Question

我有以下图表和创建该图表的数据文件。我想让Matlab为我找到以下几点：

[y，x]表示100％线所指出的峰值
[x]表示情节穿越y = 0线
[x]其中y为第1部分中50％和20％的峰值。

是否有人们知道哪些附加工具或软件包可以帮助我实现这一目标？我需要为一组地块做这个，所以合理自动化的东西是理想的。

我当然可以在Matlab中进行编程和计算部分，只需要加载数据文件，将其与曲线或函数匹配，然后找到各种[x，y]坐标。

Data plot

Answer 1

好的，这里有。据我所知，Matlab中没有任何一个例行程序来做你想做的事情;你必须自己制作一个。有几点需要注意：

显而易见，线性插值数据最容易做，应该没有问题
使用单个多项式插值也不是太难，尽管还有一些细节要处理。找到峰值应该是业务的第一顺序，这涉及找到衍生物的根（例如，使用roots）。找到峰值后，通过将多项式偏移该量，找到所有期望水平（0％，20％，50％）的多项式的根。
使用三次样条（spline）是最复杂的。对于具有完整立方体的所有子区间，应重复上面针对一般多项式概述的例程，同时考虑到最大值也可能位于子区间的边界的可能性，并且发现的任何根和极值可能位于区间之外立方有效的地方（也不要忘记x使用的spline - 偏移）。

以下是我对所有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')

这导致了这三个图：

Linear interpolation Polynomial interpolation Cubic splines interpolation

（注意在（N-1）次多项式出现问题; 20％的交叉点在结束时都是错误的。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，正如您可以清楚地看到的，如果基础数据不适合，使用单个多项式进行插值通常会引入很多问题。此外，正如您可以从这些图中清楚地看到的那样，插值方法会非常强烈地影响交叉点的位置 - 至少一些想法对于您的数据所依据的模型至关重要

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，它将为您（以及您的出版物的读者）提供对数据准确性的错误认知。使用三次样条曲线可以首先了解相交是什么以及它们的行为方式，但是一旦真实模型变得更加清晰，就会回过头来重新审视分析。当然，当用于通过数据创建更平滑，更“视觉上吸引人”的曲线时，当然不使用三次样条线发布:)

Answer 2

这不是一个完整的答案，但Matlab有内置函数，可以完成你想要做的大部分工作。

max可以帮助您找到100％行
polyfit将以最小二乘意义为您提供一组多项式拟合。如果你想让它完全通过n个点，我相信你至少需要使用n-1度。
roots将为您提供刚刚找到的多项式的过零点。您还可以通过减去常数来使用它来查找20％和50％的交叉点。如果有多个交叉点，您将需要最接近您最初找到的最大交叉点的交叉点。（你确定交叉点总是存在吗？）

Answer 3

找到全局最大使用MAX函数：

[ymax, imax] = max(y);
xmax = x(imax);
line(xlim,[ymax ymax],'Color','r')
line(xmax,ymax,'Color','r','LineStyle','o')

对于其他人，您可以使用出色的FileExchange提交 - "Fast and Robust Curve Intersections"。

y = 0处的行可以使用xlim和yline0 = [0 0];定义。然后就可以了

[x0, y0] = intersections(x,y,xlim,yline0); % function from FileExchange
x0close(1) = xmax - min(xmax-x0(x0<xmax));
x0close(2) = xmax + min(x0(x0>xmax)-xmax);
y0close = y0(ismember(x0,x0close));
line(xlim,yline0,'Color','r')
line(x0close,y0close,'Color','r','LineStyle','o')

除了

之外，对于20％和50％也可以这样做

yline20 = repmat((ymax - y0(1))*0.2,1,2);

所有这一切都假定你想要你的情节中的海峡线的交叉点，而不是插值。

在Matlab中查找绘图中的点

3 个答案: