how to use if when there is NAN

Hi all

I have a time series price matrix as follows. At each time, I need to judge if the current price is larger than the average of the last two prices in its same column, if yes, make it 1, otherwise it is 0. When there is NaN included in the if judgement, make it NaN. e.g, w(3,1) and w(4,1) are both NaN because we can not compare if w(4,1) larger than the avarage of w(3,1)and w(2,1)because p(2,1) is NaN and the comparison is therefore invalid. I tried to use for for loop but it makes the processing kind of complicated and it seems not be working

for i=1:8
    for j=1:3
        w(i,j)=(a(i,j)>mean(a(i-1:i-2,j)))
    end
end

I am seeking for a more efficient function. Thank you very much. BR, Jessie

p=
  NaN  NaN  3
  NaN  2  4
  2  1  5
  1  1  1
  2  3  2
  3  0  NaN
  1  2  NaN
  4  NaN  NaN

so the expected result is

w=
NaN  NaN  NaN
NaN  NaN  NaN
NaN  NaN  1
NaN  0  0
1  1  0
1  0  1
0  1  NaN
1  1  NaN

Answer 1

[dpb]

Yeah, compute the means first and then do the comparison over the valid regions. You can augment sizes later however you so choose...

>> wt=[0.5 0.5];             % filter weights for window 2 means
>> mn=filter(wt,1,p,nan)     % compute means w/ valid ranges only
mn =
     NaN       NaN       NaN
     NaN       NaN    3.5000
     NaN    1.5000    4.5000
  1.5000    1.0000    3.0000
  1.5000    2.0000    1.5000
  2.5000    1.5000       NaN
  2.0000    1.0000       NaN
  2.5000       NaN       NaN
>> w=double(p(3:end,:)>mn(2:end-1,:));  % weighting comparison over valid range
>> w(isnan(p(3:end,:))|isnan(mn(2:end-1,:)))=nan  % identify missing value locations
w =
 NaN   NaN     1
 NaN     0     0
   1     1     0
   1     0   NaN
   0     1   NaN
   1   NaN   NaN
>>

NB: There are only N-2 valid locations; the available data from the third to the last is compared to the first valid mean (2nd in the array) to the next-to-last since there's no observation against which to compare the last valid computed mean. This leaves a computed array of six observations from your initial eight points; as noted you can adjust this however needed to fit the remainder of the application by augmenting whichever end needs it for desired alignment.

[dpb]

BTW, I think your posited answer has a couple of mistakes in it...