How to return a 2D matrix containing the heighest non-NaN values from a 3D matrix

1 Ansicht (letzte 30 Tage)
Hi,
I have a large 3D matrix of data where in the 3rd dimension where some values are NaN and some values are non-NaN. Call it M with size m*n*p.
A 2D matrix P (size m*n) contains the number of non-NaN values held within the pages returned in the array M(m,n).
Also, take one value of P, t=P(m,n); then each of M(m,n,1:t) contains a non-NaN value, and each of M(m,n,t+1:end) contains a NaN value. That is, all the non-NaN values in M are stacked from the p=1 to p=nth pages.
An example of the data is created with this code:
P = ceil(rand(4,4) * 4);
for x = 1:4
for y = 1:4
for z = 1:4
if P(x,y) >= z
M(x,y,z) = ceil(rand*10);
else
M(x,y,z) = NaN;
end
end
end
end
For example:
M(:,:,1) =
9 10 4 4
10 4 10 3
3 3 10 9
3 8 5 9
M(:,:,2) =
9 NaN NaN 3
6 NaN 9 3
6 3 7 4
8 4 9 8
M(:,:,3) =
4 NaN NaN 9
5 NaN NaN NaN
10 NaN 10 4
8 NaN 4 4
M(:,:,4) =
NaN NaN NaN 9
7 NaN NaN NaN
1 NaN 2 NaN
2 NaN 10 NaN
I am looking for an efficient way to return the 'pth-most' values of M, which is essentially the last non-NaN value of each M(m,n) array. For the example dataset this would be:
result =
4 10 4 9
7 4 9 3
1 3 2 4
2 4 10 4
Does anyone have any clever suggestions?
Thank you Steven

Akzeptierte Antwort

Andrei Bobrov
Andrei Bobrov am 28 Okt. 2013
Bearbeitet: Andrei Bobrov am 29 Okt. 2013
s = size(M);
[i1,i2] = ndgrid(1:s(1),1:s(2));
result = M(sub2ind(s,i1,i2,s(3) - sum(isnan(M),3)));
ADD
s = size(M);
[i1,i2] = ndgrid(1:s(1),1:s(2));
i3 = s(3) - sum(isnan(M),3);
i3(i3==0)=1;
result = M(sub2ind(s,i1,i2,i3));
  3 Kommentare
Steven
Steven am 29 Okt. 2013
I wonder if you can help me further; how would you handle the situation where some of the arrays returned by M(m,n) are entirely NaN's?
In this instance, sum(isnan(M),3) == s(3), and the sub2ind function call will be badly formed?

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Matrices and Arrays finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by