试验：

Question

我在tkinter画布上有两点。我需要一个函数来确定它们之间绘制的线最接近（N，NW，W SW，S等）哪个基本方向（方向重要）？我该怎么做呢？请注意，在画布中，左上角是（0,0）。

我试过了：

def dot_product(self, v, w):
    return v[0]*w[0]+v[1]*w[1]
def inner_angle(self, v, w):
    cosx=self.dot_product(v,w)/(sqrt(v[0]**2+v[1]**2)*sqrt(w[0]**2+w[1]**2))
    rad=acos(cosx)
    return rad*180/pi
def getAngle(self, A, B):
    inner=self.inner_angle(A,B)
    det = A[0]*B[1]-A[1]*B[0]
    if det<0:
        return inner
    else:
        return 360-inner

和

def getBearing(self, pointA, pointB):

if (type(pointA) != tuple) or (type(pointB) != tuple):
    raise TypeError("Only tuples are supported as arguments")

lat1 = math.radians(pointA[0])
lat2 = math.radians(pointB[0])

diffLong = math.radians(pointB[1] - pointA[1])

x = math.sin(diffLong) * math.cos(lat2)
y = math.cos(lat1) * math.sin(lat2) - (math.sin(lat1) * math.cos(lat2) * math.cos(diffLong))

initial_bearing = math.atan2(x, y)

initial_bearing = math.degrees(initial_bearing)
compass_bearing = (initial_bearing + 360) % 360

return compass_bearing

（我用这个函数来获取方向（代码不完整，它更像是一个例子））

def findDirection(self, p1, p2):
    bearing = self.getBearing(p1, p2) # OR getAngle()
    print(bearing)
    index = [180, 0]
    closest = min(index, key=lambda x:abs(x-bearing))
    if closest == 10:
        print(str(bearing) + " : UP")
    elif closest == 360:
        print(str(bearing) + " : DOWN")
    elif closest == 0:
        print(str(bearing) + " : RIGHT")
    elif closest == 180:
        print(str(bearing) + " : LEFT")

这些都不起作用。结果似乎不够一致，无法使用。有没有更好的方法呢？

Answer 1

我希望这是有帮助的 - 我使用基于tkinter构建的Python turtle为我的方便实现了它。我将乌龟变成 logo 模式，这使得北方0度和正方向角（即东方为90度）像指南针一样。乌龟方法towards()完成了你想要的大部分工作，所以我在计算基数方向时试图模仿它：

from random import randrange
from turtle import Turtle, Screen
from math import pi, atan2, degrees

DIRECTIONS = ['N', 'NNE', 'NE', 'ENE', 'E', 'ESE', 'SE', 'SSE', 'S', 'SSW', 'SW', 'WSW', 'W', 'WNW', 'NW', 'NNW']

BUCKET = 360.0 / len(DIRECTIONS)

X, Y = 0, 1

SIZE = 500

def onclick_handler(x, y):
    # Draw random vector

    yertle.reset()
    yertle.hideturtle()
    yertle.penup()

    start = (randrange(-SIZE//2, SIZE//2), randrange(-SIZE//2, SIZE//2))
    end = (randrange(-SIZE//2, SIZE//2), randrange(-SIZE//2, SIZE//2))

    yertle.goto(start)
    yertle.dot()
    yertle.showturtle()
    yertle.pendown()
    yertle.setheading(yertle.towards(end))
    yertle.goto(end)

    # Compute vector direction

    x, y = end[X] - start[X], end[Y] - start[Y]

    angle = round(degrees(atan2(y, -x) - pi / 2), 10) % 360.0

    direction = DIRECTIONS[round(angle / BUCKET) % len(DIRECTIONS)]

    screen.title("{} degress is {}".format(round(angle, 2), direction))

yertle = Turtle()

screen = Screen()
screen.mode('logo')
screen.setup(SIZE, SIZE)
screen.onclick(onclick_handler)

onclick_handler(0, 0)

screen.mainloop()

