为 PCA 生成 95% 预测省略号

问题描述

我正在尝试编写一个函数，该函数根据前两个主成分计算 PCA 的 95% 预测椭圆的中心、宽度、高度和旋转，并使用 p

import numpy as np
import pandas as pd
from sklearn.decomposition import PCA 
from scipy.spatial.distance import mahalanobis 
from scipy.stats import chi2 

def pca_ci_out(df,ci=0.95,pci=5):
    pc = pci
    id_list = df.reset_index()['ID'].to_list()
    df_array = df.to_numpy()
    pca = PCA()
    pcs_h = pca.fit_transform(df_array)
 
    #calculate 95% confidence intervals for the mahalanobis distance from the centroid for outlier detection
    pcs_hi = pcs_h[:,:pc]
    mean = np.mean(pcs_hi,axis=0)
    inv_cov = np.linalg.inv(np.cov(pcs_hi,rowvar=False))
    dist = {id_list[i] : mahalanobis(pcs_hi[i,:],mean,inv_cov) for i in range(pcs_h.shape[0])}
    


    #we detetermine outliers as values that are very far the centroid(p = 0.001| 0.999)
    extreme_dist = np.sqrt(chi2.ppf(0.999,pc))
    outliers = [[key,*list(pcs_h[id_list.index(key),:2]),value] for key,value in dist.items() if value >= extreme_dist]
    


    # confidence interval is based on only on top two PC's since that is what we will plot
    cov_mat = np.cov(pcs_h[:,:2],rowvar=False)
    chi2_95ci = chi2.ppf(ci,2)
    eigenvalues,eigenvectors = np.linalg.eig(cov_mat)


    major = 2 * np.sqrt(chi2_95ci * eigenvalues[0])
    minor = 2 * np.sqrt(chi2_95ci * eigenvalues[1])
    


    rotation = np.arctan(eigenvectors[0][1]/eigenvectors[0][0])
    center = np.mean(pcs_h[:,axis=0)
    CI_ellipse = [tuple(center),major,minor,rotation]
    
return CI_ellipse,outliers

解决方法

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