# Substracting matrix with NaN values

2 次查看（过去 30 天）
FC93 2017-3-8

I have 4 matrix A, B, C and D. All matrix have NaN values, and matrix C and D have a lot of NaN values. Now I want to take A and substract B, C and D from it. I want that the result matrix contains only a NaN value if A has NaN or if A has a value but B, C and D do all have a NaN value. For example if a specific cell in A has a value, and the same cell has also a value in B but not in C and D (because there is an NaN) I want that matlab takes the NaN in C and B as 0. The matrix are really big so I con not do it by hand.
Example:
A=[ 1 2 3; 4 5 6; 7 NaN NaN] B =[ 1 2 3; 4 5 6; 7 2 NaN] C = [ 1 NaN 3; 4 5 6; 7 1 NaN] D = [ 1 2 3; 4 NaN 6; 7 1 NaN]
then A-B-C-D=E should be E=[ -2 -2 -6; -8 -5 -12; -14 NaN NaN]

### 采纳的回答

Walter Roberson 2017-3-8
mask = isnan(B) & isnan(C) & isnan(D);
tB = B; tB(isnan(B)) =0;
tC = C; tC(isnan(C)) =0;
tD = D; tD(isnan(D)) =0;
E = A - tB - tC - tD;
Any nan in A will be carried through.
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FC93 2017-3-8
Thank you for your help. This is exactly what I need.

### 更多回答（1 个）

KSSV 2017-3-8

A=[ 1 2 3; 4 5 6; 7 NaN NaN] ;
B =[ 1 2 3; 4 5 6; 7 2 NaN] ;
C = [ 1 NaN 3; 4 5 6; 7 1 NaN] ;
D = [ 1 2 3; 4 NaN 6; 7 1 NaN] ;
% E=[ -2 -2 -6; -8 -5 -12; -14 NaN NaN] ;
R(:,:,1) = A ;
R(:,:,2) = -B ;
R(:,:,3) = -C ;
R(:,:,4) = -D ;
iwant = nansum(R,3)
You have to cross check your E once.
Other method using loops:
A=[ 1 2 3; 4 5 6; 7 NaN NaN] ;
B =[ 1 2 3; 4 5 6; 7 2 NaN] ;
C = [ 1 NaN 3; 4 5 6; 7 1 NaN] ;
D = [ 1 2 3; 4 NaN 6; 7 1 NaN] ;
E=[ -2 -2 -6; -8 -5 -12; -14 NaN NaN] ;
R(:,:,1) = A ;
R(:,:,2) = -B ;
R(:,:,3) = -C ;
R(:,:,4) = -D ;
% iwant = nansum(R,3)
[m,n,p] = size(R) ;
iwant = NaN(m,n) ;
for i = 1:m
for j = 1:n
k = squeeze(R(i,j,:)) ;
k = k(~isnan(k)) ;
if ~isempty(k)
iwant(i,j) = sum(k) ;
end
end
end
FC93 2017-3-8

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