I have a bunch of values in a 3-dimensional matrix, and I am finding the mean value of them:
mean(mean(mat))
Now, of different reasons I have to append some rows and elements to the matrix. But I want the mean value to stay the same - as if the added elements are neutral and do not inflict in the result.
Like when you multiply a bunch of values, you can multiply additional 1's without changing the result. And with addition you can add further 0's with no inflict.
What kind of value in Matlab can I assign to the new elements in the matrix to make the elements neutral when using the mean()?
Note added
The point is, when I am calculating the mean value I only have the new resized matrix to do it from. Therefore the added elements must be neutral.
I am thinking of something like NaN, but I had no luck with that since the mean value then also end up as NaN.
Adding values equal to the mean of the matrix without the added values will leave the new mean the same. (I hope that makes sense!). Point is to fill in and not change the new mean, use the current mean.
Alternatively, you can fill in with NaN and use the nanmean function.
Add zeros to the matrix and rescale your mean to be the correct value.
i.e. if your original matrix A is n x m and you resize to B which is N x M then :
mean(mean(A)) = sum(sum(A)) / n x m
mean(mean(B)) = sum(sum(B)) / N x M
= sum(sum(A)) / N x M --- since we padded with zeros
Rearranging gives
mean(mean(A)) = mean(mean(B)) * ( ( N x M )/(n x m) )
Related
I want to reshape pixel intensity by imagesize*1(column vector).
Imvect = reshape(I,imsize,1);
But why these error comes?
Error using reshape
To RESHAPE the number of elements must not change.
Let's start with the syntax used in the documentation:
B = reshape(A,sz1,...,szN)
What reshape does is to take the matrix A, straightens it out, and gives it a new size, that's determined by the 2nd, 3rd to the Nth argument. For this to be possible, you need to have the same number of elements in the input matrix as you have in the output matrix. You can't make a 1x5 vector turn into a 2x3 vector, as one element would be missing. The number of elements in the output matrix will be proportional to the product of sz1, sz2, ..., szN. Now, if you know you want N rows, but don't know exactly how many columns you have, you might use the [] syntax, that tells MATLAB to use as many columns as necessary to make the number of elements be equal.
So reshape(A, 2, [], 3) will become a 2xNx3 matrix, where, for a matrix with 24 elements, N will be 4.
Now, in your case this is not the case. numel(I) ~= imsize. If mod(numel(I), imsize) ~= 0 then your imsize is definitely incorrect. However, if mod(numel(I), imsize) == 0, then your error might be that you want imsize number of rows, and a number of columns that makes this possible. If it's the latter, then this should work:
Imvect = reshape(I,imsize, []);
If you simply want to make you matrix I a vector of size (numel(I), 1), then you should use the colon operator :, as such:
Imvect = I(:);
An alternative, if you really want to use reshape, is to specify that you want a single column, and let MATLAB select the number of rows, as such:
Imvect = reshape(I, [], 1);
I'd like to create a function that adds several gaussian terms of various width over some specified region:
G(a,b,x) = a_1 exp(- b_1 x^2) + a_2 exp(- b_2 x^2) + ... a_N exp(-b_N x^2)
I'd like this function to output an array of length x, summing over the terms of parameters a,b provided, something like:
x = linspace(-2,2,1000);
N_gauss = #(a,b) a(:).*exp(-b(:)*x.^2);
This example actually works if a,b have only a single value, but when they become vectors it no longer works (I suppose Matlab doesn't know what should be added, multiplied or remain a vector). Is this even possible?
You can do this purely by matrix multiplication. Let's tackle the problem slowly and work our way up. You first need to form products of the elements of the vector b and scalar values stored in x. First create a 2D matrix of values where each row corresponds to the product-wise values between an element in b and an element in x. The element (i,j) in this matrix corresponds to the product of the ith element in x with the jth element in b.
