# indices when the columns are all nan

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Sagar on 16 Mar 2016
Commented: Sagar on 17 Mar 2016
I am using [m, d] = nanmax (r_all); to get the maximum values and corresponding indices in matlab. There are several columns in r_all in which all the elements are nans. Unfortunately, nanmax returns '1' in indices for those columns that have all nans. However, there are actual indices that are value '1' so it creates confusion. How can I force matlab to return 'a' or 'b' as the output instead of returning 0' or '1'? Could you please suggest other solutions to this problem?

Sagar on 16 Mar 2016
But how can I apply for loop within [m d], I cannot do m(i,j) or d(i,j) right? I can set the nan values with any other values in the matrix r_all, but the problem remains the same, i.e. nanmax still returns 1 for the maximum index in d because all the values in many columns are nans.
KSSV on 16 Mar 2016
Sagar on 16 Mar 2016

Stephen Cobeldick on 16 Mar 2016
Edited: Stephen Cobeldick on 16 Mar 2016
There is no need to waste time using slow and ugly loops and if statements. You can locate those "incorrect" indices very simply using basic MATLAB indexing:
>> mat = [0,1,2,NaN,3,NaN;4,NaN,5,NaN,6,NaN;7,8,NaN,NaN,NaN,9]
mat =
0 1 2 NaN 3 NaN
4 NaN 5 NaN 6 NaN
7 8 NaN NaN NaN 9
>> [m,idx] = nanmax(mat) % max value and its indices
m =
7 8 5 NaN 6 9
idx =
3 3 2 1 2 3
>> idx(all(isnan(mat),1)) = NaN % replace all-NaN indices with NaN
idx =
3 3 2 NaN 2 3

Guillaume on 16 Mar 2016
+1, loops and branches are completely unnecessary
A slightly simpler implementation would be:
[m, idx] = nanmax(mat);
idx(isnan(m)) = nan;
By the way, since 2015b (iirc) max ignores nan by default.
Sagar on 17 Mar 2016
Great, I like Guillaume's answer. :)
Sagar on 17 Mar 2016
Thanks Stephen, apparently I didn't know there is 'all' function which I needed. :)

KSSV on 16 Mar 2016
clc; clear all ;
[m,n] = size(data) ;
idx = zeros(n,1) ;
val = idx ;
% column wise
for c = 1:n
[i,j] = nanmax(data(:,c)) ;
if isnan(i)
j = 50 ;
end
idx(c) = j ;
val(c) = i ;
end