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How to use inerpolation option with end values? If I use interp1 command it's giving NaN as the output if input is out of the range. I want to use end values if input is out of the range?

조회 수: 108 (최근 30일)
1 120
2 5000
3 12000
My input is something like this. If i give 4 as my input currently it's giving NaN as the output but I want output to be 12000(i.e end value)

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Stephen23
Stephen23 2018년 8월 7일
편집: Stephen23 2018년 8월 7일
>> X = [1,2,3];
>> Y = [12,5000,12000];
>> interp1(X,Y,4,'nearest','extrap')
ans = 12000
>> interp1(X,Y,4,'linear','extrap')
ans = 19000
But note that you cannot pick a method like linear, spline, etc, and also limit the extrapolation to the end values:
>> interp1(X,Y,[0,1.5,4],'linear','extrap') % linear -> not end values
ans =
-4976 2506 19000
>> interp1(X,Y,[0,1.5,4],'nearest','extrap') % nearest -> not linear interpolated
ans =
12 5000 12000
But you can easily select the interpolation method and limit to the end values, by simply using max and min:
>> max(Y(1),min(Y(end),interp1(X,Y,[0,1.5,4],'linear','extrap')))
ans =
12 2506 12000
If you only need to extrapolate in one direction then you could set the extrapolation value:
>> interp1(X,Y,[1.5,4],'linear',Y(end))
ans =
2506 12000
  댓글 수: 2
Parthiban C
Parthiban C 2018년 8월 10일
Thank you for your response. This one worked for my problem. max(Y(1),min(Y(end),interp1(X,Y,[0,1.5,4],'linear','extrap'))) ans = 12 2506 12000
Christian
Christian 2021년 4월 19일
This min/max stuff does not work that way, if Y is not monotonously increasing, let's say
>> Y = [12,5000,4000];
Then I get this:
>> max(Y(1),min(Y(end),interp1(X,Y,[0,1.5,4],'linear','extrap')))
ans =
12 2506 3000
And I guess for X=4 we want to see 4000 instead of 3000.
In general monotony assumptions should only be made towards X, not towards Y.

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