I've got a matrix (n x m). And I'd like to know, for each row, the indexes of the coloums that contain the first two maximum values:
2 3 4 2
2 4 7 1
1 1 2 4
5 5 9 6
1 4 2 1
9 8 1 2
The answer should be:
2 3
2 3
3 4
3 4
2 3
1 2
How can I obtain it with matlab commands? I'd like not to use for loops. I tried with:
[x,y]=max(matrix')
y=y';
y gives me the colum indexes for the maximum elements. Now I'd set to zero these elements and repeat the instructions but I have no idea how to do. I treid:
matrix(:,y)=0;
but it doesn't work.
if A is your matrix, then sort and pick the top two indices,
[a ix]=sort(A,2)
ans= ix(:,end-1:end)
Related
i have data A=( 3,5,3,1,4 ) in a column and
B=[
4 6 9 1 3
2 7 2 5 7
7 3 1 8 2
4 1 6 9 1
2 5 8 3 6 ]
And i want: as in A first element is 3 and for this i want to get first element of column 3 row 1 from B which is 9. The second element of A is 5 and for this i want to get the the 2nd element of column 5 and row 2 from B which is 7 ,and do the process for all other elements . how to do this in matlab? the required elements are bold and underlined. The desired output is [9,7,1,4,3]
Read about linear indexes.
sub2ind will convert from [row col] to index.
Cols=[ 3,5,3,1,4 ];
Rows=1:length(Cols);
B=[
4 6 9 1 3
2 7 2 5 7
7 3 1 8 2
4 1 6 9 1
2 5 8 3 6 ];
Indexes=sub2ind(size(B),Rows,Cols);
Vals=B(Indexes)
if I've read well you want an elements replacement. It's quite simple
A(1)=B(1,3)
A(2)=B(2,5)
So after you have declared your two vectors you can handle them replacing singular components. In general when you have a 1-D vectors you can access to it's component position only declaring the position itself inside the parentheses, as I did for A.
Whe you have to face situation like in B, If you remember linear algebra and matrix in general B(a,b) means the element of matrix B placed in the line a and column b so you have to specify row and column to access that element.
Suppose that I have a matrix , let's call it A, as follows:
1 2 3 4 5 1 2 3 4 5
0 2 4 6 8 1 3 5 7 9
And I want to reshape it into a matrix like this:
1 2 3 4 5
0 2 4 6 8
1 2 3 4 5
1 3 5 7 9
So, basically, what I want to be done is that MATLAB first reads a block of size (2,5) and then splits the remaining matrix to the next row and then repeats this so on so forth until we get something like in my example.
I tried to do this using MATLAB's reshape command in several ways but I failed. Any help is appreciated. In case that it matters, my original data is larger. It's (2,1080). Thanks.
I don't believe you can do this in a single command, but perhaps someone will correct me. If speed isn't a huge concern a for loop should work fine.
Alternatively you can get your results by reshaping each row of A and then placing the results into every other row of a new matrix. This will also work with your larger data.
A = [1 2 3 4 5 1 2 3 4 5
0 2 4 6 8 1 3 5 7 9];
An = zeros(numel(A)/5, 5); % Set up new, empty matrix
An(1:2:end,:) = reshape(A(1,:), 5, [])'; % Write the first row of A to every other row of An
An(2:2:end,:) = reshape(A(2,:), 5, [])' % Write second row of A to remaining rows
An =
1 2 3 4 5
0 2 4 6 8
1 2 3 4 5
1 3 5 7 9
You may need to read more about indexing in the Matlab's documentation.
For your example, it is easy to do the following
A=[1 2 3 4 5 1 2 3 4 5; 0 2 4 6 8 1 3 5 7 9]
a1=A(:,1:5); % extract all rows, and columns from 1 to 5
a2=A(:,6:end); % extract all rows, and columns from 6 to end
B=[a1;a2] % construct a new matrix.
It is not difficult to build some sort of loops to extract the rest.
Here's a way you can do it in one line using the reshape and permute commands:
B = reshape(permute(reshape(A,2,5,[]), [1,3,2]), [], 5);
The reshape(A,2,5,[]) command reshapes your A matrix into a three-dimensional tensor of dimension 2 x 5 x nblocks, where nblocks is the number of blocks in A in the horizontal direction. The permute command then swaps the 2nd and 3rd dimensions of this 3D tensor, so that it becomes a 2 x nblocks x 5 tensor. The final reshape command then transforms the 3D tensor into a matrix of dimension (2*nblocks) x 5.
