模拟练习数据

set.seed(8888)
id <- rep(1:15,each=10)#id号，1-15重复10次
time<-rep(1:10,15)#时间（天）1-10重复15次
map.raw <- abs((round(rnorm(150,mean = 50,sd = 25))))#随机取150个符合正态分布的平均动脉压
map<-round(ifelse(map.raw>=30,map.raw,map.raw+50))#保证没有小于30的平均动脉压
lac<-round(3-map*0.06+rnorm(150,mean = 0,sd = 0.4)-0.4*time*time+4.3*time,1)#乳酸值
lac.miss.tag<-rbinom(150,1,0.3)#取30%的缺失值
lac<-ifelse(lac.miss.tag==1,NA,lac)
age<-rep(round(abs(rnorm(15,mean = 65,sd = 19))),each=10)
data<- data.frame(id,time,age,map,lac)

abs(x)取x的绝对值
round（x）取x的整数

aggr(data,numbers = TRUE, prop= FALSE)

进行插补

 A <- amelia(data,m = 5,ts = "time",cs = "id")

查看插补效果

tscsPlot(A,cs = c(3,4,5,6),var = "lac")

下图红点是插补的值，红线是其置信范围
可见置信范围较大，插补的效果不好

考虑时间因素，重新插补

A2<- amelia(data,m = 5,ts = "time",cs = "id",polytime = 2)
polytime#取0到3之间的整数，表示应在插补模型中包含多项式的幂以说明时间的影响。 设置为0将指示恒定水平，将1指示线性时间效应，将2指示平方效应，而将3指示立方时间效应。

可见插补效果很好

tscsPlot(A2,cs = c(3,4,5,6),var = "lac")

考虑缺失值前后数值影响

lags参数和leads参数

A3<- amelia(data,m = 5,ts = "time",cs = "id",lags = "lac",leads = "lac")

效果也不好

tscsPlot(A3,cs = c(3,4,5,6),var = "lac")

加入先验信息

有时候根据文献和经验，某一个缺失变量具有一定的取值范围
如lac均值为3的患者容易存活，而且其标准差为1.2

使用prior参数
You can provide Amelia with informational priors about the missing observations in your data. To specify priors, pass a four or five column matrix to the priors argument with each row specifying a different priors as such:
one.prior <- c(row, column, mean,standard deviation)
or,
one.prior <- c(row, column, minimum, maximum, confidence).

假设第4例患者存活，我们认为其乳酸水平符合经验

data[data$id==4,]
lac.prior<-matrix(c(31,32,37,38,5,5,5,5,3,3,3,3,1.2,1.2,1.2,1.2),nrow = 4,ncol = 4)
A4<- amelia(data,m = 5,ts = "time",cs = "id",priors = lac.prior)

插补值诊断：插补值是否合理

opar <- par(no.readonly=TRUE)
par(mfrow=c(2,2))
compare.density(A,var = "lac",main = "1")
compare.density(A2,var = "lac",main = "2")
compare.density(A3,var = "lac",main = "3")
compare.density(A4,var = "lac",main = "4")
par(opar)

当红蓝两条曲线重合程度越多时，证明插补越合理

opar <- par(no.readonly=TRUE)
par(mfrow=c(2,2))
overimpute(A,var = "lac",main = "1")
overimpute(A2,var = "lac",main = "2")
overimpute(A3,var = "lac",main = "3")
overimpute(A4,var = "lac",main = "4")
par(opar)

当点越靠近斜线周围时，插补越合理

最后编辑于：2020.01.10 20:17:20