How to plus two nan matrix

If my two matrix are

a=3*3,b=3*3,c=3*3

a=[1 2 3,4 5 nan,7 nan 9]

b=[nan 3 3,4 5 6,nan 8 9]

c=a+b

Is that possible that c=[1 5 6,8 10 6,7 8 18]

How to calculate?

Thanks

Answer 1

[James Tursa]

c = a + b;
c(isnan(c)) = a(isnan(c));
c(isnan(c)) = b(isnan(c));

[Cam Salzberger]

That should do it. Could also just pre-process "a" and "b" with:

a(isnan(a)) = 0;
b(isnan(b)) = 0;
c = a+b;

but then you'll end up with 0 in places that both "a" and "b" are NaN. I like James' answer better.

[min wong]

thank you!

[min wong]

But if want to average the matrix

Hoe to do it?

like c=(a+b)/2;

[James Tursa]

Depending on what you want to have happen to the "NaN" spots, I am guessing you will want either

c = (a + b)/2;
c(isnan(c)) = a(isnan(c));
c(isnan(c)) = b(isnan(c));

or

c = (a + b)/2;
c(isnan(c)) = a(isnan(c))/2;
c(isnan(c)) = b(isnan(c))/2;

[min wong]

Thanks !!

Answer 2

[Paulo Neto]

Hi James Tursa,

How I can do this: c = (a + b)/b

Regards

[James Tursa]

Did you mean element-wise division? c = (a+b)./b

[Paulo Neto]

I'm considering a and b as arrays:

a=3*3,b=3*3,c=3*3

a=[1 2 3,4 5 nan,7 nan 9]

b=[nan 3 3,4 5 6,nan 8 9]

To calculate c = a + b, I'm using your routine:

c(isnan(c)) = a(isnan(c));

c(isnan(c)) = b(isnan(c));

and results in c=[1 5 6,8 10 6,7 8 18]

But I need to calculate: (a+b)./b, can I use the same method?

[James Tursa]

What would you want the individual result to be if "a" is NaN, and if "b" is NaN?

Answer 3

[Paulo Neto]

I'm considering a and b as arrays:

a=3*3,b=3*3,c=3*3

a=[1 2 3,4 5 nan,7 nan 9]

b=[nan 3 3,4 5 6,nan 8 9]

To calculate c = a + b, I'm using your routine:

c(isnan(c)) = a(isnan(c));

c(isnan(c)) = b(isnan(c));

and results in c=[1 5 6,8 10 6,7 8 18]

But I need to calculate: (a+b)./b, can I use the same method?