I am trying to get the sort of row and column according to the value of matrix.
for example, if the matrix is
A = [3 4 7; 9 8 6; 2 1 5]
it should output
2 1
2 2
1 3
2 3
3 3
1 2
1 1
3 1
3 2
I think that should be simple, but I do not have an idea about how to handle that.
Yes it is indeed very simple.
%sort the vector instead of matrix to get linear indices
[~,ind]=sort(A(:),'descend')
%convert the linear indices to [row,col] subscripts
[I,J]=ind2sub(size(A),ind)
%display desired answer
[I J]
To delete rows which have same value in both the columns:
A(A(:,1)==A(:,2),:)=[]
Related
I have some data in 10 matrices. Each matrix has a different number of rows, but the same number of columns.
I want to combine all 10 matrices to one matrix row-wise, interleaved, meaning the rows in that matrix will look like:
row 1 from matrix 0
...
row 1 from matrix 9
row 2 from matrix 0
...
row 2 from matrix 9
...
Example (with 3 matrices):
Matrix 1: [1 2 3 ; 4 5 6; 7 8 9]
Matrix 2: [3 2 1 ; 6 5 4]
Matrix 3: [1 1 1 ; 2 2 2 ; 3 3 3]
Combined matrix will be: [1 2 3 ; 3 2 1 ; 1 1 1 ; 4 5 6 ; 6 5 4 ; 2 2 2 ; 7 8 9 ; 3 3 3]
You can download the function interleave2 here https://au.mathworks.com/matlabcentral/fileexchange/45757-interleave-vectors-or-matrices
z = interleave2(a,b,c,'row')
you can see the way the function works in the source code of course
Here's a general solution that allows you to place however many matrices you want (with matching number of columns) into the starting cell array Result:
Result = {Matrix1, Matrix2, Matrix3};
index = cellfun(#(m) {1:size(m, 1)}, Result);
[~, index] = sort([index{:}]);
Result = vertcat(Result{:});
Result = Result(index, :);
This will generate an index vector 1:m for each matrix, where m is its number of rows. By concatenating these indices and sorting them, we can get a new index that can be used to sort the rows of the vertically-concatenated set of matrices so that they are interleaved.
I have a Matrix of size M by N in which each row has some zero entries. I want to create M row vectors such that each of the vector contains the non zero elements of each row. For example if I have the following Matrix
A=[0 0 0 5;0 0 4 6;0 1 2 3;9 10 2 3]
I want four different row vectors of the following form
[5]
[4 6]
[1 2 3]
[9 10 2 3]
This can be done with accumarray using an anonymous function as fourth input argument. To make sure that the results are in the same order as in A, the grouping values used as first input should be sorted. This requires using (a linearized version of) A transposed as second input.
ind = repmat((1:size(A,2)).',1,size(A,2)).';
B = A.';
result = accumarray(ind(:), B(:), [], #(x){nonzeros(x).'});
With A = [0 0 0 5; 0 0 4 6; 0 1 2 3; 9 10 2 3]; this gives
result{1} =
5
result{2} =
4 6
result{3} =
1 2 3
result{4} =
9 10 2 3
Since Matlab doesn't support non-rectangular double arrays, you'll want to settle on a cell array. One quick way to get the desired output is to combine arrayfun with logical indexing:
nonZeroVectors = arrayfun(#(k) A(k,A(k,:)~=0),1:size(A,1),'UniformOutput',false);
I used the ('UniformOutput',false) name-value pair for the reasons indicated in the documentation (I'll note that the pair ('uni',0) also works, but I prefer verbosity). This input produces a cell array with the entries
>> nonZerosVectors{:}
ans =
5
ans =
4 6
ans =
1 2 3
ans =
9 10 2 3
I have a 3x3 matrix, populated with NaN and values of a variable:
NaN 7 NaN
5 NaN 0
NaN NaN 4
matrix = [NaN 7 NaN; 5 NaN 0; NaN NaN 4]
I would like to get the row and column numbers of non-NaN cells and put them in a matrix together with the value of the variable. That is, I would like to obtain the following matrix:
row col value
1 2 7
2 1 5
2 3 0
3 3 4
want = [1 2 7; 2 1 5; 2 3 0; 3 3 4]
Any help would be highly appreciated.
