I am trying to add a number of vectors in a Matrix where each row represents a vector, but it gives me "Subscripted assignment dimension mismatch." error. The main problem is that each vector has a different size. I tried to add zeros at the end of the short vectors but I couldn't do it. Any Help.
Example:
%signal is a vector of data.
[x(1,:),y(1,:)] = findpeaks(signal1);
[x(2,:),y(2,:)] = findpeaks(signal2); %error as the peaks count in signal 2 is not the same as in signal 1.
OK, given two vectors of unequal length,
A=rand(1,10)
B=rand(1,5)
the proper way to deal with this is to use a cell array
D={A;B}
And then you can get whatever elements you want like this, for example:
D{1}(1:3) %// A(1:3)
If you don't want to use cells, you can add rows using this little function that adds row vector M to matrix F
addRow=#(F,M) [F NaN(size(F,1),size(M,2)-size(F,2));M NaN(1,size(F,2)-size(M,2))]
you would use it like this:
F=A
F=addRow(F,B)
Related
I'm currently attempting to select a subset of 0's in a very large matrix (about 400x300 elements) and change their value to 1. I am able to do this, but it requires using a loop where each instance it selects the next value in a randperm vector. In other words, 50% of the 0's in the matrix are randomly selected, one-at-a-time, and changed to 1:
z=1;
for z=1:(.5*numberofzeroes)
A(zeroposition(rpnumberofzeroes(z),1),zeroposition(rpnumberofzeroes(z),2))=1;
z=z+1;
end
Where 'A' is the matrix, 'zeroposition' is a 2-column-wide matrix with the positions of the 0's in the matrix (the "coordinates" if you like), and 'rpnumberofzeros' is a randperm vector from 1 to the number of zeroes in the matrix.
So for example, for z=20, the code might be something like this:
A(3557,2684)=1;
...so that the 0 which appears in this location within A will now be a 1.
It performs this loop thousands of times, because .5*numberofzeroes is a very big number. This inevitably takes a long time, so my question is can this be done without using a loop? Or at least, in some way that takes less processing resources/time?
As I said, the only thing that needs to be done is an entirely random selection of 50% (or whatever proportion) of the 0's changed to 1.
Thanks in advance for the help, and let me know if I can clear anything up! I'm new here, so apologies in advance if I've made any faux pa's.
That's very easy. I'd like to introduce you to my friend sub2ind. sub2ind allows you to take row and column coordinates of a matrix and convert them into linear column-major indices so that you can access multiple values in a matrix simultaneously in a single call. As such, the equivalent code you want is:
%// First access the values in rpnumberofzeroes
vals = rpnumberofzeroes(1:0.5*numberofzeroes, :);
%// Now, use the columns of these to determine which rows and columns we want
%// to access A
rows = zeroposition(vals(:,1), 1);
cols = zeroposition(vals(:,2), 2);
%// Get linear indices via sub2ind
ind1 = sub2ind(size(A), rows, cols);
%// Now set these locations to 1
A(ind1) = 1;
The first statement gets the first half of your matrix of coordinates stored in rpnumberofzeroes. The first column is the row coordinates, the second column is the column coordinates. Notice that in your code, you wish to use the values in zeroposition to access the locations in A. As such, extract out the corresponding rows and columns from rpnumberofzeroes to figure out the right rows and columns from zeroposition. Once that's done, we wish to use these new rows and columns from zeroposition and index into A. sub2ind requires three inputs - the size of the matrix you are trying to access... so in our case, that's A, the row coordinates and the column coordinates. The output is a set of column major indices that are computed for each row and column pair.
The last piece of the puzzle is to use these to index into A and set the locations to 1.
This can be accomplished with linear indexing as well:
% find linear position of all zeros in matrix
ix=find(abs(A)<eps);
% set one half of those, selected at random, to one.
A(ix(randperm(round(numel(ix)*.5)))=1;
I'm looking for a way to select the vector that has the largest sum. Is there a simple way of doing this? I was thinking of writing a loop, but I'm not sure how to loop over a set of vectors.
Thanks for your help!
For the case in which the vectors have the same length (as stated in the comment), I think a simple loop-free way would be to build a matrix from each vector and fetch directly the row (or column) with the largest sum:
clear
clc
RandMat = rand(8,10);
[~,Ind] = max(sum(RandMat,2)); %// Get row index for largest sum. If you want the column, use 1 instead.
MaxRow = RandMat(Ind,:); %// Index in original matrix to get the vector. If you want the column, use RandMat(:,Ind);
If vectors don't have the same length then you would need to pad the missing values with NaN for example to use a regular matrix, otherwise you would need a cell array.
If you prefer a solution in which you don't have to build a matrix then you could loop through each individual vector and store the sum in a variable, then compare the sums at the end. If you would like such a solution please ask!
