replacing NaN with it previous element

 Dear all,

I have

K={
    'IT'     'ooil'    'TTT'    [  0]    [4.1583]
    'IT'     'ooil'   'TTT'    [  0]    [4.0339]
    'IT'     'ooil'    'TTT'    [  0]    [3.9389]
    'IT'     'ooil'    'TTT'    [  0]    [3.9676]
    'ITn'     'ooilb'    'TTT'    [  0]    [3.9313]
    [NaN]    [      NaN]    [NaN]    [  0]    [3.9313]
    'IT'     'ooil'    'TTT'    [  0]    [3.8720]
    'ITcv'     'ooila'     'TTT'    [  0]    [3.8829]
    [NaN]    [      NaN]    [NaN]    [  0]    [3.9313]
    'IT'     'ooil'    'TTT'    [  0]    [3.8681]
    'ITf'     'ooilg'     'TTT'    [  0]    [4.3944]
    [NaN]    [      NaN]    [NaN]    [  0]    [3.9313]}

And I want to replace the NaNs in each column with its previous element of the same column

That is,

K={
  'IT'     'ooil'    'TTT'    [  0]    [4.1583]
  'IT'     'ooil'   'TTT'    [  0]    [4.0339]
  'IT'     'ooil'    'TTT'    [  0]    [3.9389]
  'IT'     'ooil'    'TTT'    [  0]    [3.9676]
  'ITn'     'ooilb'    'TTT'    [  0]    [3.9313]
  'ITn'     'ooilb'    'TTT'    [  0]    [3.9313]
  'IT'     'ooil'    'TTT'    [  0]    [3.8720]
  'ITcv'     'ooila'     'TTT'    [  0]    [3.8829]
  'ITcv'     'ooila'     'TTT'    [  0]    [3.9313]
  'IT'     'ooil'    'TTT'    [  0]    [3.8681]
  'ITf'     'ooilg'     'TTT'    [  0]    [4.3944]
  'ITf'     'ooilg'     'TTT'    [  0]    [3.9313]}

Thanks in advance

Answer 1

[Azzi Abdelmalek]

 res=K,
 for k=1:size(K,1);
     if isnan(K{k,1});
       res(k,1:3)=K(k-1,1:3);
     end;
 end

[antonet]

thank you all!

Answer 2

[Isktaine]

Save this function then use it on K:

function K = RemoveNaNs(K)

[rows,cols]=size(K); %Works out how many rows and columns there are
for p=1:rows
    for q=1:cols
        if isnan(K{p,q})==1 %If there is a NaN entry
            K{p,q}=K{p-1,q}; %Replace it with the entry from the previous row.
        end
    end
end

end

Is this clear?

[Sean de Wolski]

What if there are two nans adjacent?

[John Petersen]

That wouldn't be a problem because you've already replaced the 1st nan. The only problem would be a nan in the very 1st line. Then p-1=0 and you would get an error. So p needs to start at 2 and there needs to be an initialization in case the first row has a nan.

[Isktaine]

Yeah John is right. I didn't know what antonet would want the value to be if there was a NaN in the first row, so I didn't address this.

Thanks Lucas! :)

Answer 3

[Thomas]

   K={

   'IT'     'ooil'    'TTT'    [  0]    [4.1583]
      'IT'     'ooil'   'TTT'    [  0]    [4.0339]
      'IT'     'ooil'    'TTT'    [  0]    [3.9389]
      'IT'     'ooil'    'TTT'    [  0]    [3.9676]
      'ITn'     'ooilb'    'TTT'    [  0]    [3.9313]
      [NaN]    [      NaN]    [NaN]    [  0]    [3.9313]
      'IT'     'ooil'    'TTT'    [  0]    [3.8720]
      'ITcv'     'ooila'     'TTT'    [  0]    [3.8829]
      [NaN]    [      NaN]    [NaN]    [  0]    [3.9313]
      'IT'     'ooil'    'TTT'    [  0]    [3.8681]
      'ITf'     'ooilg'     'TTT'    [  0]    [4.3944]
      [NaN]    [      NaN]    [NaN]    [  0]    [3.9313]}

 % needs only 3 lines of code

A=K;
[r,c]=find(cellfun(@(V) any(isnan(V(:))), A));
K(r,c)=K(r-1,c)

[Isktaine]

That's certainly more brief than my code, could you explain what V is?

[Thomas]

@V creates an anonymous function. The cellfun documentation has more on it.. http://www.mathworks.com/help/techdoc/ref/cellfun.html