a small problem with interpolation

Question

Sabbas 2012-8-11

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/45810-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?

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Oleg Komarov 2012-8-11

编辑：Oleg Komarov 2012-8-11

在 MATLAB Online 中打开

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.

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Answer 1

Oleg Komarov 2012-8-11

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/45810-a-small-problem-with-interpolation#answer_56006

编辑：Oleg Komarov 2012-8-11

在 MATLAB Online 中打开

It appears to be a bug in this contribution.

You have two options:

report this to the author on the FEX and wait for HIS answer.
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');

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Oleg Komarov 2012-8-11

在 MATLAB Online 中打开

@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 2012-8-11