程序绘制一条随机线（具有明显的起点和方向）并计算可以在窗口标题中找到的基本方向。单击窗口会生成一个新行和计算。

您应该能够通过编辑DIRECTIONS变量来处理8或32个罗盘点。

Answer 2

以下是我建议的方法，以确定最接近由其端点[A, B]和point_a定义的线段point_b所指向的指南针方向：

所有计算都是在标准的笛卡尔坐标系中完成的最后更改为屏幕坐标。这简化了方法，并使代码可以在其他地方重复使用。
首先将原点更改为point_a
秒计算带有x_axis
确定最接近的方位（标准笛卡尔坐标系列）
将标准轴承转换为屏幕坐标轴承（水平翻转）

在屏幕坐标（Y轴向下）中定义了点，请致电get_bearings(point_a, point_b)
  如果标准中定义的点数   笛卡尔坐标（Y轴向上），调用   assign_bearing_to_compass(point_a, point_b)
  （代码下面的测试显示了在标准坐标和屏幕坐标中使用点的结果。）

import math


def _change_origin_of_point_b_to_point_a(point_a, point_b):
    # uses standard Y axis orientation, not screen orientation
    return (point_b[0] - point_a[0], point_b[1] - point_a[1])

def _calc_angle_segment_a_b_with_x_axis(point_a, point_b):
    # uses standard Y axis orientation, not screen orientation
    xa, ya = point_a
    xb, yb = _change_origin_of_point_b_to_point_a(point_a, point_b)
    return math.atan2(yb, xb)

def determine_bearing_in_degrees(point_a, point_b):
    """returns the angle in degrees that line segment [point_a, point_b)]
       makes with the horizontal X axis 
    """
    # uses standard Y axis orientation, not screen orientation
    return _calc_angle_segment_a_b_with_x_axis(point_a, point_b) * 180 / math.pi

def assign_bearing_to_compass(point_a, point_b):
    """returns the standard bearing of line segment [point_a, point_b)
    """
    # uses standard Y axis orientation, not screen orientation    
    compass = {'W' : [157.5, -157.5], 
               'SW': [-157.5, -112.5], 
               'S' : [-112.5, -67.5], 
               'SE': [-67.5, -22.5], 
               'E' : [-22.5, 22.5], 
               "NE": [22.5, 67.5], 
               'N' : [67.5, 112.5], 
               'NW': [112.5, 157.5]}

    bear = determine_bearing_in_degrees(point_a, point_b)
    for direction, interval in compass.items():
        low, high = interval
        if bear >= low and bear < high:
            return direction
    return 'W'

def _convert_to_negative_Y_axis(compass_direction):
    """flips the compass_direction horizontally
    """
    compass_conversion = {'E' : 'E', 
                          'SE': 'NE', 
                          'S' : 'N', 
                          'SW': 'NW', 
                          'W' : 'W', 
                          "NW": 'SW', 
                          'N' : 'S', 
                          'NE': 'SE'}
    return compass_conversion[compass_direction]

def get_bearings(point_a, point_b):
    return _convert_to_negative_Y_axis(assign_bearing_to_compass(point_a, point_b))

试验：

（使用标准三角圆象限）

象限I：

point_a = (0, 0)
points_b = [(1, 0), (1, 3), (1, 2), (1, 1), (2, 1), (3, 1), (0, 1)]
print("point_a, point_b     Y_up     Y_down (in screen coordinates)")
for point_b in points_b:
    print(point_a, ' ', point_b, '      ', assign_bearing_to_compass(point_a, point_b), '        ', get_bearings(point_a, point_b))

结果：

point_a, point_b     Y_up     Y_down (in screen coordinates)
(0, 0)   (1, 0)        E          E
(0, 0)   (1, 3)        N          S
(0, 0)   (1, 2)        NE         SE
(0, 0)   (1, 1)        NE         SE
(0, 0)   (2, 1)        NE         SE
(0, 0)   (3, 1)        E          E
(0, 0)   (0, 1)        N          S

Quadrant II：

point_a = (0, 0)
points_b = [(-1, 0), (-1, 3), (-1, 2), (-1, 1), (-2, 1), (-3, 1), (0, 1)]
print("point_a, point_b     Y_up     Y_down (in screen coordinates)")
for point_b in points_b:
    print(point_a, ' ', point_b, '      ', assign_bearing_to_compass(point_a, point_b), '        ', get_bearings(point_a, point_b))

