好的，到此为止。据我所知，Matlab 中没有内置的例程来做你想做的事。你必须自己做一个。需要注意的几点：
应该很明显，线性插值数据是最容易做到的，应该没有问题
使用单个多项式插值也不太难，尽管还有一些细节需要处理。找到峰值应该是第一要务，这涉及到求导数的根（例如，使用roots）。找到峰值后，通过将多项式偏移这个量，在所有所需级别（0%、20%、50%）找到多项式的根。
使用三次样条 (spline) 是最复杂的。对于具有完整三次的所有子区间，应重复上面概述的一般多项式例程，考虑到最大值也可能位于子区间边界上的可能性，并且找到的任何根和极值都可能位于区间之外三次方有效（也不要忘记spline 使用的x-offsets）。
这是我对所有 3 种方法的实现：
%% Initialize
% ---------------------------
clc
clear all
% Create some bogus data
n = 25;
f = @(x) cos(x) .* sin(4*x/pi) + 0.5*rand(size(x));
x = sort( 2*pi * rand(n,1));
y = f(x);
%% Linear interpolation
% ---------------------------
% y peak
[y_peak, ind] = max(y);
x_peak = x(ind);
% y == 0%, 20%, 50%
lims = [0 20 50];
X = cell(size(lims));
Y = cell(size(lims));
for p = 1:numel(lims)
% the current level line to solve for
lim = y_peak*lims(p)/100;
% points before and after passing through the current limit
after = (circshift(y<lim,1) & y>lim) | (circshift(y>lim,1) & y<lim);
after(1) = false;
before = circshift(after,-1);
xx = [x(before) x(after)];
yy = [y(before) y(after)];
% interpolate and insert new data
new_X = x(before) - (y(before)-lim).*diff(xx,[],2)./diff(yy,[],2);
X{p} = new_X;
Y{p} = lim * ones(size(new_X));
end
% make a plot to verify
figure(1), clf, hold on
plot(x,y, 'r') % (this also plots the interpolation in this case)
plot(x_peak,y_peak, 'k.') % the peak
plot(X{1},Y{1}, 'r.') % the 0% intersects
plot(X{2},Y{2}, 'g.') % the 20% intersects
plot(X{3},Y{3}, 'b.') % the 50% intersects
% finish plot
xlabel('X'), ylabel('Y'), title('Linear interpolation')
legend(...
'Real data / interpolation',...
'peak',...
'0% intersects',...
'20% intersects',...
'50% intersects',...
'location', 'southeast')
%% Cubic splines
% ---------------------------
% Find cubic splines interpolation
pp = spline(x,y);
% Finding the peak requires finding the maxima of all cubics in all
% intervals. This means evaluating the value of the interpolation on
% the bounds of each interval, finding the roots of the derivative and
% evaluating the interpolation on those roots:
coefs = pp.coefs;
derivCoefs = bsxfun(@times, [3 2 1], coefs(:,1:3));
LB = pp.breaks(1:end-1).'; % lower bounds of all intervals
UB = pp.breaks(2:end).'; % upper bounds of all intervals
% rename for clarity
a = derivCoefs(:,1);
b = derivCoefs(:,2);
c = derivCoefs(:,3);
% collect and limits x-data
x_extrema = [...
LB, UB,...
LB + (-b + sqrt(b.*b - 4.*a.*c))./2./a,... % NOTE: data is offset by LB
LB + (-b - sqrt(b.*b - 4.*a.*c))./2./a,... % NOTE: data is offset by LB
];
x_extrema = x_extrema(imag(x_extrema) == 0);
x_extrema = x_extrema( x_extrema >= min(x(:)) & x_extrema <= max(x(:)) );
% NOW find the peak
[y_peak, ind] = max(ppval(pp, x_extrema(:)));
x_peak = x_extrema(ind);
% y == 0%, 20% and 50%
lims = [0 20 50];
X = cell(size(lims));
Y = cell(size(lims));
for p = 1:numel(lims)
% the current level line to solve for
lim = y_peak * lims(p)/100;
% find all 3 roots of all cubics
R = NaN(size(coefs,1), 3);
for ii = 1:size(coefs,1)
% offset coefficients to find the right intersects
C = coefs(ii,:);
C(end) = C(end)-lim;
% NOTE: data is offset by LB
Rr = roots(C) + LB(ii);
% prune roots
Rr( imag(Rr)~=0 ) = NaN;
Rr( Rr <= LB(ii) | Rr >= UB(ii) ) = NaN;
% insert results
R(ii,:) = Rr;
end
% now evaluate and save all valid points
X{p} = R(~isnan(R));
Y{p} = ppval(pp, X{p});
end
% as a sanity check, plot everything
xx = linspace(min(x(:)), max(x(:)), 20*numel(x));
yy = ppval(pp, xx);
figure(2), clf, hold on
plot(x,y, 'r') % the actual data
plot(xx,yy) % the cubic-splines interpolation
plot(x_peak,y_peak, 'k.') % the peak
plot(X{1},Y{1}, 'r.') % the 0% intersects
plot(X{2},Y{2}, 'g.') % the 20% intersects
plot(X{3},Y{3}, 'b.') % the 50% intersects
% finish plot
xlabel('X'), ylabel('Y'), title('Cubic splines interpolation')
legend(...
'Real data',...
'interpolation',...
'peak',...
'0% intersects',...
'20% intersects',...
'50% intersects',...
'location', 'southeast')
%% (N-1)th degree polynomial
% ---------------------------
% Find best interpolating polynomial
coefs = bsxfun(@power, x, n-1:-1:0) \ y;
% (alternatively, you can use polyfit() to do this, but this is faster)
% To find the peak, we'll have to find the roots of the derivative:
derivCoefs = (n-1:-1:1).' .* coefs(1:end-1);
Rderiv = roots(derivCoefs);
Rderiv = Rderiv(imag(Rderiv) == 0);
Rderiv = Rderiv(Rderiv >= min(x(:)) & Rderiv <= max(x(:)));
[y_peak, ind] = max(polyval(coefs, Rderiv));
x_peak = Rderiv(ind);
% y == 0%, 20%, 50%
lims = [0 20 50];
X = cell(size(lims));
Y = cell(size(lims));
for p = 1:numel(lims)
% the current level line to solve for
lim = y_peak * lims(p)/100;
% offset coefficients as to find the right intersects
C = coefs;
C(end) = C(end)-lim;
% find and prune roots
R = roots(C);
R = R(imag(R) == 0);
R = R(R>min(x(:)) & R<max(x(:)));
% evaluate polynomial at these roots to get actual data
X{p} = R;
Y{p} = polyval(coefs, R);
end
% as a sanity check, plot everything
xx = linspace(min(x(:)), max(x(:)), 20*numel(x));
yy = polyval(coefs, xx);
figure(3), clf, hold on
plot(x,y, 'r') % the actual data
plot(xx,yy) % the cubic-splines interpolation
plot(x_peak,y_peak, 'k.') % the peak
plot(X{1},Y{1}, 'r.') % the 0% intersects
plot(X{2},Y{2}, 'g.') % the 20% intersects
plot(X{3},Y{3}, 'b.') % the 50% intersects
% finish plot
xlabel('X'), ylabel('Y'), title('(N-1)th degree polynomial')
legend(...
'Real data',...
'interpolation',...
'peak',...
'0% intersects',...
'20% intersects',...
'50% intersects',...
'location', 'southeast')
这导致这三个图：
（请注意，第 (N-1) 次多项式出现问题；20% 的交叉点到最后都是错误的。因此，在复制粘贴之前，请更彻底地检查所有内容：)
正如我之前所说，并且您可以清楚地看到，如果基础数据不适合它，使用单个多项式进行插值通常会引入很多问题。此外，正如您从这些图中可以清楚地看到的那样，插值方法非常强烈地影响交叉点的位置 - 至关重要的是，您至少 了解 数据的基础模型.
对于一般情况，三次样条通常是最好的方法。但是，这是一种通用方法，会让您（以及您出版物的读者）对您的数据的准确性产生错误的认识。使用三次样条来初步了解相交是什么以及它们的行为方式，但一旦真实模型变得更加清晰，总是回来重新审视您的分析。当然，不要使用三次样条发布，因为它仅用于通过您的数据创建更平滑、更具“视觉吸引力”的曲线 :)