You can achieve this by using the outer product. Make x a column vector and b a row vector, then perform the multiplication. Also, make sure you square each of the x terms as seen in your equation.
term1 = (x(:).^2)*b(:).';
Now you can apply the exponential operator and ensure you place a negative in the exponent so you can build the right side of each term (i.e. exp(- b_i x^2)):
term2 = exp(-term1);
The last thing you need to do is multiply each of the values in the 2D matrix with the right coefficient from the a vector. You can do this by enforcing that a be a column vector and performing matrix-vector multiplication.
out = term2*a(:);
Matrix-vector multiplication is the dot product between the column vector a with each row in the 2D matrix we created before. This exactly corresponds to the summation of your equation for each value in x. As such, this achieves the Gaussian summation for each value in x and places this into a n x 1 vector where n is the total number of elements in x. Putting this all together gives us:
out = exp(-(x(:).^2)*b(:).')*a(:);
To finally abstract this into an anonymous function, do:
N_gauss = #(a,b,x) exp(-(x(:).^2)*b(:).')*a(:);
This function takes in the vectors a, b and x as per your problem.
I have a matrix A in Matlab of dimension mxn. I want to construct a vector B of dimension mx1 such that B(i)=1 if all elements of A(i,:) are equal and 0 otherwise. Any suggestion? E.g.
A=[1 2 3; 9 9 9; 2 2 2; 1 1 4]
B=[0;1;1;0]
One way with diff -
B = all(diff(A,[],2)==0,2)
Or With bsxfun -
B = all(bsxfun(#eq,A,A(:,1)),2)
Here's another example that's a bit more obfuscated, but also does the job:
B = sum(histc(A,unique(A),2) ~= 0, 2) == 1;
So how does this work? histc counts the frequency or occurrence of numbers in a dataset. What's cool about histc is that we can compute the frequency along a dimension independently, so what we can do is calculate the frequency of values along each row of the matrix A separately. The first parameter to histc is the matrix you want to compute the frequency of values of. The second parameter denotes the edges, or which values you are looking at in your matrix that you want to compute the frequencies of. We can specify all possible values by using unique on the entire matrix. The next parameter is the dimension we want to operate on, and I want to work along all of the columns so 2 is specified.
The result from histc will give us a M x N matrix where M is the total number of rows in our matrix A and N is the total number of unique values in A. Next, if a row contains all equal values, there should be only one value in this row where all of the values were binned at this location where the rest of the values are zero. As such, we determine which values in this matrix are non-zero and store this into a result matrix, then sum along the columns of the result matrix and see if each row has a sum of 1. If it does, then this row of A qualifies as having all of the same values.
Certainly not as efficient as Divakar's diff and bsxfun method, but an alternative since he took the two methods I would have used :P
Some more alternatives:
B = var(A,[],2)==0;
B = max(A,[],2)==min(A,[],2)
Which of the following statements will find the minimum difference between any pair of elements (a,b) where a is from the vector A and b is from the vector B.
A. [X,Y] = meshgrid(A,B);
min(abs(X-Y))
B. [X,Y] = meshgrid(A,B);
min(abs(min(Y-X)))
C. min(abs(A-B))
D. [X,Y] = meshgrid(A,B);
min(min(abs(X-Y)))
Can someone please explain to me?
By saying "minimum difference between any pair of elements(a,b)", I presume you mean that you are treating A and B as sets and you intend to find the absolute difference in any possible pair of elements from these two sets. So in this case you should use your option D
[X,Y] = meshgrid(A,B);
min(min(abs(X-Y)))
Explanation: Meshgrid turns a pair of 1-D vectors into 2-D grids. This link can explain what I mean to say:
http://www.mathworks.com/help/matlab/ref/meshgrid.html?s_tid=gn_loc_drop
Hence (X-Y) will give the difference in all possible pairs (a,b) such that a belongs to A and b belongs to B. Note that this will be a 2-D matrix.
abs(X-Y) would return the absolute values of all elements in this matrix (the absolute difference in each pair).
To find the smallest element in this matrix you will have to use min(min(abs(X-Y))). This is because if Z is a matrix, min(Z) treats the columns of Z as vectors, returning a row vector containing the minimum element from each column. So a single min command will give a row vector with each element being the min of the elements of that column. Using min for a second time returns the min of this row vector. This would be the smallest element in the entire matrix.
