I would like to know how to count rows in an matrix in such a way that gives an output for each colum. for example:
X=[1 1 1;
5 5 5]
I would like to find a command that when I input the matrix X the answers is [2 2 2], so that it counts the number of rows per column.
I have already found nunel(X) but the answer is a scalar numel(X)=6, whereas I need per column.
size(X,1) will give you the number of rows in the matrix (a scalar). a matrix has only one number of rows, i.e. each column has the same number of rows.
however if you still want the number of rows per each column you can use:
X = [1 1 1;
5 5 5];
nrows = size(X,1);
ncols = size(X,2);
nrowsPerCol = repmat(nrows, [1 ncols]) % [2 2 2]
Each matrix object in MATLAB has height and width property.
In other words: each column has the same number of rows.
To get this value, use MATLAB's size function:
[numOfRows, numOfCols] = size(X);
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
Having matrix A (n*2) as the source and B as a vector containing a subset of elements A, I'd like to find the row index of items.
A=[1 2;1 3; 4 5];
B=[1 5];
F=arrayfun(#(x)(find(B(x)==A)),1:numel(B),'UniformOutput',false)
gives the following outputs in a cell according to this help page
[2x1 double] [6]
indicating the indices of all occurrence in column-wise. But I'd like to have the indices of rows. i.e. I'd like to know that element 1 happens in row 1 and row 2 and element 5 happens just in row 3. If the indices were row-wise I could use ceil(F{x}/2) to have the desired output. Now with the variable number of rows, what's your suggested solution? As it may happens that there's no complete inclusion tag 'rows' in ismember function does not work. Besides, I'd like to know all indices of specified elements.
Thanks in advance for any help.
Approach 1
To convert F from its current linear-index form into row indices, use mod:
rows = cellfun(#(x) mod(x-1,size(A,1))+1, F, 'UniformOutput', false);
You can combine this with your code into a single line. Note also that you can directly use B as an input to arrayfun, and you avoid one stage of indexing:
rows = arrayfun(#(x) mod(find(x==A)-1,size(A,1))+1, B(:), 'UniformOutput', false);
How this works:
F as given by your code is a linear index in column-major form. This means the index runs down the first column of B, the begins at the top of the second column and runs down again, etc. So the row number can be obtained with just a modulo (mod) operation.
Approach 2
Using bsxfun and accumarray:
t = any(bsxfun(#eq, B(:), reshape(A, 1, size(A,1), size(A,2))), 3); %// occurrence pattern
[ii, jj] = find(t); %// ii indicates an element of B, and jj is row of A where it occurs
rows = accumarray(ii, jj, [], #(x) {x}); %// group results according to ii
How this works:
Assuming A and B as in your example, t is the 2x3 matrix
t =
1 1 0
0 0 1
The m-th row of t contains 1 at column n if the m-th element of B occurs at the n-th row of B. These values are converted into row and column form with find:
ii =
1
1
2
jj =
1
2
3
This means the first element of B ocurrs at rows 1 and 2 of A; and the second occurs at row 3 of B.
Lastly, the values of jj are grouped (with accumarray) according to their corresponding value of ii to generate the desired result.
One approach with bsxfun & accumarray -
%// Create a match of B's in A's with each column of matches representing the
%// rows in A where there is at least one match for each element in B
matches = squeeze(any(bsxfun(#eq,A,permute(B(:),[3 2 1])),2))
%// Get the indices values and the corresponding IDs of B
[indices,B_id] = find(matches)
%// Or directly for performance:
%// [indices,B_id] = find(any(bsxfun(#eq,A,permute(B(:),[3 2 1])),2))
%// Accumulate the indices values using B_id as subscripts
out = accumarray(B_id(:),indices(:),[],#(x) {x})
Sample run -
>> A
A =
1 2
1 3
4 5
>> B
B =
1 5
>> celldisp(out) %// To display the output, out
out{1} =
1
2
out{2} =
3
With arrayfun,ismember and find
[r,c] = arrayfun(#(x) find(ismember(A,x)) , B, 'uni',0);
Where r gives your desired results, you could also use the c variable to get the column of each number in B
Results for the sample input:
>> celldisp(r)
r{1} =
1
2
r{2} =
3
>> celldisp(c)
c{1} =
1
1
c{2} =
2
Suppose I have matrix like this:
a = [ 2 5 4 7; 1 2 5 8; 2 3 4 5; 4 3 1 5]
what is the function to return the maximum and its column and row's index ?
For example, in my case that function should return maximum is 8, and column index is 4, and row index is 2
You can do it by using max twice:
[m irows]=max(a)
[mm icol]=max(m)
irow=irows(icol)
a(irow,icol)
Another solution is to unroll a to be a vector with a(:), use max, which will give you a single index that you then need to convert to row and column. you can easily figure out how it works by printing b(:) with b=[1,3;2,4] for instance.
Following on from #yoh.lej, you can use ind2sub to convert the output from find into co-ordinates:
m = magic(5);
[y, i] = max(m(:));
[r, c] = ind2sub(size(m), i)
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 tried the code below to find matches, but I have gotten the error that matrix dimensions must agree.
mat = [ 8 5; 4 3; 3 5]
mat2 = [2 3; 2 3; 4 3; 4 3]
for i=1:size(mat,1)
idx= find(mat2(:,1) == mat(i,1)& mat2(:,2) == mat(i,2));
end
idx= find(mat2(:,1) == mat(:,1))
I need to find rows that match between two unequal matrices using two columns of data (x&y coordinates for example) in order to append a column of data to the correct rows in the larger matrix. I basically need to find the rows in the larger matrix where values match those values in the smaller matrix so that the column of data from the smaller matrix can be appended to the correct row.
I think you want:
>> [c,imat,imat2] = intersect( mat, mat2, 'rows')
c =
4 3
imat =
2
imat2 =
4
c are the rows that match.
imat are the indices of the matching rows in mat
imat2 are the indices of the matching rows in mat2