Looking at the results at each stage may give you a better idea of what's happening.
I need some help please. I have an array, as shown below, 6 rows and 5 columns, none of the elements in any one row repeats. The elements are all single digit numbers.
I want to find out, per row, when a number, let's say 1 appears, I want to keep of how often the other numbers of the row appear. For example, 1 shows up 3 times in rows one, three and five. When 1 shows up, 2 shows up one time, 3 shows up two times, 4 shows up two times, 5 shows up one time, 6 shows up two times, 7 shows up one time, 8 shows up three times, and 9 shows up zero times. I want to keep a vector of this information that will look like, V = [3,1,2,2,1,2,1,3,0], by starting with a vector like N = [1,2,3,4,5,6,7,8,9]
ARRAY =
1 5 8 2 6
2 3 4 6 7
3 1 8 7 4
6 5 7 9 4
1 4 3 8 6
5 7 8 9 6
The code I have below does not give the feedback I am looking for, can someone help please? Thanks
for i=1:length(ARRAY)
for j=1:length(N)
ARRAY(i,:)==j
V(j) = sum(j)
end
end
Using indices that is in A creae a zero and one 6 * 9 matrix that [i,j] th element of it is 1 if i th row of A contains j.
Then multiply the zero and one matrix with its transpose to get desirable result:
A =[...
1 5 8 2 6
2 3 4 6 7
3 1 8 7 4
6 5 7 9 4
1 4 3 8 6
5 7 8 9 6]
% create a matrix with the size of A that each row contains the row number
rowidx = repmat((1 : size(A,1)).' , 1 , size(A , 2))
% z_o a zero and one 6 * 9 matrix that [i,j] th element of it is 1 if i th row of A contains j
z_o = full(sparse(rowidx , A, 1))
% matrix multiplication with its transpose to create desirable result. each column relates to number N
out = z_o.' * z_o
Result: each column relates to N
3 1 2 2 1 2 1 3 0
1 2 1 1 1 2 1 1 0
2 1 3 3 0 2 2 2 0
2 1 3 4 1 3 3 2 1
1 1 0 1 3 3 2 2 2
2 2 2 3 3 5 3 3 2
1 1 2 3 2 3 4 2 2
3 1 2 2 2 3 2 4 1
0 0 0 1 2 2 2 1 2
I don't understand how you are approaching the problem with your sample code but here is something that should work. This uses find, any and accumarray and in each iteration for the loop it will return a V corresponding to the ith element in N
for i=1:length(N)
rowIdx = find(any(A == N(i),2)); % Find all the rows contain N(j)
A_red = A(rowIdx,:); % Get only those rows
V = [accumarray(A_red(:),1)]'; % Count occurrences of the 9 numbers
V(end+1:9) = 0; % If some numbers don't exist place zeros on their counts
end
I have a 4x8 matrix which I want to select two different columns of it then derive dot product of them and then divide to norm values of that selected columns, and then repeat this for all possible two different columns and save the vectors in a new matrix. can anyone provide me a matlab code for this purpose?
The code which I supposed to give me the output is:
A=[1 2 3 4 5 6 7 8;1 2 3 4 5 6 7 8;1 2 3 4 5 6 7 8;1 2 3 4 5 6 7 8;];
for i=1:8
for j=1:7
B(:,i)=(A(:,i).*A(:,j+1))/(norm(A(:,i))*norm(A(:,j+1)));
end
end
I would approach this a different way. First, create two matrices where the corresponding columns of each one correspond to a unique pair of columns from your matrix.
