Calculate difference between two columns but skip if column has NaN

I have a very large dataset (it's a table, name it MX for the example) for which I want to know the difference between two columns (one has a value of pre surgery, one a value of post surgery). For some rows one of the two values is missing, filled with NaN, and I want to skip the calculation if either one or both of the columns have NaN. I cannot delete the subjects with NaN from the dataset since I do need them for other calculations.

So for example the dataset is like this,

1 2

NaN 24

49 32

NaN 20

NaN NaN

33 NaN

45 16

36 42

64 18

34 NaN

49 NaN

55 34

64 16

And I want the difference between 1 - 2. I tried it with an if loop but got an error saying 'Operands to the logical and (&&) and or (||) operators must be convertible to logical scalar values (which is logic if it tries to do the calculation on the 'NaN').

Made a new column first with only zeros to fill the result of the loop:

MX.difference = zeros(length(MX.1),1);

if isnan(MX.1) && isnan(MX.2)

MX.difference(MX.1 - MX.2);

end

But got the error here as described above. Also tried it with an else statement to say that if one of the two or both conditions contains NaN, to write 'NaN' in the 'difference' column, but I still get the same error:

MX.difference = zeros(length(MX.1),1);

if isnan(MX.1) && isnan(MX.2)

MX.difference(MX.1 - MX.2);

else

MX.difference = 'NaN';

end

How do I fix this and make this work? Hope someone can help! I could create another script where I delete all the 'NaN' rows first but there must be a way to include it in this script without having to delete any rows. Thanks!

[Torsten]

Change && to &

Answer 1

[KSSV]

You simply subtract the two columns....if any of the column has NaN, your difference will be NaN.

A = [NaN                24
    49                32
   NaN             20
   NaN            NaN
    33               NaN
    45              16
    36              42
    64              18
    34              NaN
    49              NaN
    55               34
    64               16] ;
dA = A(:,1)-A(:,2)

dA = 12×1

NaN 17 NaN NaN NaN 29 -6 46 NaN NaN

You can access the NaN's using isnan

[YaaW]

I was thinking way too difficult, thanks!

[KSSV]

Thanks is accepting/ voting the answer. :)