An annoying extrapolation problem

Question

Mohammad Shojaei Arani 2023-12-4

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https://ww2.mathworks.cn/matlabcentral/answers/2056064-an-annoying-extrapolation-problem

编辑： Dyuman Joshi 2023-12-4

Hello,

I have a simple problem but it seems difficult to solve. MATLAB command interp1 has a nice extrapolation option but, unfortunately

it only considers just one extrapolation value outside of our domain. Assuming that my interval is [L R], I would like to extrapolate my function

f to be f(L) when x<L and also I want to extrapolate f with f(R) when x>R. This works but I cannot vectorise this. And, this is quite annoying. To make it clear what is my point the best is to go for a simple example. Consider the following

x1=linspace(0,pi/2,1000);y1=sin(x1);

f1=@(x)interp1(x1,y1,x,'spline');

f=@(x)interp1(x1,y1,x,'spline',0).*(x>=0 && x<=pi/2)+f1(0).*(x<0)+f1(pi/2).*(x>pi/2);

If I type f(3) or f(4) then I get 1. But, when I type f([3 4]) it, unfortunately, throws an error message. I know why it gives error message but

I cannot fix this.

Note: I COULD WRITE f1=@(x)interp1(x1,y1,x,'spline','extrap'); BUT I DO NOT LIKE TO USE THIS SINCE THIS IS NOT

APPROPRIATE FOR THE NATURE OF MY ACTUAL PROBLEM (NOT SHOWN). THE EXAMPLE ABOVE IS ONLY TO SHOW YOU MY PROBLEM.

Thanks for your help in advance!

Babak

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Dyuman Joshi 2023-12-4

编辑：Dyuman Joshi 2023-12-4

I am aware that Bruno has provided an efficient solution to the problem, but I'll make a different point nonetheless -

"I know why it gives error message but I cannot fix this."

Look into the difference between &&, || and &, | (double vs single)

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Answer 1

Bruno Luong 2023-12-4

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https://ww2.mathworks.cn/matlabcentral/answers/2056064-an-annoying-extrapolation-problem#answer_1365294

编辑：Bruno Luong 2023-12-4

在 MATLAB Online 中打开

Because you insist on using anonymous function and ugly combined expression using scalar logical expression. The trick is no longer smart.

Just give up any of those then you are OK.

Also different way

x1 = linspace(0,pi/2,1000);y1=sin(x1);
lo = min(x1); % 0
up = max(x1); % pi/2
f = @(x) interp1(x1, y1, min(max(x,lo),up), 'spline')
f = function_handle with value:
    @(x)interp1(x1,y1,min(max(x,lo),up),'spline')
f(-1:4)
ans = 1×6
         0         0    0.8415    1.0000    1.0000    1.0000