Question

我有一个排序相同的数据帧列表。更具体的是这些是在使用AmeliaII包进行多次插补后得到的估算数据帧。现在我想创建一个结构相同的新数据帧，但包含在数据帧中计算的单元格的平均值。

目前我实现这一目标的方法如下：

## do the Amelia run ------------------------------------------------------------

a.out <- amelia(merged, m=5, ts="Year", cs ="GEO",polytime=1)

## Calculate the output statistics ----------------------------------------------
left.side <- a.out$imputations[[1]][,1:2]
a.out.ncol <- ncol(a.out$imputations[[1]])

a <- a.out$imputations[[1]][,3:a.out.ncol]
b <- a.out$imputations[[2]][,3:a.out.ncol]
c <- a.out$imputations[[3]][,3:a.out.ncol]
d <- a.out$imputations[[4]][,3:a.out.ncol]
e <- a.out$imputations[[5]][,3:a.out.ncol]

# Calculate the Mean of the matrices
mean.right <- apply(abind(a,b,c,d,e,f,g,h,i,j,along=3),c(1,2),mean) 

# recombine factors with values
mean <- cbind(left.side,mean.right)

我认为使用apply，plyr等有更好的方法可以做到这一点，但作为一个R新手，我真的有点迷失在这里。你对此有什么建议吗？

Answer 1

如果我理解你的问题，那么这应该会让你走得很远：

#set up some data:
dfr1<-data.frame(a=c(1,2.5,3), b=c(9.0,9,9))
dfr2<-data.frame(a=c(5,2,5), b=c(6,5,4))
tst<-list(dfr1, dfr2)
#since all variables are numerical, use a threedimensional array
tst2<-array(do.call(c, lapply(tst, unlist)), dim=c(nrow(tst[[1]]), ncol(tst[[1]]), length(tst)))
#To see where you're at:
tst2
#rowMeans for a threedimensional array and dims=2 does the mean over the last dimension
result<-data.frame(rowMeans(tst2, dims=2))
rownames(result)<-rownames(tst[[1]])
colnames(result)<-colnames(tst[[1]])
#display the full result
result

HTH。

Answer 2

以下是使用Reduce和plyr::llply

的替代方法

dfr1 <- data.frame(a = c(1,2.5,3), b = c(9.0,9,9), c = letters[1:3])
dfr2 <- data.frame(a = c(5,2,5), b = c(6,5,4), c = letters[1:3])

tst = list(dfr1, dfr2)

require(plyr)
tst2 = llply(tst, function(df) df[,sapply(df, is.numeric)]) # strip out non-numeric cols
ans  = Reduce("+", tst2)/length(tst2)

EDIT。您可以大大简化代码并在5行R代码中完成您想要的任务。以下是使用Amelia软件包的示例。

library(Amelia)
data(africa)

# carry out imputations
a.out      = amelia(x = africa, cs = "country", ts = "year", logs = "gdp_pc") 

# extract numeric columns from each element of a.out$impuations  
tst2       = llply(a.out$imputations, function(df) df[,sapply(df, is.numeric)]) 

# sum them up and divide by length to get mean
mean.right = Reduce("+", tst2)/length(tst2)

# compute fixed columns and cbind with mean.right
left.side  = a.out$imputations[[1]][1:2]
mean0      = cbind(left.side,mean.right)

Answer 3

经过多次尝试，我发现了一种合理快速的方法来计算单元在多个数据帧中的平均值。

# First create an empty data frame for storing the average imputed values. This
# data frame will have the same dimensions of the original one

imp.df <- df

# Then create an array with the first two dimensions of the original data frame and
# the third dimension given by the number of imputations

a <- array(NA, dim=c(nrow(imp.df), ncol(imp.df), length(a.out$imputations)))

# Then copy each imputation in each "slice" of the array

for (z in 1:length(a.out$imputations)) {
a[,,z] <- as.matrix(a.out$imputations[[z]])
}

# Finally, for each cell, replace the actual value with the mean across all 
# "slices" in the array

for (i in 1:dim(a)[1]) {
  for (j in 1:dim(a)[2]) {
imp.df[i, j] <- mean(as.numeric(a[i, j,]))
    }}

计算相同数据帧的单元格的统计数据（例如平均值）

3 个答案: