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replace missing value in a matrix using intropolated values

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Skirt Zhang
Skirt Zhang 2013 年 1 月 16 日
Dear All,
I have a matrix A as below: A =
1.1093 -0.7697 1.1006 -0.6156 0.4882 -0.8045 0.1049
NaN NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN
-1.2141 1.1174 -1.4916 0.8886 1.4193 -0.2437 -0.6669
-1.1135 -1.0891 -0.7423 -0.7648 0.2916 0.2157 0.1873
-0.0068 NaN -1.0616 -1.4023 0.1978 -1.1658 -0.0825
1.5326 0.5525 NaN -1.4224 1.5877 -1.1480 NaN
I want to replace the NaN with the values generated from interpolation which can take care of both the interpolation for rows and column. So far I have in mind is take interpolation for row get value a_row; then do it for column get a_column. In the end replace nan with the mean value for the two. Can anyone help me about this? Also, I have problems with specifying y for the function : output(a)= interp1(A,y,'linear');
if I don't know y how should I specify here?
thanks a lot in advance

回答 (1 件)

Andrei Bobrov
Andrei Bobrov 2013 年 1 月 16 日
編集済み: Andrei Bobrov 2013 年 1 月 16 日
one way
t = ~isnan(A);
[x,y] = find(t);
F = TriScatteredInterp(x,y,A(t));
[ii,jj] = ndgrid(1:size(A,1),1:size(A,2));
out = F(ii,jj);
ADD variant
out(end) = mean(cellfun(@(x)interp1(1:2,x(1:2),3,'linear','extrap'),...
{out(end,end-[2 1 0])', out(end-[2 1 0],end),...
out(end - (size(out,1)+1)*(2:-1:0))'}));
  2 件のコメント
Skirt Zhang
Skirt Zhang 2013 年 1 月 16 日
Hi Bobrov, thanks for your reply. But now I still have NaNs in the output which are located in the end of the matrix. Do you have any idea to solve this problem
Andrei Bobrov
Andrei Bobrov 2013 年 1 月 16 日
Hi Skirt! See ADD part in my answer.

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