因子分析：KMO检验和巴特利球体检验-Python code

法一：采用factor_analyzer模块方法：

from factor_analyzer import factor_analyzer

# KMO值
print round(factor_analyzer.calculate_kmo(X_basic)[1],5)
# 巴特利特球形度值
print round(factor_analyzer.calculate_bartlett_sphericity(X_basic)[1], 5)

法二：自己实现代码检验：
def kmo():
    # 相关系数
    corr = X_basic.corr().values
    # print corr
    # 逆矩阵
    corr_inv = np.linalg.inv(corr)
    # 对角线取倒数，其他为0
    S2 = np.diag(corr_inv)
    S2 = np.linalg.inv(np.diag(S2))
    # 反映像协方差矩阵
    AIS = np.dot(S2, corr_inv)
    AIS = np.dot(AIS, S2)
    # 是映像协方差矩阵
    IS = corr + AIS - 2 * S2
    # 将矩阵AIS对角线上的元素开平方，并且将其余元素都变成0
    Dai = np.diag(np.sqrt(np.diag(AIS)))
    # IR是映像相关矩阵
    IR = np.dot(np.linalg.inv(Dai), IS)
    IR = np.dot(IR, np.linalg.inv(Dai))

    # AIR是反映像相关矩阵
    AIR = np.dot(np.linalg.inv(Dai), AIS)
    AIR = np.dot(AIR, np.linalg.inv(Dai))
    a = np.power((AIR - np.diag(np.diag(AIR))), 2)
    a = np.sum(a, axis=1)
    AA = np.sum(a)

    b = corr - np.identity(15)
    b = np.power(b, 2)
    b = np.sum(b, axis=1)
    BB = np.sum(b)

    MSA = b / (b + a)
    AIR = AIR - np.identity(15) + np.diag(MSA)

    kmo = BB / (AA + BB)
    return kmo

def bart():
    corr = X_basic.corr().values
    # 计算结果有问题
    # bart=st.bartlett(*corr)
    detCorr = np.linalg.det(corr)
    n = len(X_basic)
    p = len(X_basic.columns)
    statistic = -math.log(detCorr) * (n - 1 - (2 * p + 5) / 6)
    df = p * (p - 1) / 2
    # 双侧概率
    pval = (1.0 - st.chi2.cdf(statistic, df)) * 2
    return statistic,pval

转载于:https://www.cnblogs.com/laoketeng/p/11268903.html