let us consider following bit of code:
[m,n]=size(X);
if m == (n+1)
Z = vertcat(U(:,1:2:d), V(:,1:2:d));
else
Z = vertcat(U(:,[1:2:d]));
end
C=Z(:);
What I want it to do is concatenate the singular vectors into one column vector. For example, I want to concatenate the first d left and right singular vectors, but the problem is that it creates a multidimensional column, that's why I wrote C=Z(:). But, can I use vertcat to just create a 1D column vector? Thanks in advance!
It looks like you may have just flipped your row and column indexing. U(:,1:2:d) will return a row vector, which you are then vertcating with another row vector. Try this instead:
[m,n]=size(X);
if m == (n+1)
Z = vertcat(U(1:2:d,:), V(1:2:d,:));
else
Z = vertcat(U([1:2:d]), :));
end
C=Z(:);
I hope that helps.
Related
So I want to concatenate an m x n matrix to obtain a 1 x mn matrix. The matrix I want to concatenate are generated from a while loop. Although the number of columns will always be 3, I however cannot tell how many rows there will be for each iteration. Also, the row sizes for each iteration may not always be the same.
The code runs in cases where the row sizes were all equal to 6, but in cases where they aren't equal I get an error:
Error using vertcat Dimensions of matrices being concatenated are not consistent.
parts of the code are as follows:
A = [];
B = [];
searchArea = 2;
for ii = 1: numel(velocity)
Do ....
while area(ii,:) < searchArea
Do ....
% COLLATE vectors for A
A = [A; [Ax(ii), Ay(ii), Az(ii)]];
Do ...
end
%# Copy the A into new variable (B) and Reshape into row vector so as to associate each row to its corresponding velocity
B = [B; reshape(A.',1,[])];
A = [];
end
Could someone please advice me on what I am doing wrong here. I would clarify further if there be need. Thanks guys!
If it's your intent that B ends up being a row vector, then you need to change this:
B = [B; reshape(A.',1,[])]; % Does vertical concatenation
to this:
B = [B reshape(A.',1,[])]; % Does horizontal concatenation (note there's no semicolon)
so that each row vector gotten from reshaping A gets added to the end of the row instead of as a new row (as the semicolon indicates).
I have two column matrices:
size(X) = 50 x 1
size(Y) = 50 x 1
which I got from ind2sub
I want to create a structure str such that
str(i).XYZ returns [X(i), Y(i)] for i in [1,50]
I am trying to use
str = struct('XYZ', num2cell([X,Y]));
However, I believe for this to work properly, I need to modify [X,Y] to a matrix of row vectors, each row vector being [X(i), Y(i)]. Not sure if there is a better approach
You are basically on the right way, but num2cell([X,Y]) creates a 2x50 cell, which results in a 2x50 struct. You want to split your input matrix [X,Y] only among the second dimensions so each cell is a 2x1, use num2cell([X,Y],2)
str = struct('XYZ', num2cell([X,Y],2));
Concatenate the two vectors, convert from matrix to cell, and then from cell to struct array:
str = cell2struct(mat2cell([X Y], ones(1,size(X,1)), size(X,2)+size(Y,2) ).', 'XYZ');
Say I have a nxm matrix and want to treat each row as vectors in a function. So, if I have a function that adds vectors, finds the Cartesian product of vectors or for some reason takes the input of several vectors, I want that function to treat each row in a matrix as a vector.
This sounds like a very operation in Matlab. You can access the ith row of a matrix A using A(i, :). For example, to add rows i and j, you would do A(i, :) + A(j, :).
Given an nxm matrix A:
If you want to edit a single column/row you could use the following syntax: A(:, i) for the ith-column and A(i, :) for ith-row.
If you want to edit from a column/row i to a column/row j, you could use that syntax: A(:, i:j) or A(i:j, :)
If you want to edit (i.e.) from the penultimate column/row to the last one, you could you: A(:, end-1:end) or A(end-1:end, :)
EDIT:
I can't add a comment above because I don't have 50 points, but you should post the function setprod. I think you should be able to do what you want to do, by iterating the matrix you're passing as an argument, with a for-next statement.
