我需要确定Python中两个n维向量之间的角度。例如,输入可以是两个列表,如下所示:[1,2,3,4]
和[6,7,8,9]
。
import math
def dotproduct(v1, v2):
return sum((a*b) for a, b in zip(v1, v2))
def length(v):
return math.sqrt(dotproduct(v, v))
def angle(v1, v2):
return math.acos(dotproduct(v1, v2) / (length(v1) * length(v2)))
注意:当向量具有相同或相反的方向时,这将失败。正确的实现在这里:https://stackoverflow.com/a/13849249/71522
注意:如果两个向量具有相同的方向(例如,(1, 0, 0)
,(1, 0, 0)
)或相反的方向(例如,(-1, 0, 0)
,(1, 0, 0)
),则此处的所有其他答案将失败。
这是一个能够正确处理这些情况的函数:
import numpy as np
def unit_vector(vector):
""" Returns the unit vector of the vector. """
return vector / np.linalg.norm(vector)
def angle_between(v1, v2):
""" Returns the angle in radians between vectors 'v1' and 'v2'::
>>> angle_between((1, 0, 0), (0, 1, 0))
1.5707963267948966
>>> angle_between((1, 0, 0), (1, 0, 0))
0.0
>>> angle_between((1, 0, 0), (-1, 0, 0))
3.141592653589793
"""
v1_u = unit_vector(v1)
v2_u = unit_vector(v2)
return np.arccos(np.clip(np.dot(v1_u, v2_u), -1.0, 1.0))
使用numpy(强烈推荐),你会做:
from numpy import (array, dot, arccos, clip)
from numpy.linalg import norm
u = array([1.,2,3,4])
v = ...
c = dot(u,v)/norm(u)/norm(v) # -> cosine of the angle
angle = arccos(clip(c, -1, 1)) # if you really want the angle
另一种可能性是使用numpy
,它给你内角
import numpy as np
p0 = [3.5, 6.7]
p1 = [7.9, 8.4]
p2 = [10.8, 4.8]
'''
compute angle (in degrees) for p0p1p2 corner
Inputs:
p0,p1,p2 - points in the form of [x,y]
'''
v0 = np.array(p0) - np.array(p1)
v1 = np.array(p2) - np.array(p1)
angle = np.math.atan2(np.linalg.det([v0,v1]),np.dot(v0,v1))
print np.degrees(angle)
这是输出:
In [2]: p0, p1, p2 = [3.5, 6.7], [7.9, 8.4], [10.8, 4.8]
In [3]: v0 = np.array(p0) - np.array(p1)
In [4]: v1 = np.array(p2) - np.array(p1)
In [5]: v0
Out[5]: array([-4.4, -1.7])
In [6]: v1
Out[6]: array([ 2.9, -3.6])
In [7]: angle = np.math.atan2(np.linalg.det([v0,v1]),np.dot(v0,v1))
In [8]: angle
Out[8]: 1.8802197318858924
In [9]: np.degrees(angle)
Out[9]: 107.72865519428085
如果你正在使用3D矢量,你可以使用toolbelt vg简洁地完成。它是numpy顶部的一层轻薄层。
import numpy as np
import vg
vec1 = np.array([1, 2, 3])
vec2 = np.array([7, 8, 9])
vg.angle(vec1, vec2)
您还可以指定视角以通过投影计算角度:
vg.angle(vec1, vec2, look=vg.basis.z)
或者通过投影计算有符号角度:
vg.signed_angle(vec1, vec2, look=vg.basis.z)
我在上次创业时创建了这个库,它的用途就是这样:在NumPy中简单的想法是冗长或不透明的。
对于少数可能有(由于SEO并发症)在这里结束尝试计算python中两条线之间的角度,如(x0, y0), (x1, y1)
几何线,有下面的最小解决方案(使用shapely
模块,但可以很容易地修改不):
from shapely.geometry import LineString
import numpy as np
ninety_degrees_rad = 90.0 * np.pi / 180.0
def angle_between(line1, line2):
coords_1 = line1.coords
coords_2 = line2.coords
line1_vertical = (coords_1[1][0] - coords_1[0][0]) == 0.0
line2_vertical = (coords_2[1][0] - coords_2[0][0]) == 0.0
# Vertical lines have undefined slope, but we know their angle in rads is = 90° * π/180
if line1_vertical and line2_vertical:
# Perpendicular vertical lines
return 0.0
if line1_vertical or line2_vertical:
# 90° - angle of non-vertical line
non_vertical_line = line2 if line1_vertical else line1
return abs((90.0 * np.pi / 180.0) - np.arctan(slope(non_vertical_line)))
m1 = slope(line1)
m2 = slope(line2)
return np.arctan((m1 - m2)/(1 + m1*m2))
def slope(line):
# Assignments made purely for readability. One could opt to just one-line return them
x0 = line.coords[0][0]
y0 = line.coords[0][1]
x1 = line.coords[1][0]
y1 = line.coords[1][1]
return (y1 - y0) / (x1 - x0)
使用将是
>>> line1 = LineString([(0, 0), (0, 1)]) # vertical
>>> line2 = LineString([(0, 0), (1, 0)]) # horizontal
>>> angle_between(line1, line2)
1.5707963267948966
>>> np.degrees(angle_between(line1, line2))
90.0
使用numpy并处理BandGap的舍入错误:
from numpy.linalg import norm
from numpy import dot
import math
def angle_between(a,b):
arccosInput = dot(a,b)/norm(a)/norm(b)
arccosInput = 1.0 if arccosInput > 1.0 else arccosInput
arccosInput = -1.0 if arccosInput < -1.0 else arccosInput
return math.acos(arccosInput)
注意,如果其中一个向量的幅度为零(除以0),则此函数将抛出异常。
如果你想要有角度,你必须确定给定的对是右手还是左手(有关详细信息,请参阅wiki)。
我的解决方案是:
def unit_vector(vector):
""" Returns the unit vector of the vector"""
return vector / np.linalg.norm(vector)
def angle(vector1, vector2):
""" Returns the angle in radians between given vectors"""
v1_u = unit_vector(vector1)
v2_u = unit_vector(vector2)
minor = np.linalg.det(
np.stack((v1_u[-2:], v2_u[-2:]))
)
if minor == 0:
raise NotImplementedError('Too odd vectors =(')
return np.sign(minor) * np.arccos(np.clip(np.dot(v1_u, v2_u), -1.0, 1.0))
由于这个NotImplementedError
,它并不完美,但对我而言,它运作良好。这种行为可以修复(因为确定任何给定对的手势),但它需要更多我想要的代码并且必须编写。