结果：

point_a, point_b     Y_up     Y_down (in screen coordinates)
(0, 0)   (-1, 0)       W          W
(0, 0)   (-1, 3)       N          S
(0, 0)   (-1, 2)       NW         SW
(0, 0)   (-1, 1)       NW         SW
(0, 0)   (-2, 1)       NW         SW
(0, 0)   (-3, 1)       W          W
(0, 0)   (0, 1)        N          S

Quadrant III：

point_a = (0, 0)
points_b = [(-1, 0), (-1, -3), (-1, -2), (-1, -1), (-2, -1), (-3, -1), (0, -1)]
print("point_a, point_b     Y_up     Y_down (in screen coordinates)")
for point_b in points_b:
    print(point_a, ' ', point_b, '      ', assign_bearing_to_compass(point_a, point_b), '        ', get_bearings(point_a, point_b))

结果：

point_a, point_b     Y_up     Y_down (in screen coordinates)
(0, 0)   (-1, 0)        W          W
(0, 0)   (-1, -3)       S          N
(0, 0)   (-1, -2)       SW         NW
(0, 0)   (-1, -1)       SW         NW
(0, 0)   (-2, -1)       SW         NW
(0, 0)   (-3, -1)       W          W
(0, 0)   (0, -1)        S          N

Quadrant IV：

point_a = (0, 0)
points_b = [(1, 0), (1, -3), (1, -2), (1, -1), (2, -1), (3, -1), (0, -1)]
print("point_a, point_b     Y_up     Y_down (in screen coordinates)")
for point_b in points_b:
    print(point_a, ' ', point_b, '      ', assign_bearing_to_compass(point_a, point_b), '        ', get_bearings(point_a, point_b))

结果：

point_a, point_b     Y_up     Y_down (in screen coordinates)
(0, 0)   (1, 0)        E          E
(0, 0)   (1, -3)       S          N
(0, 0)   (1, -2)       SE         NE
(0, 0)   (1, -1)       SE         NE
(0, 0)   (2, -1)       SE         NE
(0, 0)   (3, -1)       E          E
(0, 0)   (0, -1)       S          N

Answer 3

要获得一个基本方向，需要一个参考相关方向的角度（在这种情况下是度数）的字典：

directions = {0:"N", 45:"NE", 90:"E", 135:"SE", 180:"S",
              225:"SW", 270:"W", 315:"NW", 360:"N"}

请注意，向北增加两次，因为在两点之间获得350度的角度会给出西北，当它应该向北时。

让Tkinter画布上的两个点a和b分别具有坐标(x1, y1)和(x2, y2)。因此，它们之间的差异（dx和dy）是x1-x2和y1-y2。

您现在可以执行dy/dx的反正切以获得角度。值得指出的是，如果dx为0，则它将除以0.您可以通过添加if not dx: return "N"来阻止此操作，如果点具有相同的x值，则返回North。

此外，如果dx大于0，那么它将返回相同的值，就好像它小于0.这是因为切线图的周期为180度。要解决此问题，您只需添加if dx > 0: angle += 180。

现在您有一个角度，您可以使用Pythons内置的directions函数min在之前定义的min(self.directions, key=lambda x: abs(x-angle))字典中引用它。这将返回字典中指定的最接近的度数。为了获得基数值，我们可以在字典中访问它。

将所有这些放在一起会产生以下功能（TLDR）：

from math import atan, degrees

...

def get_cardinal(a, b):
    dx, dy = a[0]-b[0], a[1]-b[1]
    if not dx:
        return "N"
    angle = degrees(atan(dy/dx))+90 #+90 to take into account TKinters coordinate system.
    if dx > 0:
        angle += 180
    return directions[min(directions, key=lambda x: abs(x-angle))]

这与directions词典相结合，可以为您提供答案。

Python Tkinter Canvas获得行基本方向

3 个答案:

试验：

象限I：

结果：

Quadrant II：

结果：

Quadrant III：

结果：

Quadrant IV：

结果：