Interpolating within time series

Question

Luis Eduardo Cofré Lizama 2017-4-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/336233-interpolating-within-time-series

编辑： Stephen23 2017-4-20

I need to replace NaNs in a long time series that has consecutive (varying lengths) and non-consecutive NaNs. I tried the solution in this link https://au.mathworks.com/matlabcentral/answers/160575-replacing-nan-values-with-average-values-of-the-nearest-numbers but it doesn't seem to work for all my NaNs (not familiar with the bsxfun @(x,y)). Thanks for your help,

60000000000
53000000000
NaN
85000000000
34000000000
98000000000
46000000000
29000000000
09000000000
12000000000
NaN
NaN
23000000000
20000000000
32000000000
46000000000
27000000000
50000000000
72000000000
23000000000
48000000000
55000000000
79000000000
55000000000
NaN
NaN
NaN
07000000000
54000000000

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Answer 1

Stephen23 2017-4-20

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链接

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https://ww2.mathworks.cn/matlabcentral/answers/336233-interpolating-within-time-series#answer_263704

在 MATLAB Online 中打开

>> idx = ~isnan(A);
>> interp1(find(idx),A(idx),(1:numel(A))')
ans =
       1438.6
       1446.5
       1436.7
       1426.8
       1414.3
         1420
       1418.5
       1420.3
       1426.1
       1422.1
       1436.2
       1450.2
       1464.2
       1459.2
       1463.3
       1474.5
       1460.3
       1441.5
       1442.7
       1446.2
       1443.5
       1444.6
       1443.8
       1444.6
       1454.4
       1464.3
       1474.2
       1484.1
       1482.5

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Answer 2

KSSV 2017-4-20

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/336233-interpolating-within-time-series#answer_263701

在 MATLAB Online 中打开

A = [1438.60000000000
1446.53000000000
NaN
1426.85000000000
1414.34000000000
1419.98000000000
1418.46000000000
1420.29000000000
1426.09000000000
1422.12000000000
NaN
NaN
1464.23000000000
1459.20000000000
1463.32000000000
1474.46000000000
1460.27000000000
1441.50000000000
1442.72000000000
1446.23000000000
1443.48000000000
1444.55000000000
1443.79000000000
1444.55000000000
NaN
NaN
NaN
1484.07000000000
1482.54000000000] ;
% positions 
idx = 1:length(A) ;
% get NaN positions 
idx_nan = find(isnan(A)) ;
% non nan's positions 
idx_nonnan = find(~isnan(A)) ;
% do interpolation 
A_nonnan = A(idx_nonnan) ;
A_nan = interp1(idx_nonnan,A_nonnan,idx_nan) ;
%%replace NaNs
iwant = A ;
iwant(idx_nan) = A_nan ;
% plot
plot(idx,A,'r') 
hold on
plot(idx,iwant,'b') ;
legend('with Nans', 'NaNs filled')