R语言 python aovova 方差分析，运行结果正确，能帮忙核对下么

Question

R语言转成python. 求助

R 语言如下

for(i in 1:nz){
  y<- StableM[StableM$Parameter==paste(z[i,1]),c("filename","Group","Value")]
  if( (nrow(y)>=5) && (length(unique(y$Group))>1)) 
  {z[i,2]<- summary(aov(as.numeric(y$Value) ~ y$Group, data=y )) [[1]] [[1,"Pr(>F)"]]}
}

Python 写成了下面的样子，运行应该是正确的没有问题吧

 y<- StableM[StableM$Parameter==paste(z[i,1]),c("filename","Group","Value")]
 {z[i,2]<- summary(aov(as.numeric(y$Value) ~ y$Group, data=y )) [[1]] [[1,"Pr(>F)"]]}

from statsmodels.formula.api import ols
from statsmodels.stats.anova import anova_lm  
    z = StableM['Par'].drop_duplicates()### extract abnormal
    par_list = z.tolist() 
    z = pd.DataFrame(z)
    z.loc[:,'pValue'] = 1 #df_one = pd.DataFrame(np.ones((z.size,1),dtype = np.int), columns=['pValue'])  
    nz =z.shape[0]
    #p_value= pd.DataFrame({'pValue':[]})
    
    for i in range(0,nz): #print(i) -> 0,1
        y = StableM[StableM['Par'] == par_list[i]] # paste(..., sep = " ", collapse = NULL)
        if (y.shape[0] >=5) & (len(y['Group'].unique()) > 1): 
            model = ols('Value ~  C(Group)',y).fit()
            anovat = anova_lm(model)
            aov_table = anova_lm(model, typ=2)
            z.loc[z['Par'] == par_list[i],'pValue'] = aov_table['PR(>F)'].iloc[0] # UpperHeat OutPut pvalue = 1.9472e-05
            print(anovat)    #PR(>F)   p-value