I think you're going to have to loop:
Input
M = [1 2;
3 4;
5 6];
Step 1: Generate a list of all possible row pairs (row index numbers)
n = size(M,1);
row_ind = nchoosek(1:n,2)
Step 2: Loop through these indices and generate the product set:
S{n,n} = []; //% Preallocation of cell matrix
for pair = 1:size(row_ind,1)
p1 = row_ind(pair,1);
p2 = row_ind(pair,2);
S{p1,p2} = setprod(M(p1,:), M(p2,:))
end
Transform the matrix into a list of row vectors using these two steps:
Convert the matrix into a cell array of the matrix rows, using mat2cell.
Generate a comma-separated list from the cell array, using linear indexing of the cell contents.
Example: let
v1 = [1 2];
v2 = [10 20];
v3 = [11 12];
M = [v1; v2; v3];
and let fun be a function that accepts an arbitrary number of vectors as its input. Then
C = mat2cell(M, ones(1,size(M,1)));
result = fun(C{:});
is the same as result = fun(v1, v2, v3).
i just started with matlab and stuck somewhere...consider example
X=(3:7)
Z=(2:6)
for (i=1:5)
y=abs(X(i)-Z);
dm=min(y);
D=find(y==min(y))
D1=Z(D);
end
i want D and D1 to be a column/row vector.Please help.
Currently you are storing scalar values into D and D1.
Maybe you wanted to save the values into i-th column of D and D1?
X=(3:7)
Z=(2:6)
for (i=1:5)
y=abs(X(i)-Z);
dm=min(y);
D(i)=find(y==min(y));
D1(i)=Z(D(i));
end
If you're looking to simply convert D and D1 from row vectors to column vectors, you can simply add the following lines at the end of your code:
D = D';
D1 = D1';
The ' operation simply gives you the transpose of the matrix in question.
use:
if isrow(D)
D = D.'; % .' is the transpose operator
end
BTW: you don't need to use parentheses that often.
I think the following will help too:
% convert ANY array into a column vector
D = D(:);
% convert ANY array into a row vector
D1 = D1(:).';
doing it like this will guarantee that one is column and the other row, without any performance loss.
please help me to understand this code:
x(:,i) = mean( (y(:,((i-1)*j+1):i*j)), 2 )';
i can't find it in my book. thanks.
The code you posted can be made more readable using temporary variables:
a = (i-1)*j+1;
b = i*j;
val = y(:,a:b);
x(:,i) = mean( val, 2 )'; %# =mean( val' )
What exactly you do not understand? For meaning of mean , : and ' consult matlab help.
It would help if you said exactly what you don't understand, but here are a few tips:
if you have something like a(r,c), that means matrix a, row r, column c (always in this order). In other words, you should have two elements inside the brackets separated by a comma where the first represents the row, the second the column.
If you have : by itself in one of the sides of the comma, that means "all". Thus, if you had a(r,:), then you would have matrix a, row r, all columns.
If : is not alone in one of the sides of the comma, then it will mean "to". So if you have a(r, z:y), that means matrix a, row r, columns z to y.
Mean = average. The format of the function in Matlab is M = mean(A,dim). A will be the matrix you take the average (or mean) of, M will be the place where the results are going to go. If dim = 1, you will get a row vector with each element being the average of a column. If dim = 2 (as it is in your case), then you should get a column vector, with each element being the average of a row. Be careful, though, because at the end of your code you have ', which means transpose. That means that your column vector will be transformed into a row vector.
OK, so your code:
x(:,i) = mean( (y(:,((i-1)*j+1):i*j)), 2 )';
Start with the bit inside, that is
y(:,((i-1)*j+1):i*j)
So that is saying
matrix y(r,c)
where
r (row) is :, that is, all rows
c (column) is ((i-1)j+1):ij, that is, columns going from (i-1)j+1 until ij
Your code will then get the matrix resulting from that, which I called y(r,c), and will do the following:
mean( (y(r,c), 2 )
so get the result from above and take the mean (average) of each row. As your code has the ' afterwards, that is, you have:
mean( (y(r,c), 2 )'
then it will get the column vector and transform into a row vector. Each element of this row will be the average of a row of y(r,c).
Finally:
x(:,i) = mean( (y(r,c), 2 )';
means that the result of the above will be put in column i of matrix x.
Shouldn't this be x(i,:) instead?
The i-th column of the array x is the average of the i-th group of j columns of the array y.
For example, if i is 1 and j is 3, the 1st column of x is the average of the first three columns of y.