I want to compare two image histograms. They are as follows:
h1 --> double valued 1 dimension vector .4096 in length.
h2 --> double valued 1 dimension vector .4096 in length.
I am using this matlab function here:
http://clickdamage.com/sourcecode/code/compareHists.m
It is as follows:
% s = compareHists(h1,h2)
% returns a histogram similarity in the range 0..1
%
% Compares 2 normalised histograms using the Bhattacharyya coefficient.
% Assumes that sum(h1) == sum(h2) == 1
%
function s = compareHists(h1,h2)
s = sum(sum(sum(sqrt(h1).*sqrt(h2))));
My question is :
Is there a need for multiple sums?
Even if there is only one sum in the above equation, it would suffice..right?
like this: sum(sqrt(h1).*sqrt(h2)) --> ?
Can some one please explain the code above? Also, tell me if I use a single sum will it be all right?
I tried both ways and got the same answer for two image histograms. I did this with only two histograms not more and hence want to be sure.
Thanks!
In general, sum does the sum along one dimension only. If you want to sum along multiple dimensions you either
use sum several times; or
use linear indexing to reduce to a single dimension and then use sum once: sum(sqrt(h1(:)).*sqrt(h2(:))).
In your case, if there's only one dimension, yes, a single sum would suffice.
You are right. Only one sum is needed. However, if either h1 or h2 is a multidimensional matrix, then you may want to sum as many as the dimensions. For example:
A=magic(4); % a 4 by 4 matrix of magic numbers.
sum(A) % returns [34,34,34,34], i.e. the sum of elements in each column.
sum(sum(A)) % returns 136, i.e. the sum of all elements in A.
I believe the code you downloaded originaly was written to handle multiple histograms stacked as columns of a matrix. This is (IMHO) the reason for the multiple sums.
In your case you can leave it with only one sum.
You can do even better - without any sum
Hover here to see the answer
s = sqrt(h1(:)')*sqrt(h2(:));
The trick is to use vector multiplication!
I don't see any points in 3 sums too, but if you have not a vector with histogram but a matrix you will need 2 sums like this sum(sum(sqrt(h1).*sqrt(h2))) to compare them. First one will calculate the sum of the rows, the second - the sum of the columns.
Related
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)
Suppose I have matrix A of 100x200x300. Third dimension is called "page" in Matlab and this matrix has 300 pages then.
Now I want to calculate standard deviation within each page and get a result matrix of 1x1x300.
I can't just do
std(std(A,0,1),0,2)
because normalization will be wrong as I think.
You need to collapse the first two dimensions into one (i.e. into columns) using reshape; and then compute std along each column:
Ar = reshape(A, size(A,1)*size(A,2), size(A,3));
result = std(Ar);
This will give you a 1x300 vector as result. If you really need it to be 1x1x300, use
result = shiftdim(result, -1);
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.