sum two variables ignoring NaN

I have table T

Var1 Var2
--------------
1      NaN
NaN  2

I want to add var1 and var2 ignoring NaN:

sum 
---
1
2

Answer 1

[Image Analyst]

Try this:

var1 = [1;2;3;nan];
var2 = [4;nan;6; 15];
t = table(var1, var2)
% Initialize with var1
theSum = t.var1;
% If any are nan, replace with value from var2.
badRows = isnan(theSum)
theSum(badRows) = t.var2(badRows)
% If both are not nan, make it the sum
goodRows1 = ~isnan(t.var1)
goodRows2 = ~isnan(t.var2)
goodRows = goodRows1 & goodRows2;
theSum(goodRows) = t.var1(goodRows) + t.var2(goodRows)

You get:

t =

4×2 table

var1 var2

____ ____

1 4

2 NaN

3 6

NaN 15

theSum =

5

2

9

15

If one is nan, you get the other one which is not a nan.

If both are nans, you get a nan.

If both are not nan, you get the sum.

[Image Analyst]

You said in your comment to David that my solution also gives 0 instead of nan (which you want) when both are nan. That is not true. Look with this new sample data where there is a row where both are nan.:

var1 = [1;2;3;nan; nan];
var2 = [4;nan;6; 15; nan];
t = table(var1, var2)
% Initialize with var1
theSum = t.var1;
% If any are nan, replace with value from var2.
badRows = isnan(theSum)
theSum(badRows) = t.var2(badRows)
% If both are not nan, make it the sum
goodRows1 = ~isnan(t.var1)
goodRows2 = ~isnan(t.var2)
goodRows = goodRows1 & goodRows2;
theSum(goodRows) = t.var1(goodRows) + t.var2(goodRows)

You get:

theSum =

5

2

9

15

NaN

and you can see the last row, where both are nan, has a nan for the 4th element of the output vector.

[alpedhuez]

I meant

var1 = [1; 2; 3; nan; nan];
var2 = [4; nan; 6; 15; nan];
Var1 = t.var1
Var2 = t.var2
theSum = sum([Var1, Var2], 2, 'omitnan')

does not provide NaN.

Answer 2

[David Goodmanson]

Hi alpedhuez,

sum(Var1,Var2,2,'omitnan')

Here the '2' indicates that you are summing rows rather than columns (the default is summing columns). If all the entries in a row are nan, then you get 0.

[Image Analyst]

Didn't work. Forgot brackets. I think you meant:

var1 = [1; 2; 3; nan; nan];
var2 = [4; nan; 6; 15; nan];
Var1 = t.var1
Var2 = t.var2
theSum = sum([Var1, Var2], 2, 'omitnan')

The difference from mine is that if both are nan's, mine will give a nan, while yours gives a zero.

[alpedhuez]

No I think Image Analyst also gives 0. I want NaN when both are NaNs.