I have a matrix having rows with repeated numbers. I want to find those rows and replace them with a dummy row so as to keep the number of rows of the matrix constant.
Dummy_row = [1 2 3]
(5x3) Matrix A
A = [2 3 6;
4 7 4;
8 7 2;
1 3 1;
7 8 2]
(5x3) Matrix new_A
new_A = [2 3 6;
1 2 3;
8 7 2;
1 2 3;
7 8 2]
I tried the following which deleted the rows having repeated numbers.
y = [1 2 3]
w = sort(A,2)
v = all(diff(t,1,2)~=0|w(:,1:2)==0,2) % When v is zero, the row has repeated numbers
z = A(w,:)
Can you please help?
bsxfun based solution -
%// Create a row mask of the elements that are to be edited
mask = any(sum(bsxfun(#eq,A,permute(A,[1 3 2])),2)>1,3);
%// Setup output variable and set to-be-edited rows as copies of [1 2 3]
new_A = A;
new_A(mask,:) = repmat(Dummy_row,sum(mask),1)
Code run -
A =
2 3 6
4 7 4
8 7 2
1 3 1
7 8 2
new_A =
2 3 6
1 2 3
8 7 2
1 2 3
7 8 2
You could use the following:
hasRepeatingNums = any(diff(sort(A, 2), 1, 2)==0, 2);
A(hasRepeatingNums,:) = repmat(Dummy_row, nnz(hasRepeatingNums), 1);
See if this works for you,
A= [ 2 3 6;
4 7 4;
8 7 2;
5 5 5;
1 8 8;
1 3 1;
7 8 2 ];
Dummy_row = [1 2 3];
b = diff(sort(A,2),1,2);
b = sum(b == 0,2);
b = b > 0;
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c;
gives,
newA =
2 3 6
1 2 3
8 7 2
1 2 3
1 2 3
1 2 3
7 8 2
Edit
Not much change is needed, try this,
Dummy_row = [1 2 3];
b = sum(A == 0,2);
b = b > 0;
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c;
Related
I am trying to index my matrix based on two conditions, I'll explain.
Let's say I have two matrices:
a = [7 3 4; 5 6 7; 4 8 0];
b = [1 9 8; 2 4 6; 6 1 6];
And a third matrix to index:
c = [1 2 3; 4 5 6; 7 8 9];
My aim is to index c in a way that I get a 3x3 matrix in which only the values of c are copied over for whose indexes the following conditions are met and the rest are zeros.
a <= 5, b >= 6
Resulting matrix:
result = [0 2 3; 0 0 0; 7 0 9]
I hope I was able to explain my problem.
Given
a = [7 3 4; 5 6 7; 4 8 0];
b = [1 9 8; 2 4 6; 6 1 6];
c = [1 2 3; 4 5 6; 7 8 9];
result = zeros(size(c);
Using logical indexing,
>> d = (a <= 5) & (b >= 6)
d =
0 1 1
0 0 0
1 0 1
>> result(d) = c(d)
result =
0 2 3
0 0 0
7 0 9
Loop throw rows and columns and set the result.
