NAN add works strange

Hi,

I got some question about Nan. As you can see in my code, NaN+1=Nan, but when I add two matrix value (one is Nan while the other one is -31),the sum is 0.

Could anyone help me out? I hope it could be either -31 or Nan (Or the best would be I can choose whether to get Nan or -31)

>> Temp_abs(1,1)

ans =

    -31

>> mean(Temp_anomali(1,1,:))

ans =

   NaN

>> mean(Temp_anomali(1,1,:))+Temp_abs(1,1)

ans =

      0

>> nan+1

ans =

   NaN

>> nan-31

ans =

   NaN

Many thanks

PS. The Temp_anomali(1,1,:) are all Nan.

[Jon]

Can you provide the values of Temp_anomali(1,1,:)?

[Star Strider]

What MATLAB version are you using? I cannot reproduce your results with this code in R2015b:

Temp_anomali(1,1,:) = [1:10 NaN 11:15];
Temp_abs(1,1) = -31;
Q1 = mean(Temp_anomali(1,1,:))
Q2 = mean(Temp_anomali(1,1,:))+Temp_abs(1,1)

Q1 =
   NaN

Q2 =
   NaN

Answer 1

[Steven Lord]

The class of Temp_abs is one of the four signed integer types (int8, int16, int32, or int64.) When you add a value of an integer type and a scalar double, the result is of the integer type. Since there are no bit patterns available to represent NaN in an integer type (all the bit patterns of the appropriate length are used in storing values between INTMIN and INTMAX for the class inclusive) the NaN gets converted into a zero of that integer class.

 x = NaN;
 y = int8(-32);
 z = x+y
 class(z)

You will see that z is 0 and its class is int8.

 w = -32;
 v = x+w
 class(v)

You will see that v is NaN and its class is double, since double plus double results in a double.

[SQ]

Thank you very much! This is the answer I need :)

Temp_abs is int 16 in my original document, which caused the problem!

I am just wondering, if it is possible to add Nan with a number, and the mean is the number?

For example, if Nan+3+5, mean =4

Thanks again!

[Steven Lord]

Are you using release R2015a or later of MATLAB?

 mean([NaN, 3, 5], 'omitnan')

If not, do you have Statistics and Machine Learning Toolbox (or Statistics Toolbox, if you have an older release?)

 nanmean([NaN, 3, 5])