CODEWITHSUNDEEP
matlab
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Reputation: 2299

Checking equality of row elements in Matlab?

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]

Upvotes: 1

Views: 166

Answers (3)

Luis Mendo
Luis Mendo

Reputation: 112769

Some more alternatives:

B = var(A,[],2)==0;
B = max(A,[],2)==min(A,[],2)

Upvotes: 2

rayryeng
rayryeng

Reputation: 104555

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

Upvotes: 2

Divakar
Divakar

Reputation: 221714

One way with diff -

B = all(diff(A,[],2)==0,2)

Or With bsxfun -

B = all(bsxfun(@eq,A,A(:,1)),2)

Upvotes: 2

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