This can be done without loops:
[jj, ii, kk] = find((~isnan(matrix).*(reshape(1:numel(matrix), size(matrix)))).');
result = [ii jj matrix(kk)];
The trick is to multiply ~isnan(matrix) by a matrix of indices so that the third output of find gives the linear index of non-NaN entries. The transpose is needed to have the same order as in the question.
The following should work!
[p,q]=find(~isnan(matrix)) % Loops through matrix to find indices
want = zeros(numel(p),3) % three columns you need with same number of rows as p
for i=1:numel(p)
want[i,:] = [p(i) q(i) matrix(p(i), matrix(i))]
end
Should give you the correct result which is:
2 1 5
1 2 7
2 3 0
3 3 4
If you don't mind the ordering of the rows, you can use a simplified version of Luis Mendo's answer:
[row, col] = find(~isnan(matrix));
result = [row(:), col(:), matrix(~isnan(matrix))];
Which will result in:
2 1 5
1 2 7
2 3 0
3 3 4
In Matlab I have a big matrix containing the coordinates (x,y,z) of many points (over 200000). There is an extra column used as identification. I have written this code in order to sort all coordinate points. My final goal is to find duplicated points (rows with same x,y,z). After sorting the coordinate points I use the diff function, two consecutive rows of the matrix with the same coordinates will take value [0 0 0], and then with ismember I can find which rows of that matrix resulting from applying "diff" have the [0 0 0] row. With the indices returned from ismember I can find which points are repeated.
Back to my question...This is the code I wrote to sort properly my coordintes+id matrix. I guess It could be done better. Any suggestion?
%coordinates are always positive
a=[ 1 2 8 4; %sample matrix
1 0 5 6;
2 4 7 1;
3 2 1 0;
2 3 5 0;
3 1 2 8;
1 2 4 8];
b=a; %for checking purposes
%sorting first column
a=sortrows(a,1);
%sorting second column
for i=0:max(a(:,1));
k=find(a(:,1)==i);
if not(isempty(k))
a(k,:)=sortrows(a(k,:),2);
end
end
%Sorting third column
for i=0:max(a(:,2));
k=find(a(:,2)==i);
if not(isempty(k))
%identifying rows with same value on first column
for j=1:length(k)
[rows,~] = ismember(a(:,1:2), [ a(k(j),1),i],'rows');
a(rows,3:end)=sortrows(a(rows,3:end),1);
end
end
end
%Checking that rows remain the same
m=ismember(b,a,'rows');
if length(m)~=sum(m)
disp('Error while sorting!');
end
Why don't you just use unique?
[uniqueRows, ii, jj] = unique(a(:,1:3),'rows');
Example
a = [1 2 3 5
3 2 3 6
1 2 3 9
2 2 2 8];
gives
uniqueRows =
1 2 3
2 2 2
3 2 3
and
jj =
1
3
1
2
meaning third row equals first row.
If you need the full unique rows, including the fourth column: use ii to index a:
fullUniqueRows = a(ii,:);
which gives
fullUniqueRows =
1 2 3 9
2 2 2 8
3 2 3 6
Trying to sort a based on the fourth column? Do this -
a=[ 1 2 8 4; %sample matrix
1 0 5 6;
2 4 7 1;
3 2 1 0;
2 3 5 0;
3 2 1 8;
1 2 4 8];
[x,y] = sort(a(:,4))
sorted_a=a(y,:)
Trying to get the row indices having repeated x-y-z coordinates being represented by the first three columns? Do this -
out = sum(squeeze(all(bsxfun(#eq,a(:,1:3),permute(a(:,1:3),[3 2 1])),2)),2)>1
and use it similarly for sorted_a.
I have a 100 by 2 matrix. And I'm trying to figure out how to divide all terms in the second column by a constant.
For example, let's say I have this matrix.
[1 2;
3 4;
5 6]
I want to divide the 2nd column by 2.
[1 2/2;
3 4/2;
5 6/2]
So my final matrix will be.
[1 1;
3 2;
5 3]
Thank you.
If your matrix is M then:
M(:,2)=M(:,2)./2;
will divide all terms in the second column by a constant (2). By the way, because the value you divide with is a constant you can also write / instead of ./
If you'd like to assemble a new matrix and not overwrite the first one just write something like this:
A=[M(:,1) M(:,2)./2]
I'm not sure how natan 's equations should be read, but I'd multiply the first matrix
1 2
3 4
5 6
by the matrix
1 0
0 .5
The resulting matrix is
1 1
3 2
5 3