Find the last non-nan in large array

I have very large 4-dimensional arrays with dimensions being time,depth, lat, lon. I want to find the linear index in the depth dimension to the last non-nan data value to I can evaluate the values at the "bottom". I am looking for the output to be 3-D (time, lat, lon) with linear indexes to the last depth with a value (non-nan).
Any idea how to do this efficiently? I have a bunch of very large files to run this on.

 Accepted Answer

One way:
%m: a 4D matrix (time x depth x lat x lon
[~, idx] = max(cumsum(~isnan(m), 2), [], 2)
This relies on the fact that max returns the index of the first max value if several are identical. The first max value of cumsum(~isnan) is the last number before zero or more nan.

3 Comments

Thanks, that works! I knew there had to be a way without looping through each lat/lon/time triplet.
Hi, I was looking exactly the same, but when I run this code I get a 4D array and when plt is the number of values before the last NaN, but I want to plot the value,
what I am doing wrong?
[~, idx] = max(cumsum(~isnan(m), 3), [], 3) %??
Thanks
I am not clear as to what size your m is ? You should expect the output, idx, to be the same size as m, except with the third dimension now scalar.

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