从绘制的图获取x，y坐标并获取与其他图的交点 [英] getting x,y coordinates from a plotted graph and get the intersection with an other graph

问题描述

我是matlab的新手，所以这个问题可能很简单.我尝试在Matlab中获取绘制图上点的[x,y]坐标.

I'm some kind of new to matlab,so the question may be elementary. I try to get the [x,y] coordinates of the points on a plotted graph in matlab.

该图很简单.我已经有了某些点(xb,yb)的坐标，并使用line(xb, yb)绘制了图形.

The graph is simple. I already had the coordinates of some points (xb,yb), and plotted the graph using line(xb, yb).

现在，我需要找到绘制线上所有点的坐标，以便获得这些线与另一个已定义图形的交点.

Now I need to find the coordinates of all the points on the plotted lines in order to get the intersection points of these lines with another defined graph.

我使用的代码是:

line(Xb,Yb) 
hold on
X=min(Xb):.001:max(Xb); 
y=0.03;
plot(X,y);

Xb和Yb是1 * 38数组，它们在我正在研究的实际问题中塑造区域的边界.但是我没有边界上所有的x和Y点.我需要用plot(X,y)绘制的水平线与边界的交点坐标.

The Xb and Yb are 1*38 arrays, and shape the boundary of a region in real problem which I'm studying. But I don't have all the x, Y of the points on the boundary. I need to the intersection coordinates of the boundary with horizontal line which is plotted by plot(X,y).

Xb和Yb是:

 Xb = [-0.0400 -0.0550 -0.0700 -0.0850 -0.1000 -0.1000 -0.1000 -0.1000 -0.1000 -0.1000 -0.0800 -0.0600 -0.0400 -0.0200 0 0.0200 0.0400 0.0600 0.0800 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.0850 0.0700 0.0550 0.0400 0.0380 0.0324 0.0235 0.0124 0 -0.0124 -0.0235 -0.0324 -0.0380 -0.0400 ];'
 Yb = [0 0 0 0 0 0.0200 0.0400 0.0600 0.0800 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.0800 0.0600 0.0400 0.0200 0 0 0 0 0 0.0124 0.0235 0.0324 0.0380 0.0400 0.0380 0.0324 0.0235 0.0124 0];'

非常感谢您的帮助.

推荐答案

如果将函数指定为数组x1,y1和x2,y2，则可以使用它们的线性内插器找到它们的交点:

If your functions are given as arrays x1,y1 and x2,y2, then you can find their crossing using their linear interpolators:

%dummy input
x1=[0 1 2 3]; 
y1=[1 4 2 0];
x2=[-1 3 4 5];
y2=[-1 2 5 3];

x0 = (max(min(x1),min(x2))+min(max(x1),max(x2)))/2;
fun1 = @(x) interp1(x1,y1,x,'linear');
fun2 = @(x) interp1(x2,y2,x,'linear');
difffun = @(x) fun1(x)-fun2(x);
crossing = fzero(difffun,x0);

您的x0应该非常接近实际的交叉点，我选择了两个区间内的一个点.匿名函数fun1和fun2使用interp1创建一个可调用函数，该函数只能告诉您特定x坐标处的插值.如果具有曲线拟合工具箱"，则可以使用fit一步获得一个函数.那么difffun是由两个输入函数之差给出的函数，其零将为您提供两个函数的交点. fzero会做到这一点:从给定的起点找到一个零的函数.

where your x0 should be fairly close to the actual crossing point, I chose a point inside both of your intervals. The anonymous functions fun1 and fun2 create a callable function using interp1, which can only tell you the interpolant at specific x coordinates. If you have the Curve Fitting Toolbox, you can get a function in a single step using fit. Then difffun is a function given by the difference of the two input functions, the zero(s) of which will give you the crossing of the two functions. fzero will do just that: find a zero of a function from a given starting point.

如果起始x0离得很远，您可能会遇到麻烦；如果有多个零，则一次只能找到一个.在您的情况下，插值函数的行为不太好，因此选择正确的x0(可能是手动设置值)是最重要的.

You can run into troubles if the starting x0 is way off, and in case of multiple zeros you will only find one at a time. In your case, the interpolating functions are not too well-behaved, so choosing a right x0 (probably manually setting a value) is paramount.

使用上面的虚拟输入的结果:

Results with the dummy input above:

输出:

crossing =

   2.272727272727273

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