Index of NaN Values
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Tiago Dias
am 23 Jul. 2018
Bearbeitet: Walter Roberson
am 23 Jul. 2018
Hello,
My objective is to know the index of NaN values per column.
y = [1,2;NaN,4;5,NaN;7,8;9,NaN]
idx = isnan(y)
for j = 1:m
for i = 1:n
if idx(i,j) == 1
idx_all(i,j) = i
end
end
end
I get a result of
idx_all = 0 0
2 0
0 3
0 0
0 5
I would like the result to be something like
idx = [2 3
0 5]
Because the 1st column has just one NaN on row 2, and the 2nd column has NaN for row 3 and 5.
Thanks.
2 Kommentare
jonas
am 23 Jul. 2018
Is it really necessary to save the indices in that form? You should read about the find function.
Akzeptierte Antwort
Walter Roberson
am 23 Jul. 2018
A loop is easiest.
nc = size(y,2);
idx = zeros(0, nc);
for C = 1 : nc
nidx = find(isnan(y(:,C)));
idx(1:length(nidx), C) = nidx;
end
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