python中求两个向量的夹角方式

import numpy as np
x=np.array([3,5])
y=np.array([4,2])
# 两个向量
Lx=np.sqrt(x.dot(x))
Ly=np.sqrt(y.dot(y))
#相当于勾股定理，求得斜线的长度
cos_angle=x.dot(y)/(Lx*Ly)
#求得cos_sita的值再反过来计算，绝对长度乘以cos角度为矢量长度，初中知识。。
print(cos_angle)
angle=np.arccos(cos_angle)
angle2=angle*360/2/np.pi
#变为角度
print(angle2)
#x.dot(y) =  y=∑(ai*bi)

>> a=[1,2,3;4,5,6;7,8,9];
>> b=[6,6,6;6,6,6;6,6,6];
>> dot(a,
ans =
    72    90   108
1*6+4*6+7*6=72
#三个相加
#dot 是矩阵相乘
#print(np.dot(x,x))
#34
# a=np.linalg.norm(x-y)
# print(a)
costheta=x.dot(y)/(np.linalg.norm(x)*np.linalg.norm(y))
#范数
# #[-1  3]
# a=np.square(x-y)
# #[1,9]
# print(np.sum(a))
# #10
# print(np.sqrt(10))

In [25]: a = np.array([1,2])                                                    
In [26]: b = np.array([2, 1])                                                   
In [27]: dot = 0                                                                
In [28]: for e, f in zip(a, b): 
    ...:     dot += e*f 
    ...:                                                                        
In [29]: dot                                                                    
Out[29]: 4
In [30]: a*b                                                                    
Out[30]: array([2, 2])
In [31]: np.sum(a*b)                                                            
Out[31]: 4
In [32]: (a*b).sum()                                                            
Out[32]: 4
In [33]: np.dot(a,b)                                                            
Out[33]: 4
In [34]: a.dot(b)                                                               
Out[34]: 4
In [35]: b.dot(a)                                                               
Out[35]: 4
In [36]: amag = np.sqrt((a*a).sum())                                            
In [37]: amag                                                                   
Out[37]: 2.23606797749979
In [38]: amag = np.linalg.norm(a)                                               
In [39]: amag                                                                   
Out[39]: 2.23606797749979
In [40]: cosangle = a.dot(b)/(np.linalg.norm(a) * np.linalg.norm(b))            
In [41]: cosangle                                                               
Out[41]: 0.7999999999999998
In [42]: angle = np.arccos(cosangle)                                            
In [43]: angle                                                                  
Out[43]: 0.6435011087932847