Easiest way I can think of is to create all possible combinations of pairs, and eliminate the duplicates. You can do this by creating a meshgrid of values where the outputs X and Y give you a pairing of each pair of vectors and only selecting out the lower triangular part of each matrix offsetting by 1 to get the main diagonal just one below the diagonal.... so do this:
num_columns = size(A,2);
[X,Y] = meshgrid(1:num_columns);
X = X(tril(ones(num_columns),-1)==1); Y = Y(tril(ones(num_columns),-1)==1);
In your case, here's what the grid of coordinates looks like:
>> [X,Y] = meshgrid(1:num_columns)
X =
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
Y =
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4
5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8
As you can see, if we select out the lower triangular part of each matrix excluding the diagonal, you will get all combinations of pairs that are unique, which is what I did in the last parts of the code. Selecting the lower-part is important because by doing this, MATLAB selects out values column-wise, and traversing the columns of the lower-triangular part of each matrix gives you the exact orderings of each pair of columns in the right order (i.e. 1-2, 1-3, ..., 1-7, 2-3, 2-4, ..., etc.)
The point of all of this is that can then use X and Y to create two new matrices that contain the columns located at each pair of X and Y, then use dot to apply the dot product to each matrix column-wise. We also need to divide the dot product by the multiplication of the magnitudes of the two vectors respectively. You can't use MATLAB's built-in function norm for this because it will compute the matrix norm for matrices. As such, you have to sum over all of the rows for each column respectively for each of the two matrices then multiply both of the results element-wise then take the square root - this is the last step of the process:
matrix1 = A(:,X);
matrix2 = A(:,Y);
B = dot(matrix1, matrix2, 1) ./ sqrt(sum(matrix1.^2,1).*sum(matrix2.^2,1));
I get this for B:
>> B
B =
Columns 1 through 11
1 1 1 1 1 1 1 1 1 1 1
Columns 12 through 22
1 1 1 1 1 1 1 1 1 1 1
Columns 23 through 28
1 1 1 1 1 1
Well.. this isn't useful at all. Why is that? What you are actually doing is finding the cosine angle between two vectors, and since each vector is a scalar multiple of another, the angle that separates each vector is in fact 0, and the cosine of 0 is 1.
You should try this with different values of A so you can see for yourself that it works.
To make this code compatible for copying and pasting, here it is:
%// Define A here:
A = repmat(1:8, 4, 1);
%// Code to produce dot products here
num_columns = size(A,2);
[X,Y] = meshgrid(1:num_columns);
X = X(tril(ones(num_columns),-1)==1); Y = Y(tril(ones(num_columns),-1)==1);
matrix1 = A(:,X);
matrix2 = A(:,Y);
B = dot(matrix1, matrix2, 1) ./ sqrt(sum(matrix1.^2,1).*sum(matrix2.^2,1));
Minor Note
If you have a lot of columns in A, this may be very memory intensive. You can get your original code to work with loops, but you need to change what you're doing at each column.
You can do something like this:
num_columns = nchoosek(size(A,2),2);
B = zeros(1, num_columns);
counter = 1;
for ii = 1 : size(A,2)
for jj = ii+1 : size(A,2)
B(counter) = dot(A(:,ii), A(:,jj), 1) / (norm(A(:,ii))*norm(A(:,jj)));
counter = counter + 1;
end
end
Note that we can use norm because we're specifying vectors for each of the inputs into the function. We first preallocate a matrix B that will contain the dot products of all possible combinations. Then, we go through each pair of combinations - take note that the inner for loop starts from the outer most for loop index added with 1 so you don't look at any duplicates. We take the dot product of the corresponding columns referenced by positions ii and jj and store the results in B. I need an external counter so we can properly access the right slot to place our result in for each pair of columns.
Is there an efficient way to produce square multi-diagonal matricies such as these:
[[1,2,3],
[2,1,2],
[3,2,1]]
[[1,2,3,4,5],
[2,1,2,3,4],
[3,2,1,2,3],
[4,3,2,1,2],
[5,4,3,2,1]]
My efforts so far have produced the following:
t=10
sum=zeros(t,t)
for i=1:t
sum+=diag(ones(1,i)*(t-i)+1,t-i);
end
sum
sum+sum'-diag(ones(1,10),0)
The command toeplitz does exactly what you want:
toeplitz([1,2,3,4,5,6])
ans =
1 2 3 4 5 6
2 1 2 3 4 5
3 2 1 2 3 4
4 3 2 1 2 3
5 4 3 2 1 2
6 5 4 3 2 1
What you are looking for is called a symmetric (Hermitian) toeplitz matrix.
I'm not familiar with matlab, but I found this documentation on mathworks: