How to compact an array with repeated data and NaNs?

Hi!

I have two vectors with measurement set points and an array with measurement values. The vectors and array contain repeated values and the array also contains NaNs. I would like to remove the repeated data and for the measurement set points where there are both a NaN and a value, I would like the value to remain.

For example, if the two vectors are called x and y and the array is called z.

x = [0 0.5 0.5 1 2]

y = [0 1 2 2 ]

z = [1 NaN 2 5 10; 7 3 3 NaN 2; 8 4 4 NaN NaN; 8 NaN 4 8 NaN]

I would like this to become

x = [0 0.5 1 2]

y = [0 1 2]

z = [1 2 5 10; 7 3 NaN 2; 8 4 8 NaN]

How do I do this in an easy way (I have quite a large array with data…)?

Thanks!

[James Tursa]

So you want the unique values of x and y to remain, and then delete the corresponding column(s) and row(s) of z based on what got deleted from x and y, but do so in a way that replaces any NaN's with values if there was a non-NaN available for that spot?

Can we assume that x and y are sorted as in your example?

[Konstantinos Sofos]

And what will happen in the case that all the values of a row are NaN or the dimensions are not consistent? In your example with Z matrix the output of the last row has 2 times the 8....

[Elin]

James: That's exactly what I want to do. And, yes, you can assume that x any y are sorted.

Konstantinos: I have no measurement data where all elements in either a row or column is NaN and the dimensions are consistent, so that is not a problem. The reason the Z matrix has the value 8 two times in the last row is because both measurement set points (x = 0, y = 2) and (x = 1, y = 2) give the measurement value z = 8.

Answer 1

[James Tursa]

Brute force using loops (if there are different non-NaN values available for a particular spot, picks the max of them):

[m,n] = size(z);
for k=1:n-1
    if( x(k) == x(k+1) )
        z(:,k) = max(z(:,k:k+1),[],2);
    end
end
for k=1:m-1
    if( y(k) == y(k+1) )
        z(k,:) = max(z(k:k+1,:),[],1);
    end
end
[x,iax,~] = unique(x);
[y,iay,~] = unique(y);
z = z(iay,iax);

[Elin]

Thank you!