python - 旋转二维物体的功能？

Question

是否可以在 python 中编写一个函数来旋转任何二维结构，参数仅为结构中点的坐标 (x,y)？将包括轴、速度和方向的附加参数。

据我所知，只有通过计算对称点和轴的点距离才有可能，因此它总是会变化，因此除了由标准形状(三角形、矩形、正方形等)组成的二维结构外，这是不可能的

我们将不胜感激。

Answer 1

首先，我们需要一个函数来围绕原点旋转一个点。

当我们将一个点 (x,y) 围绕原点旋转 theta 度时，我们得到坐标:

(x*cos(theta)-y*sin(theta), x*sin(theta)+y*cos(theta))

如果我们想围绕原点以外的点旋转它，我们只需要移动它，使中心点成为原点。现在，我们可以编写以下函数:

from math import sin, cos, radians

def rotate_point(point, angle, center_point=(0, 0)):
    """Rotates a point around center_point(origin by default)
    Angle is in degrees.
    Rotation is counter-clockwise
    """
    angle_rad = radians(angle % 360)
    # Shift the point so that center_point becomes the origin
    new_point = (point[0] - center_point[0], point[1] - center_point[1])
    new_point = (new_point[0] * cos(angle_rad) - new_point[1] * sin(angle_rad),
                 new_point[0] * sin(angle_rad) + new_point[1] * cos(angle_rad))
    # Reverse the shifting we have done
    new_point = (new_point[0] + center_point[0], new_point[1] + center_point[1])
    return new_point

一些输出:

print(rotate_point((1, 1), 90, (2, 1)))
# This prints (2.0, 0.0)
print(rotate_point((1, 1), -90, (2, 1)))
# This prints (2.0, 2.0)
print(rotate_point((2, 2), 45, (1, 1)))
# This prints (1.0, 2.4142) which is equal to (1,1+sqrt(2))

现在，我们只需要使用之前的函数旋转多边形的每个角:

def rotate_polygon(polygon, angle, center_point=(0, 0)):
    """Rotates the given polygon which consists of corners represented as (x,y)
    around center_point (origin by default)
    Rotation is counter-clockwise
    Angle is in degrees
    """
    rotated_polygon = []
    for corner in polygon:
        rotated_corner = rotate_point(corner, angle, center_point)
        rotated_polygon.append(rotated_corner)
    return rotated_polygon

示例输出:

my_polygon = [(0, 0), (1, 0), (0, 1)]
print(rotate_polygon(my_polygon, 90))
# This gives [(0.0, 0.0), (0.0, 1.0), (-1.0, 0.0)]