How to fix error about interp1?
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Hello,
I have some data and i want to run this code:
xi = [85];
yi = interp1(y,x,xi)
but I have this error:
Error using griddedInterpolant
The grid vectors are not strictly monotonic increasing.
Error in interp1 (line 183)
F = griddedInterpolant(X,V,method);
Error in Q4 (line 19)
yi = interp1(y,x,xi)
How can I fix it?
采纳的回答
Star Strider
2017-3-12
编辑：Star Strider
2017-3-12
One way (without seeing your data, so this is a guess on my part):
y85 = find(y <= 85, 1, 'last');
yi = interp1(y(y85-1:y85+1),x(y85-1:y85+1),xi, 'linear','extrap');
That is how I would do it. This assumes there is only 1 where ‘y’ is approximately 85.
Note — This is obviously UNTESTED CODE. It should work, if there is one ‘y’ near 85.
EDIT — Added the 'extrap' option to make my approach more robust.
4 个评论
Ghazal Hnr
2017-3-12
编辑：Ghazal Hnr
2017-3-12
Star Strider
2017-3-12
编辑：Star Strider
2017-3-12
Without your data, I cannot tell.
Try this:
yi = interp1(y(y85-1:y85),x(y85-1:y85),xi, 'linear','extrap')
or this:
yi = interp1(y(y85:y85+1),x(y85:y85+1),xi, 'linear','extrap')
One of them should work.
Also, be certain you are interpolating the correct value. If you want the value of ‘y’ at ‘x=85’, do this instead:
yi = interp1(x, y, xi, 'linear','extrap')
You may have misinterpreted the documation for interp1.
Ghazal Hnr
2017-3-13
Star Strider
2017-3-13
My pleasure.
更多回答（1 个）
John D'Errico
2017-3-12
0 个投票
The fix is easy. Don't use interp1 to interpolate data that is not strictly monotone in the independent variable.
After all, how can interp1 know which value to predict on a function that is apparently multi-valued? For example, consider points around the perimeter of acircle. For any given x, there are TWO solutions. Interp1 requires a single valued function for any given x.
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