2

考虑以下具有 4 列的数据框：

df = data.frame(A = rnorm(10), B = rnorm(10), C = rnorm(10), D = rnorm(10))

A、B、C、D 列属于不同的组，这些组在单独的数据框中定义：

groups = data.frame(Class = c("A","B","C","D"), Group = c("G1", "G2", "G2", "G1"))

#> groups
#  Class Group
#1     A    G1
#2     B    G2
#3     C    G2
#4     D    G1

我想平均属于同一组的列的元素，并得到类似于：

#> res
#            G1          G2
#1  -0.30023039 -0.71075139
#2   0.53053443 -0.12397126
#3   0.21968567 -0.46916160
#4  -1.13775100 -0.61266026
#5   1.30388130 -0.28021734
#6   0.29275876 -0.03994522
#7  -0.09649998  0.59396983
#8   0.71334020 -0.29818438
#9  -0.29830924 -0.47094084
#10 -0.36102888 -0.40181739

其中 G1 的每个单元格是 A 和 D 的相关单元格的平均值，G2 的每个单元格是 B 和 C 的相关单元格的平均值，以此类推。

我能够达到这个结果，但是以一种相当蛮力的方式：

l = levels(groups$Group)
res = data.frame(matrix(nc = length(levels), nr = nrow(df)))
for(i in 1:length(l)) {
    df.sub = df[which(groups$Group == l[i])]
    res[,i] = apply(df.sub, 1, mean)
}
names(res) <- l

有没有更好的方法来做到这一点？实际上，我有 20 多个专栏和 10 多个小组。

谢谢！

4

2 回答 2

3

使用数据表

library(data.table)
groups <- data.table(groups, key="Group")
DT <- data.table(df)

groups[, rowMeans(DT[, Class, with=FALSE]), by=Group][, setnames(as.data.table(matrix(V1, ncol=length(unique(Group)))), unique(Group))]

             G1         G2
 1: -0.13052091 -0.3667552
 2:  1.17178729 -0.5496347
 3:  0.23115841  0.8317714
 4:  0.45209516 -1.2180895
 5: -0.01861638 -0.4174929
 6: -0.43156831  0.9008427
 7: -0.64026238  0.1854066
 8:  0.56225108 -0.3563087
 9: -2.00405840 -0.4680040
10:  0.57608055 -0.6177605



# Also, make sure you have characters, not factors, 
groups[, Class := as.character(Class)]
groups[, Group := as.character(Group)]

简单的基础： 

 tapply(groups$Class, groups$Group, function(X) rowMeans(df[, X]))

使用sapply：

 sapply(unique(groups$Group), function(X) 
     rowMeans(df[, groups[groups$Group==X, "Class"]]) )
于 2013-10-23T16:19:53.077 回答
0

我个人会选择 Ricardo 的解决方案，但另一种选择是merge首先使用您的两个数据集，然后使用您首选的聚合方法。

library(reshape2)

## Retain the "rownames" so we can aggregate by row
temp <- merge(cbind(id = rownames(df), melt(df)), groups, 
              by.x = "variable", by.y = "Class")
head(temp)
#   variable id      value Group
# 1        A  1 -0.6264538    G1
# 2        A  2  0.1836433    G1
# 3        A  3 -0.8356286    G1
# 4        A  4  1.5952808    G1
# 5        A  5  0.3295078    G1
# 6        A  6 -0.8204684    G1

## This is the perfect form for `dcast` to do its work
dcast(temp, id ~ Group, value.var="value", mean)
#    id          G1          G2
# 1   1  0.36611287  1.21537927
# 2  10  0.22889368  0.50592144
# 3   2  0.04042780  0.58598977
# 4   3 -0.22397850 -0.27333780
# 5   4  0.77073788 -2.10202579
# 6   5 -0.52377589  0.87237833
# 7   6 -0.61773147 -0.05053117
# 8   7  0.04656955 -0.08599288
# 9   8  0.33950565 -0.26345809
# 10  9  0.83790336  0.17153557

（以上数据使用set.seed(1)您的样本“df”。

于 2013-10-23T16:33:12.943 回答