将向量 w 投影到向量 v 上并绘制垂直线 - 准备 PCA

Question

我想做矢量投影来为 PCA 做准备，我按照This 教程计算矢量投影。 w 是“指向”数据点的向量，v 是跨越应该投影 w 的线的向量。

代码是：

import numpy as np
import matplotlib.pyplot as plt
from matplotlib import style
style.use('fivethirtyeight')
from sklearn.preprocessing import StandardScaler

# Normalize the input data
A = np.array([[10,8],[1,2],[7,5],[3,5],[7,6],[8,7],[9,9],[4,5],[6,5],[6,8],
             [1,9],[10,2],[6,3],[2,5],[1,14],[8,8],[9,5],[4,4],[5,6],[8,8],
             [11,9],[10,12],[6,4],[5,2],[10,2],[8,3],[6,9],[0,4],[13,6],[9,6]])

A = StandardScaler(with_std=False,copy=False).fit_transform(A)


fig = plt.figure(figsize=(15,10))
ax0 = fig.add_subplot(111)
ax0.set_ylim(bottom=min(A[:,1])-3,top=max(A[:,1])+3)

ax0.scatter(A[:,0],A[:,1])

# Initialize a first vector a

v = np.array([1,0.5])




# Plot the vector v
#ax0.arrow(0,0,a[0],a[1],length_includes_head=True,width=0.03,color='green')


# Plot the line y=alpha*v defined by the vector a and passing the origin
ax0.plot(np.linspace(min(A[:,0])-3,max(A[:,0])+3),np.linspace(min(A[:,0])-3,max(A[:,0])+3)*(v[1]/v[0]),
         'k--',linewidth=1.5,zorder=0)

# Run through all datapoints

coordinates_on_ba_run = [] # Store the coordinates of the projected points on a 

for i in range(len(A[:,0])):
    # Plot the vector v
    #ax0.arrow(0,0,v[0],v[1],length_includes_head=True,width=0.03,color='green')


    # Point on one of the datapoints and denote this vector with w
    w = np.array([A[i][0],A[i][1]])
    #ax0.arrow(0,0,w[0],w[1],length_includes_head=True,width=0.03,color='blue')

    # Caclculate c and the projection vector cv. Additionally, test if the dot product of v and (w-cv) is zero

    c = np.dot(w,v.reshape(2,1))/np.dot(v,v.reshape(2,1))
    print(np.dot((w-c*v),v)) #This must be zero for each projection!
    cv = c*v

 

    # Draw a line from the datappoint in A to the tip of the vector cv. 


    ax0.plot([w[0],cv[0]],[w[1],cv[1]],linewidth=1,color='red',linestyle='--',zorder=0)



    
plt.show()

这给出了以下结果：

2.22044604925e-16
-2.22044604925e-16
0.0
0.0
2.77555756156e-17
-5.55111512313e-17
1.11022302463e-16
2.22044604925e-16
0.0
0.0
0.0
0.0
0.0
-2.22044604925e-16
0.0
-2.22044604925e-16
0.0
1.11022302463e-16
0.0
-2.22044604925e-16
0.0
-4.4408920985e-16
0.0
0.0
0.0
0.0
0.0
-2.22044604925e-16
-4.4408920985e-16
-2.22044604925e-16

因此代码正在运行，并且每次转换必须为零的“控制”计算 (np.dot((w-c*v),v)) 为零......因此结果应该是正确的......但是，正如你可以看到的裸体眼睛，虚线不垂直于向量 v 跨越的线。那么这只是一个可视化问题还是代码中有错误？感谢任何帮助

Answer 1

找到错误了...如果你看一下轴的比例，你会发现它们不相等，即x轴有限制（-10,10）而y轴有限制（- 6,10)... 因此，这会扭曲视图，并且肉眼看，红色虚线与 v 跨越的线之间的角度不是 90 度，而是取决于比率。这也解释了为什么np.dot((w-c*v),v) 的计算返回零表示结果是正确的。

这是工作代码：

import numpy as np
import matplotlib.pyplot as plt
from matplotlib import style
style.use('fivethirtyeight')
from sklearn.preprocessing import StandardScaler


# Normalize the input data
A = np.array([[10,8],[1,2],[7,5],[3,5],[7,6],[8,7],[9,9],[4,5],[6,5],[6,8],
             [1,9],[10,2],[6,3],[2,5],[1,14],[8,8],[9,5],[4,4],[5,6],[8,8],
             [11,9],[10,12],[6,4],[5,2],[10,2],[8,3],[6,9],[0,4],[13,6],[9,6]])

A = StandardScaler(with_std=False,copy=False).fit_transform(A)

fig = plt.figure(figsize=(10,10))
ax0 = fig.add_subplot(111)
ax0.set_aspect('equal')
ax0.set_xlim((-10,10))
ax0.set_ylim((-10,10))

ax0.scatter(A[:,0],A[:,1])


# Run through all the data

for i in range(len(A[:,0])):

    # v
    v = np.array([3,2])
    ax0.plot(np.linspace(-10,10),np.linspace(-10,10)*(v[1]/v[0]),color='black',linestyle='--',linewidth=1.5)   

    # w
    w = np.array([A[i][0],A[i][1]])
    #ax0.arrow(0,0,w[0],w[1],length_includes_head=True,width=0.01,color='green')

    # cv
    cv = (np.dot(w,v))/np.dot(v,np.transpose(v))*v
    #ax0.arrow(0,0,cv[0],cv[1],length_includes_head=True,width=0.005,color='black')
    print(cv)

    # line between w and cv
    ax0.plot([w[0],cv[0]],[w[1],cv[1]],'r--',linewidth=1.5)


    # Check the result
    print(np.dot((w-cv),cv))

plt.show()