我试图用R.中的systemfit软件包来完成看似不相关的回归(SUR),然而,用多个推测的数据(用鼠标包)来完成这些分析并不简单。
在googling这个问题时,我发现有一个关于相同问题的被删除的帖子,它似乎使用了下面的例子(归功于海报,小编辑)。
library(systemfit)
library(mice)
nhanes2
r1 <- bmi ~ hyp
r2 <- bmi ~ age
system <- list( r1, r2 )
imp <- mice(nhanes2, m = 5)
m=5
completed=lapply(1:5,function(i)complete(imp,i))
fit.model <- systemfit(system, data= completed[[1]])以上为每个估算的数据集生成完整的系统匹配输出。这是很好的,但我剩下的任务是汇集SUR产生的全部输出。
我还尝试在zelig中运行该函数,但没有成功:
completed.mi=do.call(Zelig:mi,completed)
system=list(r1= bmi ~ hyp,r2=bmi~age)
z.out=zelig(formula= system,model="sur",data=completed.mi)
>Error: sur is not a supported model type.最后,调用计算数据的长形式会产生很大的自由度,它不反映每个估算数据集中的实际案例数(示例不包括在内)。
我的问题是:
1)系统是否支持MI数据?
2)如果是这样的话,您是否能够将输出集中在所有推测的数据集中?
( 3)在R中完成SUR是否有替代的一揽子方案(系统适配除外)?
4)如果3为否,是否有类似的分析将实现同样的目标,是否有不同的一揽子(如均方根)可能支持集中管理信息数据?
发布于 2019-02-01 12:03:10
我不认为老鼠支持汇集SUR的结果。您必须使用Rubin的规则手动汇总结果。我可以用你的例子讲到一个特定的点,也许你可以从那里学到。
library(systemfit)
library(mice)
nhanes2
# add two imputation as example
imp <- mice(nhanes2, m = 2)
m=2
# create a data set with all the
# complete imputed data sets
df<-mice::complete(imp, action="long", include=FALSE)
#create separate data frames for each mi
for(i in (df$.imp)) {
nam <- paste0("df_", i)
assign(nam, df[df$.imp==i,])
}
# Store the coefficients and se of each
# sur in the M imputed data sets
M <-2 # number of imputations
M2 <- M*2 #number of total sur regressions
results <- as.data.frame(matrix(NA, nrow=M2, ncol = 4)) # will store here coefficients and se
colnames(results)<-c("coef_r1", "coef_r2", "se_r1", "se_r2")
# perform sur
r1 <- bmi ~ hyp
r2 <- bmi ~ age
system <- list( r1, r2 )
# start with first data set
fitsur1 <- systemfit(list( r1= r1, r2 = r2),
data=df_1)
# start with second data set
fitsur2 <- systemfit(list( r1= r1, r2 = r2),
data=df_2)
# this can be warped in a loop
# but I could not do it...
# Extract coefficients
# Note I extract the coefficient
# from only one age-group of r2,
# Use same approach for the other
# extract coef from fitsur1
a <- as.data.frame(summary(fitsur1 )$coefficients[2,1])
colnames(a)<-c("coef_r1")
b <- as.data.frame(summary(fitsur1 )$coefficients[4,1])
colnames(b)<-c("coef_r2")
ab <- cbind(a,b)
# extract coef from fitsur2
c <- as.data.frame(summary(fitsur2 )$coefficients[2,1])
colnames(c)<-c("coef_r1")
d <- as.data.frame(summary(fitsur2 )$coefficients[4,1])
colnames(d)<-c("coef_r2")
cd <- cbind(c,d)
# Follow same approach to extract SE
# I cannot manage to extract them from
# the 'fitsur' list ...
# merge extracted coef and se
results <- rbind(ab, cd)
# Then bind the coefficients and se
# from all imputed regressions
# Calculate the mean of pooled coefficients
pooled.coef_r1 <- mean(results$coef_r1)
pooled.coef_r2 <- mean(results$coef_r2)计算集合SE更复杂--我使用了这个post https://stats.stackexchange.com/questions/327237/calculating-pooled-p-values-manually
# example for se_r1
(betweenVar <- mean(results[,3])) # mean of variances
(withinVar <- sd(results[,1])^2) # variance of variances
(dfCorrection <- (nrow(results)+1)/(nrow(results))) # dfCorrection
(totVar <- betweenVar + withinVar*dfCorrection) # total variance
(pooledSE <- sqrt(totVar)) # standard error我还没有查看p值,但是现在应该更容易了。
https://stackoverflow.com/questions/49889123
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