I want to add one value at the beginning of a matrix for example, my matrix is
0,0,0,0,0,0
0,1,1,1,0,0
1,0,0,1,0,0
1,1,1,0,0,0
then I want to add '1' or '0' at the first column therefore it will become like this
1,0,0,0,0,0,0
1,0,1,1,1,0,0
1,1,0,0,1,0,0
1,1,1,1,0,0,0
0,0,0,0,0,0,0
0,0,1,1,1,0,0
0,1,0,0,1,0,0
0,1,1,1,0,0,0
how can I do that in MATLAB?
You can concatenate matrices without creating another one from scratch like this:
% your matrix
A = [ 0,0,0,0,0,0;
0,1,1,1,0,0;
1,0,0,1,0,0;
1,1,1,0,0,0 ];
A_with_zeros = [zeros(size(A,1),1) A]
A_with_ones = [ones(size(A,1),1) A]
% Output:
% A_with_zeros = [ 0 0 0 0 0 0 0
% 0 0 1 1 1 0 0
% 0 1 0 0 1 0 0
% 0 1 1 1 0 0 0 ]
%
% A_with_ones = [ 1 0 0 0 0 0 0 0
% 1 0 0 1 1 1 0 0
% 1 0 1 0 0 1 0 0
% 1 0 1 1 1 0 0 0 ]
Documentation about concatenating matrices:
horzcat
vertcat
These two documentation pages describe in details the existing methods that let you concatenate arrays horizontally (what I did in the example above) and vertically (if you wanted to add lines to your matrix).
Use the array concatenation syntax - [a b] to glue matrices together horizontally, [a; b] to glue them together vertically.
Like so:
>> a = ones(3,1)
a =
1
1
1
>> b = magic(3)
b =
8 1 6
3 5 7
4 9 2
>> c = [a b]
c =
1 8 1 6
1 3 5 7
1 4 9 2
Vertically:
>> d = ones(1,3)
d =
1 1 1
>> e = [d; b]
e =
1 1 1
8 1 6
3 5 7
4 9 2
If you mismatch the dimensions, MATLAB will give you an error. Don't do this:
>> f = ones(1,10)
f =
1 1 1 1 1 1 1 1 1 1
>> g = magic(3)
g =
8 1 6
3 5 7
4 9 2
>> [f; g]
??? Error using ==> vertcat
CAT arguments dimensions are not consistent.
Create a new matrix with an extra column, copy the old matrix in, then put the data for the new column in at (:, 1).
output = zeros(size(input, 1), size(input, 2) + 1);
output(:, 2:end) = input;
output(:, 1) = new_column;
or if you mean that you want to get two matricies, one with a column of 1s and one with a column of 0s:
output0 = zeros(size(input, 1), size(input, 2) + 1);
output0(:, 2:end) = input;
output1 = ones(size(input, 1), size(input, 2) + 1);
output1(:, 2:end) = input;
Related
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.
I'm attempting to find a critical point in a matrix. The value at index (i,j) should be greater than or equal to all elements in its row, and less than or equal to all elements in its column.
Here is what I have (it's off but I'm close):
function C = critical(A)
[nrow ncol] = size(A);
C = [];
for i = 1:nrow
for j = 1:ncol
if (A(i,j) >= A(i,1:end)) && (A(i,j) <= A(1:end,j))
C = [C ; A(i,j)]
end
end
end
You can use logical indexing.
minI = min(A,[],1);
maxI = max(A,[],2);
[row,col] = find(((A.'==maxI.').' & A==minI) ==1)
Details
Remember that Matlab is column major. We therefore transpose A and maxI.
A = [
3 4 1 1 2
2 4 2 1 4
4 3 2 1 2
3 3 1 1 1
2 3 0 2 1];
A.'==maxI.'
ans =
0 0 1 1 0
1 1 0 1 1
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
Then do the minimum
A==minI
ans =
0 0 0 1 0
1 0 0 1 0
0 1 0 1 0
0 1 0 1 1
1 1 1 0 1
And then multiply the two
((A.'==maxI.').' & A==minI)
ans =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
0 1 0 0 0
Then find the rows and cols
[row,col] = find(((A.'==maxI.').' & A==minI) ==1)
row =
4
5
col =
2
2
Try this vectorised solution using bsxfun
function [ r,c,criP ] = critical( A )
%// finding the min and max values of each col & row resptly
minI = min(A,[],1);
maxI = max(A,[],2);
%// matching all the values of min & max for each col and row resptly
%// getting the indexes of the elements satisfying both the conditions
idx = find(bsxfun(#eq,A,maxI) & bsxfun(#eq,A,minI));
%// getting the corresponding values from the indexes
criP = A(idx);
%// Also getting corresponding row and col sub
[r,c] = ind2sub(size(A),idx);
end
Sample Run:
r,c should be a vector of equal length which represents the row and column subs of each Critical point. While val is a vector of same length giving the value of the critical point itself
>> A
A =
3 4 1 1 2
2 4 2 1 4
4 3 2 1 2
3 3 1 1 1
2 3 0 2 1
>> [r,c,val] = critical(A)
r =
4
5
c =
2
2
val =
3
3
I think there is a simpler way with intersect:
>> [~, row, col] = intersect(max(A,[],2), min(A));
row =
4
col =
2
UPDATE:
With intersect, in case you have multiple critical points, it will only give you the first one. To have all the indicies, there is also another simple way:
>> B
B =
3 4 1 4 2 5
2 5 2 4 4 4
4 4 2 4 2 4
3 4 1 4 1 4
2 5 4 4 4 5
>> row = find(ismember(max(B,[],2),min(B)))
row =
3
4
>> col = find(ismember(min(B),max(B,[],2)))
col =
2 4 6
Note that the set of critical points now should be the combination of row and col, means you have total 6 critical points in this example: (3,2),(4,2),(3,4),(4,4),(3,6),(4,6).
Here you can find how to export such combination.
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
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.
Suppose I have an (m x n) matrix Q, and a row vector r, e.g.
Q = [ 1 2 3 ; 4 2 3 ; 5 6 7 ; 1 2 3 ; 1 2 3 ; 1 2 5 ];
r = [ 1 2 3 ];
What is the easiest way to obtain a logical vector (of length m) that indicates which of the rows in Q are identical (for all elements) to the specified row r?
In the sample case above, that should be
[ 1 0 0 1 1 0 ];
You can use ismember and do it in a single line:
>> ismember(Q,r,'rows')'
ans =
1 0 0 1 1 0
all(bsxfun(#eq, r, Q),2)'
bsxfun(#eq, r, Q) compares each row and returns a matrix with same size as Q:
>> bsxfun(#eq, r, Q)
ans =
1 1 1
0 1 1
0 0 0
1 1 1
1 1 1
1 1 0
the all function computes if the result of bsxfun is all true along each row separately. Thus it returns:
>> all(ans,2)'
ans =
1 0 0 1 1 0
and yeah, there is also a transpose operator ' to match your desired row output
a = [1 1 1; 2 2 2; 3 3 3];
b = a(1:2,;);
[temp locb] = ismember(a,b,'rows');
b(locb(locb~=0),:)
ans =
1 1 1
2 2 2
Easier way with repmat:
a = [1 2 3; 4 5 6; 7 8 9];
t = [4 5 6];
[x,y] = size(a);
r = all(a==repmat(t,y,1), 2)'