两条相交线之间的像素 [英] Pixels between 2 intersecting lines

问题描述

我需要找到2行交叉之间的像素值。下图显示了我想要的点，即棕色区域。

I need to find pixel values that are between the intersection of 2 lines. The following image shows the points that I want namely the brown region.

这4个坐标可以改变，不一定是角点。

These 4 co-ordinates can change and are not necessarily the corner points.

获取像素值的最快方法是什么？是否有任何功能可以给我必要的掩码。

What is the fastest way to get the pixel values ? Is there any function that can give me the necessary mask.

推荐答案

你应该计算每个点，它是否在线上方或以下。如果该行以公式形式 Ax + By + C 给出，那么就像根据你的点计算这个表达式的符号一样简单 （X，Y）。如果您的行以任何其他形式给出，您应首先计算上面的表格。 （参见此处和这里）

You should calculate for each point, whether it is above the line or below. If the line is given in its equation form Ax+By+C, then it is as simple as calculating the sign of this expression, per your point (x,y). If your lines are given in any other form, you should first calculate the form above. (See here and here)

设 L1 是第一行以下所有点的集合， L2 第二行以下所有点的集合。然后，你的套装 X = Xor（L1，L2）

Let L1 be the set of all points below the first line, and L2 the set of all points below the second line. Then, your set is X = Xor(L1,L2)

[ < img src =https://i.stack.imgur.com/WJEay.pngalt =在此处输入图像说明> ] Xor [ ]

等于：

这是一个Matlab代码，根据我所描述的解决方案解决角点问题。您可以在代码中调整线方程。

Here is a Matlab code that solves you problem for the corner points, based on the solution that I've described. You can adjust the line equations in your code.

function CreateMask()
    rows = 100;
    cols = 200;
    [X,Y] = ndgrid(1:cols,1:rows);

    belowFirstLine  = X*(1/cols)  + Y*(-1/rows) + 0 < 0;
    belowSecondLine = X*(-1/cols) + Y*(-1/rows) + 1 < 0;

    figure;imshow( transpose(xor(belowSecondLine,belowFirstLine)));
end

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