for row=1:size(a,1)
for col=1:size(a,2)
if(a(row,col)> b(row,col))
result(row,col) = 0
else
result(row,col) = c(row,col)
end
end
end
If I have matrix size(mat)= X*Y*6
let call mat(:,:,1)=A
and mat(:,:,2)=B and so on
how do i rezise mat to X*Y*12
where
mat(:,:,1)=mat(:,:,2)= A
mat(:,:,3)=mat(:,:,4)=B
and so on
You can use the following syntax:
%defines input matrix (in your case it is already defined)
m = 500;
n = 400;
z = 6;
mat = rand(m,n,z);
%initialize output matrix
newMat = zeros(m,n,z*2);
%assign old matrix values into the new matrix
newMat(:,:,1:2:end) = mat;
newMat(:,:,2:2:end) = mat;
If you have Matlab 2015a or newer you can use repelem:
N = 2; %// number of times to repeat
result = repelem(mat, 1, 1, N); %// repeat N times along 3rd dim
For older Matlab versions you can do it manually as follows:
N = 2; %// number of times to repeat
ind = ceil(1/N:1/N:size(mat,3)); %// build index with repetitions
result = mat(:,:,ind); %// apply index along desired dim
Example:
>> %// Data
>> mat = randi(9,2,4,2)
mat(:,:,1) =
5 8 9 2
7 3 1 5
mat(:,:,2) =
5 7 1 1
1 8 8 2
>> %// First approach
>> N = 2;
>> result = repelem(mat, 1, 1, N)
result(:,:,1) =
5 8 9 2
7 3 1 5
result(:,:,2) =
5 8 9 2
7 3 1 5
result(:,:,3) =
5 7 1 1
1 8 8 2
result(:,:,4) =
5 7 1 1
1 8 8 2
>> %// Second approach
>> N = 2;
>> ind = ceil(1/N:1/N:size(mat,3));
>> result = mat(:,:,ind)
result(:,:,1) =
5 8 9 2
7 3 1 5
result(:,:,2) =
5 8 9 2
7 3 1 5
result(:,:,3) =
5 7 1 1
1 8 8 2
result(:,:,4) =
5 7 1 1
1 8 8 2
I am trying to match 1st column of A with 1st to 3rd columns of B, and append corresponding 4th column of B to A.
For example,
A=
1 2
3 4
B=
1 2 4 5 4
1 2 3 5 3
1 1 1 1 2
3 4 5 6 5
I compare A(:,1) and B(:, 1:3)
1 and 3 are in A(:,1)
1 is in the 1st, 2nd, 3rd rows of B(:, 1:3), so append B([1 2 3], 4:end)' to A's 1st row.
3 is in the 2nd and 4th rows of B(:,1:3), so append B([2 4], 4:end)' to A's 2nd row.
So that it becomes:
1 2 5 4 5 3 1 2
3 4 5 3 6 5 0 0
I could code this using only for and if.
clearvars AA A B mem mem2 mem3
A = [1 2 ; 3 4]
B = [1 2 4 5 4; 1 2 3 5 3; 1 1 1 1 2; 3 4 5 6 5]
for n=1:1:size(A,1)
mem = ismember(B(:,[1:3]), A(n,1));
mem2 = mem(:,1) + mem(:,2) + mem(:,3);
mem3 = find(mem2>0);
AA{n,:} = horzcat( A(n,:), reshape(B(mem3,[4,5])',1,[]) ); %'
end
maxLength = max(cellfun(#(x)numel(x),AA));
out = cell2mat(cellfun(#(x)cat(2,x,zeros(1,maxLength-length(x))),AA,'UniformOutput',false))
I am trying to make this code efficient, by not using for and if, but couldn't find an answer.
Try this
a = A(:,1);
b = B(:,1:3);
z = size(b);
b = repmat(b,[1,1,numel(a)]);
ab = repmat(permute(a,[2,3,1]),z);
row2 = mat2cell(permute(sum(ab==b,2),[3,1,2]),ones(1,numel(a)));
AA = cellfun(#(x)(reshape(B(x>0,4:end)',1,numel(B(x>0,4:end)))),row2,'UniformOutput',0);
maxLength = max(cellfun(#(x)numel(x),AA));
out = cat(2,A,cell2mat(cellfun(#(x)cat(2,x,zeros(1,maxLength-length(x))),AA,'UniformOutput',false)))
UPDATE Below code runs in almost same time as the iterative code
a = A(:,1);
b = B(:,1:3);
z = size(b);
b = repmat(b,[1,1,numel(a)]);
ab = repmat(permute(a,[2,3,1]),z);
df = permute(sum(ab==b,2),[3,1,2])';
AA = arrayfun(#(x)(B(df(:,x)>0,4:end)),1:size(df,2),'UniformOutput',0);
AA = arrayfun(#(x)(reshape(AA{1,x}',1,numel(AA{1,x}))),1:size(AA,2),'UniformOutput',0);
maxLength = max(arrayfun(#(x)(numel(AA{1,x})),1:size(AA,2)));
out2 = cell2mat(arrayfun(#(x,i)((cat(2,A(i,:),AA{1,x},zeros(1,maxLength-length(AA{1,x}))))),1:numel(AA),1:size(A,1),'UniformOutput',0));
How about this:
%# example data
A = [1 2
3 4];
B = [1 2 4 5 4
1 2 3 5 3
1 1 1 1 2
3 4 5 6 5];
%# rename for clarity & reshape for algorithm's convenience
needle = permute(A(:,1), [2 3 1]);
haystack = B(:,1:3);
data = B(:,4:end).';
%# Get the relevant rows of 'haystack' for each entry in 'needle'
inds = any(bsxfun(#eq, haystack, needle), 2);
%# Create data that should be appended to A
%# All data and functionality in this loop is local and static, so speed
%# should be optimal.