This can help:
http://www.mathworks.com/help/matlab/ref/min.html?searchHighlight=min
Options C is correct if you treat A and B as vectors and not sets. In this case you won't be considering all possible pairs. You'll end up finding the minimum of (a-b) where a,b are both in the same position in their corresponding vectors (pair-wise difference).
D. [X,Y] = meshgrid(A,B);
min(min(abs(X-Y)))
meshgrid will generate two grids - X and Y - from the vectors, which are arranged so that X-Y will generate all combinations of ax-bx where ax is in a and bx is in b.
The rest of the expression just gets the minimum absolute value from the array resulting from the subtraction, which is the value you want.
CORRECT ANSWER IS D
Let m = size(A) and n = size(B)
You want to subtract each pair of (a,b) such that a is from vector A and b is from vector B.
meshgrid(A,B) creates two matrices X Y both of size nxm where X have rows sames have vector A while Yhas columns same as vector B .
Hence , Z = X-Y will give you a matrix with n*m values corresponding to the difference between each pair of values taken from A and B . Now all you have to do is to find the absolute minimum among all values of Z.
You can do that by
req_min = min(min(abs(z)))
The whole code is
[X Y ] = meshgrid(A,B);
Z= X-Y;
Z = abs(Z);
req_min = min(min(Z));
You could also use bsxfun instead of meshgrid:
min(min(abs(bsxfun(#minus, A(:), B(:).'))))
Or use pdist2:
min(min(pdist2(A(:),B(:))))
I would like to have a program that makes the following actions:
Read several matrices having the same size (1126x1440 double)
Select the most occuring value in each cell (same i,j of the matrices)
write this value in an output matrix having the same size 1126x1440 in the corresponding i,j position, so that this output matrix will have in each cell the most occurent value from the same position of all the input matrices.
Building on #angainor 's answer, I think there is a simpler method using the mode function.
nmatrices - number of matrices
n, m - dimensions of a single matrix
maxval - maximum value of an entry (99)
First organize data into a 3-D matrix with dimensions [n X m X nmatrices]. As an example, we can just generate the following random data in a 3-D form:
CC = round(rand(n, m, nmatrices)*maxval);
and then the computation of the most frequent values is one line:
B = mode(CC,3); %compute the mode along the 3rd dimension
Here is the code you need. I have introduced a number of constants:
nmatrices - number of matrices
n, m - dimensions of a single matrix
maxval - maximum value of an entry (99)
I first generate example matrices with rand. Matrices are changed to vectors and concatenated in the CC matrix. Hence, the dimensions of CC are [m*n, nmatrices]. Every row of CC holds individual (i,j) values for all matrices - those you want to analyze.
CC = [];
% concatenate all matrices into CC
for i=1:nmatrices
% generate some example matrices
% A = round(rand(m, n)*maxval);
A = eval(['neurone' num2str(i)]);
% flatten matrix to a vector, concatenate vectors
CC = [CC A(:)];
end
Now we do the real work. I have to transpose CC, because matlab works on column-based matrices, so I want to analyze individual columns of CC, not rows. Next, using histc I find the most frequently occuring values in every column of CC, i.e. in (i,j) entries of all matrices. histc counts the values that fall into given bins (in your case - 1:maxval) in every column of CC.
% CC is of dimension [nmatrices, m*n]
% transpose it for better histc and sort performance
CC = CC';
% count values from 1 to maxval in every column of CC
counts = histc(CC, 1:maxval);
counts have dimensions [maxval, m*n] - for every (i,j) of your original matrices you know the number of times a given value from 1:maxval is represented. The last thing to do now is to sort the counts and find out, which is the most frequently occuring one. I do not need the sorted counts, I need the permutation that will tell me, which entry from counts has the highest value. That is exactly what you want to find out.
% sort the counts. Last row of the permutation will tell us,
% which entry is most frequently found in columns of CC
[~,perm] = sort(counts);
% the result is a reshaped last row of the permutation
B = reshape(perm(end,:)', m, n);
B is what you want.