Please see the following issue:
P=rand(4,4);
for i=1:size(P,2)
for j=1:size(P,2)
[r,p]=corr(P(:,i),P(:,j))
end
end
Clearly, the loop will cause the number of correlations to be doubled (i.e., corr(P(:,1),P(:,4)) and corr(P(:,4),P(:,1)). Does anyone have a suggestion on how to avoid this? Perhaps not using a loop?
Thanks!
I have four suggestions for you, depending on what exactly you are doing to compute your matrices. I'm assuming the example you gave is a simplified version of what needs to be done.
First Method - Adjusting the inner loop index
One thing you can do is change your j loop index so that it only goes from 1 up to i. This way, you get a lower triangular matrix and just concentrate on the values within the lower triangular half of your matrix. The upper half would essentially be all set to zero. In other words:
for i = 1 : size(P,2)
for j = 1 : i
%// Your code here
end
end
Second Method - Leave it unchanged, but then use unique
You can go ahead and use the same matrix like you did before with the full two for loops, but you can then filter the duplicates by using unique. In other words, you can do this:
[Y,indices] = unique(P);
Y will give you a list of unique values within the matrix P and indices will give you the locations of where these occurred within P. Note that these are column major indices, and so if you wanted to find the row and column locations of where these locations occur, you can do:
[rows,cols] = ind2sub(size(P), indices);
Third Method - Use pdist and squareform
Since you're looking for a solution that requires no loops, take a look at the pdist function. Given a M x N matrix, pdist will find distances between each pair of rows in a matrix. squareform will then transform these distances into a matrix like what you have seen above. In other words, do this:
dists = pdist(P.', 'correlation');
distMatrix = squareform(dists);
Fourth Method - Use the corr method straight out of the box
You can just use corr in the following way:
[rho, pvals] = corr(P);
corr in this case will produce a m x m matrix that contains the correlation coefficient between each pair of columns an n x m matrix stored in P.
Hopefully one of these will work!
this works ?
for i=1:size(P,2)
for j=1:i
Since you are just correlating each column with the other, then why not just use (straight from the documentation)
[Rho,Pval] = corr(P);
I don't have the Statistics Toolbox, but according to http://www.mathworks.com/help/stats/corr.html,
corr(X) returns a p-by-p matrix containing the pairwise linear correlation coefficient between each pair of columns in the n-by-p matrix X.
I have two 50 x 6 matrices, say A and B. I want to assign weights to each element of columns in matrix - more weight to elements occurring earlier in a column and less weight to elements occurring later in the same column...likewise for all 6 columns. Something like this:
cumsum(weight(row)*(A(row,col)-B(row,col)); % cumsum is for cumulative sum of matrix
How can we do it efficiently without using loops?
If you have your weight vector w as a 50x1 vector, then you can rewrite your code as
cumsum(repmat(w,1,6).*(A-B))
BTW, I don't know why you have the cumsum operating on a scalar in a loop... it has no effect. I'm assuming that you meant that's what you wanted to do with the entire matrix. Calling cumsum on a matrix will operate along each column by default. If you need to operate along the rows, you should call it with the optional dimension argument as cumsum(x,2), where x is whatever matrix you have.
I am trying to take a matrix and normalize the values in each cell around the average for that column. By normalize I mean subtract the value in each cell from the mean value in that column i.e. subtract the mean for Column1 from the values in Column1...subtract mean for ColumnN from the values in ColumnN. I am looking for script in Matlab. Thanks!
You could use the function mean to get the mean of each column, then the function bsxfun to subtract that from each column:
M = bsxfun(#minus, M, mean(M, 1));
Additionally, starting in version R2016b, you can take advantage of the fact that MATLAB will perform implicit expansion of operands to the correct size for the arithmetic operation. This means you can simply do this:
M = M-mean(M, 1);
Try the mean function for starters. Passing a matrix to it will result in all the columns being averaged and returns a row vector.
Next, you need to subtract off the mean. To do that, the matrices must be the same size, so use repmat on your mean row vector.
a=rand(10);
abar=mean(a);
abar=repmat(abar,size(a,1),1);
anorm=a-abar;
or the one-liner:
anorm=a-repmat(mean(a),size(a,1),1);
% Assuming your matrix is in A
m = mean(A);
A_norm = A - repmat(m,size(A,1),1)
As has been pointed out, you'll want the mean function, which when called without any additional arguments gives the mean of each column in the input. A slight complication then comes up because you can't simply subtract the mean -- its dimensions are different from the original matrix.
So try this:
a = magic(4)
b = a - repmat(mean(a),[size(a,1) 1]) % subtract columnwise mean from elements in a
repmat replicates the mean to match the data dimensions.