KMO检验和巴特莱特球形检验
一、R语言
Bartlett球形检验： library(psych) cortest.bartlett(cor(dataframe), n=sample_size) 或 cortest.bartlett(dataframe)
KMO检验： https://stat.ethz.ch/pipermail/r-help/2007-August/138049.html # KMO Kaiser-Meyer-Olkin Measure of Sampling Adequacy kmo = function( data ){ library(MASS) X <- cor(as.matrix(data)) iX <- ginv(X) S2 <- diag(diag((iX^-1))) AIS <- S2%*%iX%*%S2 # anti-image covariance matrix IS <- X+AIS-2*S2 # image covariance matrix Dai <- sqrt(diag(diag(AIS))) IR <- ginv(Dai)%*%IS%*%ginv(Dai) # image correlation matrix AIR <- ginv(Dai)%*%AIS%*%ginv(Dai) # anti-image correlation matrix a <- apply((AIR - diag(diag(AIR)))^2, 2, sum) AA <- sum(a) b <- apply((X - diag(nrow(X)))^2, 2, sum) BB <- sum(b) MSA <- b/(b+a) # indiv. measures of sampling adequacy AIR <- AIR-diag(nrow(AIR))+diag(MSA) # Examine the anti-image of the # correlation matrix. That is the # negative of the partial correlations, # partialling out all other variables. kmo <- BB/(AA+BB) # overall KMO statistic # Reporting the conclusion if (kmo >= 0.00 && kmo < 0.50){ test <- 'The KMO test yields a degree of common variance unacceptable for FA.' } else if (kmo >= 0.50 && kmo < 0.60){ test <- 'The KMO test yields a degree of common variance miserable.' } else if (kmo >= 0.60 && kmo < 0.70){ test <- 'The KMO test yields a degree of common variance mediocre.' } else if (kmo >= 0.70 && kmo < 0.80){ test <- 'The KMO test yields a degree of common variance middling.' } else if (kmo >= 0.80 && kmo < 0.90){ test <- 'The KMO test yields a degree of common variance meritorious.' } else { test <- 'The KMO test yields a degree of common variance marvelous.' } ans <- list( overall = kmo, report = test, individual = MSA, AIS = AIS, AIR = AIR ) return(ans) } # end of kmo() kmo(dataframe) ## R Software ## R language
二、python
链接：https://www.cnblogs.com/caiyishuai/p/12421034.html
import pandas as pd import numpy as np import math as math import numpy as np from numpy import * from scipy.stats import bartlett from factor_analyzer import * import numpy.linalg as nlg from sklearn.cluster import KMeans from matplotlib import cm import matplotlib.pyplot as plt def main(): df=pd.read_csv("./data/applicant.csv") # print(df) df2=df.copy() print("\n原始数据:\n",df2) del df2['ID'] # print(df2) # 皮尔森相关系数 df2_corr=df2.corr() print("\n相关系数:\n",df2_corr) #热力图 cmap = cm.Blues # cmap = cm.hot_r fig=plt.figure() ax=fig.add_subplot(111) map = ax.imshow(df2_corr, interpolation='nearest', cmap=cmap, vmin=0, vmax=1) plt.title('correlation coefficient--headmap') ax.set_yticks(range(len(df2_corr.columns))) ax.set_yticklabels(df2_corr.columns) ax.set_xticks(range(len(df2_corr))) ax.set_xticklabels(df2_corr.columns) plt.colorbar(map) plt.show() # KMO测度 def kmo(dataset_corr): corr_inv = np.linalg.inv(dataset_corr) nrow_inv_corr, ncol_inv_corr = dataset_corr.shape A = np.ones((nrow_inv_corr, ncol_inv_corr)) for i in range(0, nrow_inv_corr, 1): for j in range(i, ncol_inv_corr, 1): A[i, j] = -(corr_inv[i, j]) / (math.sqrt(corr_inv[i, i] * corr_inv[j, j])) A[j, i] = A[i, j] dataset_corr = np.asarray(dataset_corr) kmo_num = np.sum(np.square(dataset_corr)) - np.sum(np.square(np.diagonal(A))) kmo_denom = kmo_num + np.sum(np.square(A)) - np.sum(np.square(np.diagonal(A))) kmo_value = kmo_num / kmo_denom return kmo_value print("\nKMO测度:", kmo(df2_corr)) # 巴特利特球形检验 df2_corr1 = df2_corr.values print("\n巴特利特球形检验:", bartlett(df2_corr1[0], df2_corr1[1], df2_corr1[2], df2_corr1[3], df2_corr1[4], df2_corr1[5], df2_corr1[6], df2_corr1[7], df2_corr1[8], df2_corr1[9], df2_corr1[10], df2_corr1[11], df2_corr1[12], df2_corr1[13], df2_corr1[14])) # 求特征值和特征向量 eig_value, eigvector = nlg.eig(df2_corr) # 求矩阵R的全部特征值，构成向量 eig = pd.DataFrame() eig['names'] = df2_corr.columns eig['eig_value'] = eig_value eig.sort_values('eig_value', ascending=False, inplace=True) print("\n特征值\n：",eig) eig1=pd.DataFrame(eigvector) eig1.columns = df2_corr.columns eig1.index = df2_corr.columns print("\n特征向量\n",eig1) # 求公因子个数m,使用前m个特征值的比重大于85%的标准，选出了公共因子是五个 for m in range(1, 15): if eig['eig_value'][:m].sum() / eig['eig_value'].sum() >= 0.85: print("\n公因子个数:", m) break # 因子载荷阵 A = np.mat(np.zeros((15, 5))) i = 0 j = 0 while i < 5: j = 0 while j < 15: A[j:, i] = sqrt(eig_value[i]) * eigvector[j, i] j = j + 1 i = i + 1 a = pd.DataFrame(A) a.columns = ['factor1', 'factor2', 'factor3', 'factor4', 'factor5'] a.index = df2_corr.columns print("\n因子载荷阵\n", a) fa = FactorAnalyzer(n_factors=5) fa.loadings_ = a # print(fa.loadings_) print("\n特殊因子方差:\n", fa.get_communalities()) # 特殊因子方差，因子的方差贡献度 ，反映公共因子对变量的贡献 var = fa.get_factor_variance() # 给出贡献率 print("\n解释的总方差（即贡献率）:\n", var) # 因子旋转 rotator = Rotator() b = pd.DataFrame(rotator.fit_transform(fa.loadings_)) b.columns = ['factor1', 'factor2', 'factor3', 'factor4', 'factor5'] b.index = df2_corr.columns print("\n因子旋转:\n", b) # 因子得分 X1 = np.mat(df2_corr) X1 = nlg.inv(X1) b = np.mat(b) factor_score = np.dot(X1, b) factor_score = pd.DataFrame(factor_score) factor_score.columns = ['factor1', 'factor2', 'factor3', 'factor4', 'factor5'] factor_score.index = df2_corr.columns print("\n因子得分：\n", factor_score) fa_t_score = np.dot(np.mat(df2), np.mat(factor_score)) print("\n应试者的五个因子得分：\n",pd.DataFrame(fa_t_score)) # 综合得分 wei = [[0.50092], [0.137087], [0.097055], [0.079860], [0.049277]] fa_t_score = np.dot(fa_t_score, wei) / 0.864198 fa_t_score = pd.DataFrame(fa_t_score) fa_t_score.columns = ['综合得分'] fa_t_score.insert(0, 'ID', range(1, 49)) print("\n综合得分：\n", fa_t_score) print("\n综合得分：\n", fa_t_score.sort_values(by='综合得分', ascending=False).head(6)) plt.figure() ax1=plt.subplot(111) X=fa_t_score['ID'] Y=fa_t_score['综合得分'] plt.bar(X,Y,color="#87CEFA") # plt.bar(X, Y, color="red") plt.title('result00') ax1.set_xticks(range(len(fa_t_score))) ax1.set_xticklabels(fa_t_score.index) plt.show() fa_t_score1=pd.DataFrame() fa_t_score1=fa_t_score.sort_values(by='综合得分',ascending=False).head() ax2 = plt.subplot(111) X1 = fa_t_score1['ID'] Y1 = fa_t_score1['综合得分'] plt.bar(X1, Y1, color="#87CEFA") # plt.bar(X1, Y1, color='red') plt.title('result01') plt.show() if __name__ == '__main__': main()
