Matlab matrix sort

Matlab matrix sort - matlab

I 've a specific problem of sorting rows in matlab.
This is mine example entry matrix:
A =
[0 1 1;
0 1 2;
1 0 3;
1 0 4;
1 1 5;
0 1 6;]
and this is "sorting vector"
V=
1
4
6
2
3
5
How to get an output matrix like this:
B=
[0 1 1;
1 0 4;
0 1 6;
0 1 2;
1 0 3;
1 1 5]
?
First I've added vector V to matrix A (last column) but the next step I don't know how it should look. I'm stuck.
In advance, thanks for your time and help :)

To rearrange or select any desired rows:
B = A(V,:);
The same concept could be used for columns and for rearranging, selecting or repeating any desired row or column:
V2 = [3 1 3];
B2 = A(:,V2);
B2 =
1 0 1
2 0 2
3 1 3
4 1 4
5 1 5
6 0 6
Learn about colon operator(:) here:
http://www.mathworks.com/help/matlab/ref/colon.html

This might be the answer:
B = A(V(:),:);

Related

in matlab,how can i put all zeros in the end of the rows in a matrix? [duplicate]

I have a matrix in MATLAB with zeroes and I would like to get another matrix with the first N non-zero elements in each row. Let's say for example N = 3, and the matrix is
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3]
I'd like the result to be:
B = [2 6 7;
3 2 4;
1 3 4;
1 2 1]
I have a huge matrix so I would like to do it without a loop, could you please help me? Thanks a lot!

Since MATLAB stores a matrix according to column-major order, I first transpose A, bubble up the non-zeros, and pick the first N lines, and transpose back:
N = 3;
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3];
Transpose and preallocate output B
At = A';
B = zeros(size(At));
At =
0 3 0 1
0 2 1 2
2 4 0 0
0 7 3 0
6 0 4 0
7 0 8 1
9 6 6 3
Index zeros
idx = At == 0;
idx =
1 0 1 0
1 0 0 0
0 0 1 1
1 0 0 1
0 1 0 1
0 1 0 0
0 0 0 0
Bubble up the non-zeros
B(~sort(idx)) = At(~idx);
B =
2 3 1 1
6 2 3 2
7 4 4 1
9 7 8 3
0 6 6 0
0 0 0 0
0 0 0 0
Select first N rows and transpose back
B(1:N,:)'
You can do the bubbling in row-major order, but you would need to retrieve the row and column subscripts with find, and do some sorting and picking there. It becomes more tedious and less readable.

Using accumarray with no loops:
N = 3;
[ii,jj] = find(A); [ii,inds]=sort(ii); jj = jj(inds);
lininds = ii+size(A,1)*(jj-1);
C = accumarray(ii,lininds,[],#(x) {A(x(1:N)')}); %' cell array output
B = vertcat(C{:})
B =
2 6 7
3 2 4
1 3 4
1 2 1

Usually I don't go with a for loop solution, but this is fairly intuitive:
N = 3;
[ii,jj] = find(A);
B = zeros(size(A,1),N);
for iRow = 1:size(A,1),
nzcols = jj(ii==iRow);
B(iRow,:) = A(iRow,nzcols(1:N));
end
Since you are guaranteed to have more than N nonzeros per row of A, that should get the job done.

One-liner solution:
B = cell2mat(cellfun(#(c) c(1:N), arrayfun(#(k) nonzeros(A(k,:)), 1:size(A,1), 'uni', false), 'uni', false)).'
Not terribly elegant or efficient, but so much fun!

N = 3;
for ii=1:size(A,1);
B(ii,:) = A( ii,find(A(ii,:),N) );
end

Actually , you can do it like the code blow:
N=3
for n=1:size(A,1)
[a b]=find(A(n,:)>0,N);
B(n,:)=A(n,transpose(b));
end
Then I think this B matrix will be what you want.

Find position of vector in matrix [duplicate]

This question already has answers here:
General method to find submatrix in matlab matrix
(3 answers)
Closed 8 years ago.
I have the vector:
1 2 3
and the matrix:
4 1 2 3 5 5
9 8 7 6 3 1
1 4 7 8 2 3
I am trying to find a simple way of locating the vector [1 2 3] in my matrix.
A function returning either coordinates (Ie: (1,2:4)) or a matrix of 1s where there is a match a 0s where there isn't, Ie:
0 1 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
So far, the only function I've found is is 'ismember', which however only tells me if the individual components of the vector appear in the matrix. Suggestions?

