a small problem with interpolation

Dear all,

I have

   A={ [NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [4.8455]    [0.8076]    [0.0864]    [ 0.1917]    [ 0.5184]    [ 0.4275]
    [5.1224]    [0.8537]    [0.3679]    [ 0.7887]    [ 2.2073]    [ 1.8883]
    [5.1932]    [0.8656]    [0.6473]    [ 1.3217]    [ 3.8835]    [ 3.3626]
    [5.2023]    [0.8671]    [0.8252]    [ 1.6275]    [ 4.9510]    [ 4.2929]
    [5.1816]    [0.8636]    [0.7828]    [ 1.6663]    [ 4.6970]    [ 4.0567]
    [5.1818]    [0.8636]    [0.6551]    [ 1.4768]    [ 3.9308]    [ 3.3937]
    [5.1937]    [0.8656]    [0.6070]    [ 1.4121]    [ 3.6422]    [ 3.1528]
    [5.1665]    [0.8611]    [0.6443]    [ 1.5050]    [ 3.8656]    [ 3.3284]
    [5.1786]    [0.8631]    [0.6824]    [ 1.5289]    [ 4.0944]    [ 3.5341]
    [5.0321]    [0.8386]    [0.6997]    [ 1.5754]    [ 4.1985]    [ 3.5209]
    [4.9381]    [0.8231]    [0.7431]    [ 1.6090]    [ 4.4583]    [ 3.6695]
    [4.9326]    [0.8221]    [0.7708]    [ 1.7676]    [ 4.6245]    [ 3.8018]
    [4.7815]    [0.7970]    [0.8851]    [ 2.0084]    [ 5.3107]    [ 4.0625]
    [3.9136]    [0.6523]    [2.5109]    [ 5.3401]    [15.0656]    [ 9.7846]
    [3.8419]    [0.6403]    [4.3700]    [ 8.2200]    [26.2199]    [16.7885]
    [3.8559]    [0.6426]    [4.3385]    [ 7.9125]    [26.0312]    [16.7235]
    [3.8803]    [0.6467]    [3.9299]    [ 7.9730]    [23.5793]    [15.2499]
    [3.8741]    [0.6457]    [3.8246]    [ 8.2150]    [22.9478]    [14.8167]
    [3.8733]    [0.6456]    [3.9232]    [ 8.2825]    [23.5394]    [15.1959]
    [3.8604]    [0.6434]    [3.9364]    [ 8.4879]    [23.6182]    [15.1961]
    [3.8052]    [0.6342]    [3.9610]    [ 8.4783]    [23.7659]    [15.0720]
    [3.8177]    [0.6363]    [4.0807]    [ 8.3607]    [24.4842]    [15.5765]
    [3.7615]    [0.6269]    [5.4267]    [10.9853]    [32.5598]    [20.4115]
    [3.6900]    [0.6150]    [5.8236]    [12.4549]    [34.9416]    [21.4899]
    [3.7249]    [0.6208]    [6.0688]    [12.3084]    [36.4125]    [22.6080]
    [3.7621]    [0.6270]    [6.4914]    [12.5395]    [38.9481]    [24.4177]
    [3.7485]    [0.6248]    [7.1611]    [12.7942]    [42.9669]    [26.8452]
    [3.7492]    [0.6249]    [7.1063]    [12.8555]    [42.6379]    [26.6462]
    [3.7029]    [0.6171]    [6.6088]    [12.9124]    [39.6529]    [24.4730]
    [3.7172]    [0.6195]    [6.3574]    [12.7681]    [38.1441]    [23.6259]
    [3.8148]    [0.6358]    [6.2090]    [12.7283]    [37.2539]    [23.6851]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]}

I want to extrapolate the values in the last row.

I tried

 out3=[];
 outtx=CELL2MAT(A)
 for c = 1:size(outtx,2)

     outtxa=inpaint_nans(outtx(:,c)',2);
     outtxa=outtxa';
end
out3=[out3       outtxa ];

but I get the following error

 ?? Subscript indices must either be real positive integers or logicals.

 Error in ==> inpaint_nans at 233
    fda(n,[n, n-1,n+n])=[-2 1 1];

thanks

PS: is there also a way to find the values in the first 4 rows?

[Oleg Komarov]

Please, store your A in a double matrix.

A = cell2mat(A);

And keep it as double. Storing a double matrix in that ways wastes memory, it's annoying since you always have to convert with cell2mat and it's harder to work with.

Answer 1

[Oleg Komarov]

It appears to be a bug in this contribution.

You have two options:

1. report this to the author on the FEX and wait for HIS answer.
2. Use interp1() with the 'extrap' option

% From my previous answer to your post

xi  = (1:size(A,1))';

% Index the non NaN

idx = ~isnan(A(:,1));

out = interp1(xi(idx),A(idx,:),xi,'linear','extrap');

[John D'Errico]

It looks like vector input seems to be causing problems, though it was not a problem in earlier releases that I knew of, so it may have been version related. The release I sent in just now fixes the problem.

[Sabbas]

Where can I get that release? Because I think that I still get the same problem

[Sabbas]

I also tried

 A=cell2mat(A);
full=A;
for c = 1:size(A,2)

 idx =  isnan( A(:,c));
 full(idx,c) = interp1(A(~idx,1),A(~idx,c),A(idx,1),'linear','extrap');
end

 but I get no extrapolated values

[Oleg Komarov]

@Sabbas: which A are you using, not the same in the example?

I suppose you're looping because NaNs appear at different locations, if not, then no need to loop.

Here follows the polished code from your first comment:

A = cell2mat(A);
xi  = (1:size(A,1))';
for c = 1:size(A,2)
    idx        = ~isnan(outtx(:,c));
    outtx(:,c) = interp1(xi(idx),A(idx,c),xi,'linear','extrap');
end

It doesn't work because some of your series could be all NaN or have just one data point.

When it error check the index of the loop and control that the column has at least 2 data points. However, I would recommend checking this BEFORE applying interpolation. Generally, series which do not have enough observations are discarded in toto.

[Sabbas]

Ok, I see the problem now. thanks