concatenated both together:
>> disp([x(:) y(:)])
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Concatenating both vectors together into a 2D matrix is important for the next line of code. This is a vital step in order to achieve what you want. After the second line of code, the goal will be to make each element of this concatenated matrix into an individual cell in a cell array, which is what mat2cell is doing in the end. By running this last line of code, then displaying the contents of b, this is what we get:
>> format compact;
>> celldisp(b)
b{1} =
1 1
b{2} =
1 2
b{3} =
1 3
b{4} =
2 1
b{5} =
2 2
b{6} =
2 3
b{7} =
3 1
b{8} =
3 2
b{9} =
3 3
b will be a 9 element cell array and within each cell is another cell array that is 1 x 2 which stores one row of the concatenated matrix as individual cells.
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)
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 have multiple matrices of the same size and want to compare them.
As a result I need a matrix which gives me the biggest of the 3 for every value.
I will clarify what i mean with an example:
I have 3 matrices with data of 3 persons.
I would like to compare these 3 and get a matrix as result.
In that matrix every cell/value should be the name of the matrix who had the highest value for that cell. So if in the 3 matrices the first value (1 colum, 1 row) is accordingly 2, 5, 8 the first value of the result matrix should be 3 (or the name of the 3 matrix).
If the three matrices are A, B, C, do this:
[~, M] = max(cat(3,A,B,C),[],3);
It creates a 3D "matrix" and maximizes across the third dimension.
Concatenate them on the 3rd dimension, and the use the SECOND output from max to get exactly what you want
A = rand(3,3);
B = rand(3,3);
C = rand(3,3);
D = cat(3, A, B, C)
[~, Solution] = max(D, [], 3)
e.g.:
D =
ans(:,:,1) =
0.70101 0.31706 0.83874
0.89421 0.33783 0.55681
0.68520 0.11697 0.45631
ans(:,:,2) =
0.268715 0.213200 0.124450
0.869847 0.999649 0.153353
0.345447 0.023523 0.338099
ans(:,:,3) =
0.216665 0.297900 0.604734
0.103340 0.767206 0.660668
0.127052 0.430861 0.021584
Solution =
1 1 1
1 2 3
1 3 1
edit
As I did not know about the second argument of the max-function, here is what you should NOT use:
old
Well, quick&dirty:
x=[2 5 8];
w=max(x)
[~,loc] = ismember(w,x)