Use strfind with a linearized version of the matrix, and then convert linear indices to subindices. Care should be taken to remove matches of the vector spanning different rows.
mat = [1 2 3 1 2 3 1 2;
3 0 1 2 3 5 4 4]; %// data
vec = [1 2 3]; %// data
ind = strfind(reshape(mat.',[],1).', vec);
[col row] = ind2sub(fliplr(size(mat)), ind);
keep = col<=size(mat,2)-length(vec)+1; %// remove result split across rows
row = row(keep);
col = col(keep);
Result for this example:
>> row, col
row =
1 1 2
col =
1 4 3
meaning the vector appears three times: row 1, col 1; row 1, col 4; row 2, col 3.
The result can be expressed in zero-one form as follows:
result = zeros(fliplr(size(mat)));
ind_ones = bsxfun(#plus, ind(keep).', 0:numel(vec)-1);
result(ind_ones) = 1;
result = result.';
which gives
>> result
result =
1 1 1 1 1 1 0 0
0 0 1 1 1 0 0 0

One way to get the starting location of the vector in the matrix is using colfilt:
>> A = [1 2 3 1 2 3 1 2; ...
3 0 1 2 3 5 4 4]; % matrix from Luis Mendo
>> T = [1 2 3];
>> colFun = #(x,t) all(x==repmat(t,1,size(x,2)),1);
>> B = colfilt(A,size(T),'sliding',colFun,T(:))
B =
0 1 0 0 1 0 0 0
0 0 0 1 0 0 0 0
That gives a mask of the center points, which translate to (row,col) coordinates:
>> [ii,jj]=find(B);
>> locs = bsxfun(#minus,[ii jj],floor((size(T)-1)/2))
locs =
1 1
2 3
1 4

Find the first N non-zero elements in each row of a matrix

I have a matrix in MATLAB with zeroes and I would like to get another matrix with the first N non-zero elements in each row. Let's say for example N = 3, and the matrix is
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3]
I'd like the result to be:
B = [2 6 7;
3 2 4;
1 3 4;
1 2 1]
I have a huge matrix so I would like to do it without a loop, could you please help me? Thanks a lot!

Since MATLAB stores a matrix according to column-major order, I first transpose A, bubble up the non-zeros, and pick the first N lines, and transpose back:
N = 3;
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3];
Transpose and preallocate output B
At = A';
B = zeros(size(At));
At =
0 3 0 1
0 2 1 2
2 4 0 0
0 7 3 0
6 0 4 0
7 0 8 1
9 6 6 3
Index zeros
idx = At == 0;
idx =
1 0 1 0
1 0 0 0
0 0 1 1
1 0 0 1
0 1 0 1
0 1 0 0
0 0 0 0
Bubble up the non-zeros
B(~sort(idx)) = At(~idx);
B =
2 3 1 1
6 2 3 2
7 4 4 1
9 7 8 3
0 6 6 0
0 0 0 0
0 0 0 0
Select first N rows and transpose back
B(1:N,:)'
You can do the bubbling in row-major order, but you would need to retrieve the row and column subscripts with find, and do some sorting and picking there. It becomes more tedious and less readable.

Using accumarray with no loops:
N = 3;
[ii,jj] = find(A); [ii,inds]=sort(ii); jj = jj(inds);
lininds = ii+size(A,1)*(jj-1);
C = accumarray(ii,lininds,[],#(x) {A(x(1:N)')}); %' cell array output
B = vertcat(C{:})
B =
2 6 7
3 2 4
1 3 4
1 2 1

Usually I don't go with a for loop solution, but this is fairly intuitive:
N = 3;
[ii,jj] = find(A);
B = zeros(size(A,1),N);
for iRow = 1:size(A,1),
nzcols = jj(ii==iRow);
B(iRow,:) = A(iRow,nzcols(1:N));
end
Since you are guaranteed to have more than N nonzeros per row of A, that should get the job done.

One-liner solution:
B = cell2mat(cellfun(#(c) c(1:N), arrayfun(#(k) nonzeros(A(k,:)), 1:size(A,1), 'uni', false), 'uni', false)).'
Not terribly elegant or efficient, but so much fun!

N = 3;
for ii=1:size(A,1);
B(ii,:) = A( ii,find(A(ii,:),N) );
end

Actually , you can do it like the code blow:
N=3
for n=1:size(A,1)
[a b]=find(A(n,:)>0,N);
B(n,:)=A(n,transpose(b));
end
Then I think this B matrix will be what you want.

If same value in two matrices, keep the value in the first matrix

how can I achieve the matrix C with following conditions:
A(i,j) is any B(k) , then A(i,j) = B(k)
else A(i,j) = 0
example:
A = [1 0 3 6 3 4; 2 0 3 1 8 2];
B = [1;2;3];
C = [1 0 3 0 3 0; 2 0 3 1 0 2]
Thank you!

A.*ismember(A,B)
Well not much to explain, ismember fits exactly your condition. Thus ismember(A,B) is 1 for all values you want to copy.

If your concern is to avoid products, you may try:
A(~ismember(A,B))=0;

depending on what you're looking for, from the matlab documentation
[LIA,LOCB] = ismember(A,B) also returns an array LOCB containing the
highest absolute index in B for each element in A which is a member of
B and 0 if there is no such index.
a = [9 9 8 8 7 7 7 6 6 6 5 5 4 4 2 1 1 1]
b = [1 1 1 3 3 3 3 3 4 4 4 4 4 9 9 9]
[lia1,locb1] = ismember(a,b)
% returns
lia1 = [1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1]
locb1 = [16 16 0 0 0 0 0 0 0 0 0 0 13 13 0 3 3 3]

Another possibility, without ismember:
A .* any(bsxfun(#eq, A, permute(B,[3 2 1])),3)

Thank you guys!
I followed your advise and extended the code for my task. I guess my question was unclear formulated:
Following your advice, the Matrix C will give only 1 or 0.
I need the actual value of A in it.
In my following solution the code just rewrites A in the end. I gues there are many smarter solutions, but it works.
A = [1 0 3 6 3 4; 2 0 3 1 8 2];
B = [1;2;3];
[N M] = size (A);
C = ismember(A,B);
for i = 1:N
for j = 1:M
if C(i,j) == 0;
A(i,j) = 0;
end
end
end

How to find one value from a matrix

0I have on matrix-
A=[1 2 2 3 5 5;
1 5 5 8 8 7;
2 9 9 3 3 5];
From matrix i need to count now many nonzero elements ,how any 1,how many 2 and how many 3 in each row of given matrix"A".For these i have written one code like:
[Ar Ac]=size(A);
for j=1:Ar
for k=1:Ac
count(:,j)=nnz(A(j,:));
d(:,j)=sum(A(j,:)== 1);
e(:,j)=sum(A(j,:)==2);
f(:,j)=sum(A(j,:)==3);
end
end
but i need to write these using on loop i.e. here i manually use sum(A(j,:)== 1),sum(A(j,:)== 2) and sum(A(j,:)== 3) but is there any option where i can only write sum(A(j,:)== 1:3) and store all the values in the different row i.e, the result will be like-
b=[1 2 1;
1 0 0;
0 1 2];
Matlab experts need your valuable suggestions

Sounds like you're looking for a histogram count:
U = unique(A);
counts = histc(A', U)';
b = counts(:, ismember(U, [1 2 3]));
Example
%// Input matrix and vector of values to count
A = [1 2 2 3 5 5; 1 5 5 8 8 7; 2 9 9 3 3 5];
vals = [1 2 3];
%// Count values
U = unique(A);
counts = histc(A', U)';
b = counts(:, ismember(U, vals));
The result is:
b =
1 2 1
1 0 0
0 1 2

Generalizing the sought values, as required by asker:
values = [ 1 2 3 ]; % or whichever values are sought
B = squeeze(sum(bsxfun(#(x,y) sum(x==y,2), A, shiftdim(values,-1)),2));

Here is a simple and general way. Just change n to however high you want to count. n=max(A(:)) is probably a good general value.
result = [];
n = 3;
for col= 1:n
result = [result, sum(A==col, 2)];
end
result
e.g. for n = 10
result =
1 2 1 0 2 0 0 0 0 0
1 0 0 0 2 0 1 2 0 0
0 1 2 0 1 0 0 0 2 0

Why not use this?
B=[];
for x=1:size(A,1)
B=[B;sum(A(x,:)==1),sum(A(x,:)==2),sum(A(x,:)==3)];
end

I'd do this way:
B = [arrayfun(#(i) find(A(i,:) == 1) , 1:3 , 'UniformOutput', false)',arrayfun(#(i) find(A(i,:) == 2) , 1:3 , 'UniformOutput', false)',arrayfun(#(i) find(A(i,:) == 3) , 1:3 , 'UniformOutput', false)'];
res = cellfun(#numel, B);

Here is a compact one:
sum(bsxfun(#eq, permute(A, [1 3 2]), 1:3),3)
You can replace 1:3 with any array.
you can make an anonymous function for it
rowcnt = #(M, R) sum(bsxfun(#eq, permute(M, [1 3 2]), R),3);
then running it on your data returns
>> rowcnt(A,1:3)
ans =
1 2 1
1 0 0
0 1 2
and for more generalized case
>> rowcnt(A,[1 2 5 8])
ans =
1 2 2 0
1 0 2 2
0 1 1 0