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)
Related
I am looking for an easy way to obtain the column-wise average of a subset of values in a matrix (indexed by a logical matrix), preferably without having to use a loop. The problem that I am experiencing is that because the number of values in each column is different, matlab collapses the values-matrix and its column-wise mean becomes the total mean (of the entire matrix). Is there a specific function or easy workaround for this problem? See below for an example.
%% define value matrix and logical indexing matrix
values=[1 2 3 4; 5 6 7 8; 9 10 11 12];
indices=[1 0 0 1; 0 1 0 1; 1 1 0 0];
indices=indices==1; %convert to logical
%% calculate column-wise average
mean(values(indices),1)
accumarray-based approach
Use the column index as a grouping variable for accumarray:
[~, col] = find(indices);
result = accumarray(col(:), values(indices), [size(values,2) 1], #mean, NaN).';
Note that:
The second line uses (:) to force the first input to be a column vector. This is needed because the first line may produce col as a row or column vector, depending on the size of indices.
NaN is used as fill value. This is specified as the fifth input to accumarray.
The third input to accumarray defines output size. It is necessary to specify it explicitly (as opposed to letting accumarray figure it out) in case the last columns of indices only contain false.
Hacky approach
Multiply and divide element-wise by indices. That will turn unwanted entries into NaN, which can then be ignored by mean using the 'omitnan' option:
result = mean(values.*indices./indices, 1, 'omitnan');
Manual approach
Multiply element-wise by indices, sum each column, and divide element-wise by the sum of each column of indices:
result = sum(values.*indices, 1) ./ sum(indices, 1);
Generating evenly space can use linspace, but I wonder if I can vectorize it. What I mean is as follow:
Given an input vector, say [1 2], I want to generate a 2X6 matrix such that:
in the first row, the entries are [0:0.2:1]
the entries for the second row are [0:0.4:2]
In general, the input vector may not be known, it can change from [1 2] to [1:3:10] or other vectors. However, the first column much be a zero vector and the number of columns can be treated to be the known in advanced.
I do not want to write it using a for loop if possible.
Assuming A = [x, y], you can generate the desired 2 x 6 matrix M as follows:
B = A/A(1);
M = B.'*linspace(0, A(1), 6);
In an attempt to create my own covariance function in MatLab I need to perform matrix multiplication on a row to create a matrix.
Given a matrix D where
D = [-2.2769 0.8746
0.6690 -0.4720
-1.0030 -0.9188
2.6111 0.5162]
Now for each row I need manufacture a matrix. For example the first row R = [-2.2770, 0.8746] I would want the matrix M to be returned where M = [5.1847, -1.9915; -1.9915, 0.7649].
Below is what I have written so far. I am asking for some advice to explain how to use matrix multiplication on a rows to produce matrices?
% Find matrices using matrix multiplication
for i=1:size(D, 1)
P1 = (D(i,:))
P2 = transpose(P1)
M = P1*P2
end
You are trying to compute the outer product of each row with itself stored as individual slices in a 3D matrix.
Your code almost works. What you're doing instead is computing the inner product or the dot product of each row with itself. As such it'll give you a single number instead of a matrix. You need to change the transpose operation so that it's done on P1 not P2 and P2 will now simply be P1. Also you are overwriting the matrix M at each iteration. I'm assuming you'd like to store these as individual slices in a 3D matrix. To do this, allocate a 3D matrix where each 2D slice has an equal number of rows and columns which is the number of columns in D while the total number of slices is equal to the total number of rows in D. Then just index into each slice and place the result accordingly:
M = zeros(size(D,2), size(D,2), size(D,1));
% Find matrices using matrix multiplication
for ii=1:size(D, 1)
P = D(ii,:);
M(:,:,ii) = P.'*P;
end
We get:
>> M
M(:,:,1) =
5.18427361 -1.99137674
-1.99137674 0.76492516
M(:,:,2) =
0.447561 -0.315768
-0.315768 0.222784
M(:,:,3) =
1.006009 0.9215564
0.9215564 0.84419344
M(:,:,4) =
6.81784321 1.34784982
1.34784982 0.26646244
Depending on your taste, I would recommend using bsxfun to help you perform the same operation but perhaps doing it faster:
M = bsxfun(#times, permute(D, [2 3 1]), permute(D, [3 2 1]));
In fact, this solution is related to a similar question I asked in the past: Efficiently compute a 3D matrix of outer products - MATLAB. The only difference is that the question wanted to find the outer product of columns instead of the rows.
The way the code works is that we shift the dimensions with permute of D so that we get two matrices of the sizes 2 x 1 x 4 and 1 x 2 x 4. By performing bsxfun and specifying the times function, this allows you to efficiently compute the matrix of outer products per slice simultaneously.
I have vector with 5 numbers in it, and a matrix of size 6000x20, so every row has 20 numbers. I want to count how many of the 6000 rows contain all values from the vector.
As the vector is a part of a matrix which has 80'000'000 rows, each containing unique combinations, I want a fast solution (which doesn't take more than 2 days).
Thanks
With the sizes you have, a bsxfun-based approach that builds an intermediate 6000x20x5 3D-array is affordable:
v = randi(9,1,5); %// example vector
M = randi(9,6000,20); %// example matrix
t = bsxfun(#eq, M, reshape(v,1,1,[]));
result = sum(all(any(t,2),3));
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.