延伸的线段装配到一个边界框 [英] Extending a line segment to fit into a bounding box

问题描述

我有两个的PointF 已定义的线段，随着二维边框。我想尽可能延长线段在两个方向，使得该段是与边界框的壁齐平。下面是我想要做的一些例子：

I have a line segment defined by two pointFs, along with a 2D bounding rectangle. I want to extend the line segment as much as possible in both directions so that the segment is flush with the walls of the bounding box. Here are some examples of what I'm trying to do:

有没有人对如何做到这一点有什么建议？

Does anyone have any suggestions on how to do this?

推荐答案

下面是一个Python code例如：

Here is an code example in python:

def extend(xmin, ymin, xmax, ymax, x1, y1, x2, y2):
    if y1 == y2:
        return (xmin, y1, xmax, y1)
    if x1 == x2:
        return (x1, ymin, x1, ymax)

    # based on (y - y1) / (x - x1) == (y2 - y1) / (x2 - x1)
    # => (y - y1) * (x2 - x1) == (y2 - y1) * (x - x1)
    y_for_xmin = y1 + (y2 - y1) * (xmin - x1) / (x2 - x1)
    y_for_xmax = y1 + (y2 - y1) * (xmax - x1) / (x2 - x1)

    x_for_ymin = x1 + (x2 - x1) * (ymin - y1) / (y2 - y1)
    x_for_ymax = x1 + (x2 - x1) * (ymax - y1) / (y2 - y1)

    if ymin <= y_for_xmin <= ymax:
        if xmin <= x_for_ymax <= xmax:
            return (xmin, y_for_xmin, x_for_ymax, ymax)
        if xmin <= x_for_ymin <= xmax:
            return (xmin, y_for_xmin, x_for_ymin, ymin)
    if ymin <= y_for_xmax <= ymax:
        if xmin <= x_for_ymin <= xmax:
            return (x_for_ymin, ymin, xmax, y_for_xmax)
        if xmin <= x_for_ymax <= xmax:
            return (x_for_ymax, ymax, xmax, y_for_xmax)

def test():
    assert (2, 1,  2, 5) == extend(1, 1,  5, 5,  2, 3,  2, 4)
    assert (1, 2,  4, 5) == extend(1, 1,  5, 5,  2, 3,  3, 4)
    assert (1, 3,  5, 3) == extend(1, 1,  5, 5,  3, 3,  4, 3)
    assert (1, 1,  5, 5) == extend(1, 1,  5, 5,  2, 2,  3, 3)
    assert (3, 1,  5, 5) == extend(1, 1,  5, 5,  3.5, 2,  4, 3)

if __name__ == '__main__':
    test()

有不检查该段包含在矩形和也应该工作，如果它是外部以它（返回None -implicit-如果没有实际的段交叉口）。

It doesn't check that the segment is contained in the rectangle and should work also if it is exterior to it (returns None -implicit- if there is no actual segment intersection).

它是基于该矩形具有的节段平行的轴的假设。

It is based on the assumption that the rectangle has the segments parallel with the axes.

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