Matlab interp2外推

Question

我正在使用2-D插值interp2.对于某些数据值,interp2命令返回NaN,因为其中一个维度超出了已知值向量定义的范围.

可以使用interp1命令进行推断.但是,有没有办法做到这一点interp2？

谢谢

这是我使用interp2命令的代码:

function [Cla] = AirfoilLiftCurveSlope(obj,AFdata,Rc,M)

% Input:
% AFdata: Airfoil coordinates.
% Rc: Local Reynolds number.
% M: Mach number for Prandtle Glauert compressibility correction.

% Output: 
% Cla: 2 dimensional lift curve slopea applicable to linear region of lift polar.

load('ESDU84026a.mat');

xi = size(AFdata);

if mod(xi(1,1),2) == 0
    %number is even
    AFupper = flipud(AFdata(1:(xi(1,1)/2),:));
    AFlower = AFdata(((xi(1,1)/2)+1):end,:);
else
    %number is odd
    AFupper = flipud(AFdata(1:floor((xi(1,1)/2)),:));
    AFlower = AFdata((floor(xi(1,1)/2)+1):end,:);
end


t_c = Airfoil.calculateThickness(AFdata(:,2));

Y90 = ((interp1(AFupper(:,1),AFupper(:,2),0.9,'linear')) - (interp1(AFlower(:,1),AFlower(:,2),0.9,'linear')))*100;

Y99 = ((interp1(AFupper(:,1),AFupper(:,2),0.99,'linear')) - (interp1(AFlower(:,1),AFlower(:,2),0.99,'linear')))*100;

Phi_TE = (2 * atan( ( (Y90/2) - (Y99/2) )/9))*180/pi;                       % Degrees
Tan_Phi_Te = ( (Y90/2) - (Y99/2) )/9;

Cla_corr = interp2(Tan_Phi,Rc_cla,cla_ratio,Tan_Phi_Te,Rc,'linear');

beta =sqrt((1-M^2));                                                        % Prandtle Glauert correction
Cla_theory = 2*pi + 4.7*t_c*(1+0.00375 * Phi_TE);                           % per rad 
Cla = (1.05/beta) * Cla_corr * Cla_theory;                                  % per rad

if isnan(Cla) == 1 %|| Cla > 2*pi
    Cla = 2*pi;
end

end

Answer 1

是的,interp2根据文档,有两种方法可以将有意义的值返回到界限之外.

1. 使用'spline'插值方法.与选项#2不同,这实际上将根据样条曲线的边界条件推断数据.
2. 指定最终extrapval参数.将返回此常量,而不是NaN所有其他插值方法.

不幸的是,似乎没有办法指定类似"网格上的最近邻居"之类的东西.如果越界元素靠近边缘,也许你可以只扩展输入数组.例如这样:

x = [x(1, 1), x(1, :), x(1, end); ...
     x(:, 1), x, x(:, end); ...
     x(end, 1), x(end, :), x(end, end)]