Question

我在壁炉上绘制了一张图图,用:

plot(x,y)

我的图表有不同的斜坡, 我如何在每一个斜坡上绘制正切点, 并计算斜坡的系数?

Answer 1

如果您没有为绘图点设定明确的函数, 您可以使用 < a href=" http:// en.wikipedia.org/wiki/Finite_ difference" rel=“ nofollow” >unite difference 来估计衍生物。以下是适合数据范围以外的点 :

plot(x,y);
hold all;

% first sort the points, so x is monotonically rising
[x, sortidx] = sort(x);
y = y(sortidx);

% this is the x point for which you want to compute the slope
xslope = (x(1)+x(end))/2;

idx_a = find(x<xslope,1, last );
idx_b = find(x>xslope,1, first );
% or even simpler:
idx_b = idx_a+1;
% this assumes min(x)<xslope<max(x)

xa = x(idx_a);
xb = x(idx_b);
slope = (y(idx_b) - y(idx_a))/(xb - xa);

现在绘制这个斜坡, 取决于你想要什么: 短线 :

yslope = interp1(x,y,xslope);
ya_sloped = yslope + (xa-xslope)*slope;
yb_sloped = yslope + (xb-xslope)*slope;
line([xa;xb],[ya_sloped;yb_sloped]);

或长线线

yslope = interp1(x,y,xslope);
xa = xa + 4*(xa-xslope);
xb = xb + 4*(xb-xslope);
ya_sloped = yslope + (xa-xslope)*slope;
yb_sloped = yslope + (xb-xslope)*slope;
line([xa;xb],[ya_sloped;yb_sloped]);

我敢肯定,在这个代码里没有虫子, 但我会测试出来,当我周围有木板的时候;)

Answer 2

您将需要在任何您有兴趣使用的点( y2-y1) / (x2- x1) 找到斜度, 然后用绘图() 绘制一条有该斜度的直线。要绘制这条线, 您需要 y 拦截, 并且因为您知道该线上至少一个点的坐标( 这是您想要绘制正切点), 您可以在 y=mx+b 中为 b 解析。

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