Taking weighted means over matrices with missing data
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I have a matrix with some NaN entries. I want to calculate the weighted averages of each row, according to a supplied row vector of weights. Where a row has one or more missing values, I simply want to scale up the remaining weights. This is easy enough to do as a for loop over all rows; but I was wondering if there's a simple way to do so using just matrix operations.
Illustration: say my matrix is:
[4 NaN 1 NaN; 5 3 8 NaN; 1 6 2 4; 8 4 7 2]
The weight vector is:
[0.4 0.3 0.2 0.1]
I want to return the column vector:
[3; 5; 3; 6]
For the last row, 6 = 8*0.4 + 4*0.3 + 7*0.2 + 2*0.1. For the first row, 3 = (4*0.4 + 1*0.2)/(0.4 + 0.2). And so on.
Is this possible using matrix operations over the whole matrix at once, or will it have to be done row by row in a loop?
0 Kommentare
Antworten (4)
Teja Muppirala
am 14 Mai 2013
M = [4 NaN 1 NaN; 5 3 8 NaN; 1 6 2 4; 8 4 7 2];
w = [0.4 0.3 0.2 0.1];
W = bsxfun(@times,~isnan(M),w);
W = bsxfun(@rdivide,W,sum(W,2));
M(isnan(M)) = 0;
sum(M.*W,2)
ans =
3.0000
5.0000
3.0000
6.0000
1 Kommentar
Teja Muppirala
am 14 Mai 2013
A bit simpler:
W = bsxfun(@times,~isnan(M),w);
M(isnan(M)) = 0;
sum(M.*W,2)./sum(W,2)
David Sanchez
am 14 Mai 2013
I think it is not possible, but you can write your own function to do it.
0 Kommentare
Yao Li
am 14 Mai 2013
A=[4 NaN 1 NaN; 5 3 8 NaN; 1 6 2 4; 8 4 7 2];
A_u=A./A;
A(isnan(A))=0;
A_u(isnan(A_u))=0;
weight=[0.4 0.3 0.2 0.1];
y=weight*(A')./(weight*A_u')
1 Kommentar
Teja Muppirala
am 14 Mai 2013
This will have some trouble if any of the elements in A are zero, because 0./0 equals NaN.
Andrei Bobrov
am 14 Mai 2013
M = [4 NaN 1 NaN; 5 3 8 NaN; 1 6 2 4; 8 4 7 2];
w = [0.4 0.3 0.2 0.1]';
t = isnan(M);
M(t) = 0;
out = (M*w)./(~t*w);
0 Kommentare
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!