How to plus two nan matrix
显示 更早的评论
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
采纳的回答
James Tursa
2017-5-16
c = a + b;
c(isnan(c)) = a(isnan(c));
c(isnan(c)) = b(isnan(c));
5 个评论
Cam Salzberger
2017-5-16
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
2017-5-17
min wong
2017-5-17
James Tursa
2017-5-17
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
2017-5-18
更多回答（2 个）
Paulo Neto
2018-11-28
0 个投票
3 个评论
James Tursa
2018-11-28
Did you mean element-wise division? c = (a+b)./b
Paulo Neto
2018-11-28
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
2018-11-28
What would you want the individual result to be if "a" is NaN, and if "b" is NaN?
Paulo Neto
2018-11-28
0 个投票
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?
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