How to interpolate data

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hj lee
hj lee 2020-11-9
回答： Mathieu NOE 2020-11-9
Hello,
P=[NaN NaN NaN 163.626652810406 140.245221578496 143.291946126177 158.198843187782 179.073274059170 206.654075204109 255.461365237779 318.955295177489 396.993670297888 486.124439415540 587.818128810315 703.723462252509 831.059610098512 974.884947235331 1136.56020938783 NaN NaN] : it is 1*20 double
V= [148.527280000000 166.471365263158 184.415450526316 202.359535789474 220.303621052632 238.247706315790 256.191791578947 274.135876842105 292.079962105263 310.024047368421 327.968132631579 345.912217894737 363.856303157895 381.800388421053 399.744473684211 417.688558947368 435.632644210526 453.576729473684 471.520814736842 489.464900000000] : it is also 1*20 double
I want to know the specific V data when P is 500.
I try to use
V_specific = interp1(P,V,500);
but there is error message, I don't know what is the problem.
is there any methods to solve this problem?
Thanks

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回答（2 个）

Sylvain
Sylvain 2020-11-9
Input Arguments
collapse all
x — Sample points
vector
Sample points, specified as a row or column vector of real numbers. The values in x must be distinct. The length of x must conform to one of the following requirements:
  • If v is a vector, then length(x) must equal length(v).
  • If v is an array, then length(x) must equal size(v,1).
Example: [1 2 3 4 5 6 7 8 9 10]
Example: 1:10
Example: [3 7 11 15 19 23 27 31]'
Data Types: single | double | duration | datetime
Thus remove the NaN values:
V_specific = interp1(P(4:end-3),V(4:end-3),500)
Mathieu NOE
Mathieu NOE 2020-11-9
hi
get rid of the NaN
ind_not_NaN = ~isnan(P);
V_specific = interp1(P(ind_not_NaN),V(ind_not_NaN),500);

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