append = zeros( size(A,1), numel(data) );
for ii = 1:size(inds,3)
newrow = data(:,inds(:,:,ii));
append(ii,1:numel(newrow)) = newrow(:);
end
%# Now append to A, stripping unneeded zeros
A = [A append(:, ~all(append==0,1))]
I need to transform a matrix:
X = [1 2; 3 4; 5 6; 7 8]
X = 1 2
3 4
5 6
7 8
to
X = [1 2; 5 6; 3 4; 7 8]
X = 1 2
5 6
3 4
7 8
and do this operation for a matrix with any number of rows. So that in a matrix with 200 rows, row 101 will become row 2, row 102 will become row 4 and so on.
How can I achieve this in MATLAB?
For arrays with an even number of rows, you do the following:
nRows = size(X,1);
idx = [1:nRows/2;nRows/2+1:nRows];
X_rearranged = X(idx(:),:);
For arrays with odd number of rows, you add 1 to nRows, and use idx(1:end-1) instead of idx(:)
You can use:
X = [1 2; 3 4; 5 6; 7 8]
Y = [a(1,:); a(3,:); a(2,:); a(4,:)]
Say I have a matrix A with 3 columns c1, c2 and c3.
1 2 9
3 0 7
3 1 4
And I want a new matrix of dimension (3x3n) in which the first column is c1, the second column is c1^2, the n column is c1^n, the n+1 column is c2, the n+2 column is c2^2 and so on. Is there a quickly way to do this in MATLAB?
Combining PERMUTE, BSXFUN, and RESHAPE, you can do this quite easily such that it works for any size of A. I have separated the instructions for clarity, you can combine them into one line if you want.
n = 2;
A = [1 2 9; 3 0 7; 3 1 4];
[r,c] = size(A);
%# reshape A into a r-by-1-by-c array
A = permute(A,[1 3 2]);
%# create a r-by-n-by-c array with the powers
A = bsxfun(#power,A,1:n);
%# reshape such that we get a r-by-n*c array
A = reshape(A,r,[])
A =
1 1 2 4 9 81
3 9 0 0 7 49
3 9 1 1 4 16
Try the following (don't have access to Matlab right now), it should work
A = [1 2 9; 3 0 7; 3 1 4];
B = [];
for i=1:n
B = [B A.^i];
end
B = [B(:,1:3:end) B(:,2:3:end) B(:,3:3:end)];
More memory efficient routine:
A = [1 2 9; 3 0 7; 3 1 4];
B = zeros(3,3*n);
for i=1:n
B(3*(i-1)+1:3*(i-1)+3,:) = A.^i;
end
B = [B(:,1:3:end) B(:,2:3:end) B(:,3:3:end)];
Here is one solution:
n = 4;
A = [1 2 9; 3 0 7; 3 1 4];
Soln = [repmat(A(:, 1), 1, n).^(repmat(1:n, 3, 1)), ...
repmat(A(:, 2), 1, n).^(repmat(1:n, 3, 1)), ...
repmat(A(:, 3), 1, n).^(repmat(1:n, 3, 1))];