手动实施 PCA 会产生错误的图，其中特征向量不正交

Question

我需要绘制我这样计算的特征向量：

def fit(self, X):
    
    '''
    fits sorted eigenvalues and eigenvectors to class attributes. same goes for variance and explained variance.
    '''
    
    n_samples = X.shape[0]
    # We center the data and compute the sample covariance matrix.
    X -= np.mean(X, axis=0)
    self.cov_matrix_ = np.dot(X.T, X) / (n_samples-1)
    #test = np.cov(X)
    
    #Negative values are ignored with eigh
    (self.eigvalues_, self.components_) = np.linalg.eigh(self.cov_matrix_)
    
    idx = self.eigvalues_.argsort()[::-1]   
    self.eigvalues_ = self.eigvalues_[idx]
    self.components_ = self.components_[:,idx]
    self.variance_ = np.sum(self.eigvalues_)
    self.explained_variance_ = self.eigvalues_ / self.variance_
    
def transform(self, X):
    #project data onto eigenvectors
    print(self.components_.shape, X.shape)
    self.projected_ = X @ self.components_.T
    return self.projected_

进入我的数据集的前 2 个特征的图表。

我的 self.components_ 的形状是我的 100x240 数据集的 240 个特征向量，形状为 240x240。在绘制了我的 2 个具有最大特征值的特征向量的前两个值之后，结果如下：

pca = PCA()

pca.fit(subsample)

#pca.transform(subsample)

plt.scatter(subsample[:,0], subsample[:,1], edgecolor='none', alpha=0.5)
plt.quiver(pca.components_[0,0], pca.components_[0,1], 
       angles='xy', scale_units='xy', scale=1, width=0.002 )
plt.quiver(pca.components_[1,0], pca.components_[1,1], 
       angles='xy', scale_units='xy', scale=1, width=0.002 )

我做错了什么？

Answer 1

您应该按行而不是列对特征向量进行排序，即

self.components_ = self.components_[:,idx]

应该是

self.components_ = self.components_[idx]

此外，您应该确保以相同的纵横比绘制，因为箭袋可能未对齐：

plt.gca().set_aspect('equal')

在您的代码中包含一个最小的工作示例是一种很好的做法，所以下次请记住:)。我不得不推断你的代码的其余部分可能是什么以获得最小的工作示例。无论如何，这是我建议的代码：

import numpy as np 
from matplotlib import pyplot as plt

class PCA:
    def fit(self, X):
        
        '''
        fits sorted eigenvalues and eigenvectors to class attributes. same goes for variance and explained variance.
        '''
        
        n_samples = X.shape[0]
        # We center the data and compute the sample covariance matrix.
        X -= np.mean(X, axis=0)
        self.cov_matrix_ = np.dot(X.T, X) / (n_samples-1)
        #test = np.cov(X)
        
        #Negative values are ignored with eigh
        (self.eigvalues_, self.components_) = np.linalg.eigh(self.cov_matrix_)
        
        idx = self.eigvalues_.argsort()[::-1]   
        self.eigvalues_ = self.eigvalues_[idx]
        self.components_ = self.components_[idx]
        self.variance_ = np.sum(self.eigvalues_)
        self.explained_variance_ = self.eigvalues_ / self.variance_
        
    def transform(self, X):
        #project data onto eigenvectors
        print(self.components_.shape, X.shape)
        self.projected_ = X @ self.components_.T
        return self.projected_

pca = PCA()

# Generate some dummy data
subsample = np.random.randn(69,2)*0.1 
subsample[:,0] = subsample[:,0]*8 
subsample[:,1] = subsample[:,0]*2 + subsample[:,1] # Add some correlations

pca.fit(subsample)

plt.scatter(subsample[:,0], subsample[:,1], edgecolor='none', alpha=0.5)
plt.quiver(pca.components_[0,0]*2, pca.components_[0,1]*2, # *2 to make arrows larger
       angles='xy', scale_units='xy', scale=1, width=0.006)
plt.quiver(pca.components_[1,0]*2, pca.components_[1,1]*2, 
       angles='xy', scale_units='xy', scale=1, width=0.006)
plt.gca().set_aspect('equal')
plt.show()

手动实施 PCA 会产生错误的图，其中特征向量不正交

Manual Implementation of PCA produces a wrong plot, where eigenvectors are not orthogonal

python

numpy

machine-learning

pca

covariance