计算两点之间的角度(顺时针)

本文介绍了计算两点之间的角度(顺时针)的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我很久没有使用数学了，这应该是一个简单的问题.

I have been not using math for a long time and this should be a simple problem to solve.

假设我有两个点A:(1，0)和B:(1，-1).

Suppose I have two points A: (1, 0) and B: (1, -1).

我想使用一个程序(Python或任何一种编程语言)来计算A，原点(0，0)和B之间的顺时针角度.它将是这样的:

I want to use a program (Python or whatever programming language) to calculate the clockwise angle between A, origin (0, 0) and B. It will be something like this:

angle_clockwise(point1, point2)

请注意，参数的顺序很重要.由于角度计算将是顺时针方向:

Note that the order of the parameters matters. Since the angle calculation will be clockwise:

• 如果我调用angle_clockwise(A，B)，它将返回45.
• 如果我调用angle_clockwise(B，A)，它将返回315.

换句话说，算法是这样的:

In other words, the algorithm is like this:

在第一个参数点与(0，0)之间绘制一条线(线1).

在第二个点参数与(0，0)之间绘制一条线(第2条线).

将第1行绕着(0，0)顺时针旋转，直到与第2行重叠.

行进的角距离线1将是返回的角度.

有什么办法可以编码此问题?

Is there any way to code this problem?

推荐答案

使用两个向量的内积和行列式.如果您想了解其工作原理，那么这实际上就是您应该了解的内容.您需要了解/阅读矢量数学才能理解.

Use the inner product and the determinant of the two vectors. This is really what you should understand if you want to understand how this works. You'll need to know/read about vector math to understand.

请参阅: en.wikipedia/wiki/Dot_product 和 en.wikipedia/wiki/Determinant

from math import acos from math import sqrt from math import pi def length(v): return sqrt(v[0]**2+v[1]**2) def dot_product(v,w): return v[0]*w[0]+v[1]*w[1] def determinant(v,w): return v[0]*w[1]-v[1]*w[0] def inner_angle(v,w): cosx=dot_product(v,w)/(length(v)*length(w)) rad=acos(cosx) # in radians return rad*180/pi # returns degrees def angle_clockwise(A, B): inner=inner_angle(A,B) det = determinant(A,B) if det<0: #this is a property of the det. If the det < 0 then B is clockwise of A return inner else: # if the det > 0 then A is immediately clockwise of B return 360-inner

在行列式计算中，您将两个向量连接起来以形成2 x 2矩阵，并为此计算行列式.

In the determinant computation, you're concatenating the two vectors to form a 2 x 2 matrix, for which you're computing the determinant.