8

这是我的数据的一个可重复的小示例:

> mydata <- structure(list(subject = c(1, 1, 1, 2, 2, 2), time = c(0, 1, 2, 0, 1, 2), measure = c(10, 12, 8, 7, 0, 0)), .Names = c("subject", "time", "measure"), row.names = c(NA, -6L), class = "data.frame")

> mydata

subject  time  measure
1          0      10
1          1      12
1          2       8
2          0       7
2          1       0
2          2       0

我想生成一个新变量,其中包含该measure特定主题的平均值,因此:

subject  time  measure  mn_measure
1          0      10      10
1          1      12      10
1          2       8      10
2          0       7      2.333
2          1       0      2.333
2          2       0      2.333

有没有一种简单的方法可以做到这一点,除了以编程方式循环遍历所有记录或首先重塑为宽格式?

4

3 回答 3

14

使用基本 R 函数ave(),尽管它的名称令人困惑,但它可以计算各种统计数据,包括mean

within(mydata, mean<-ave(measure, subject, FUN=mean))

  subject time measure      mean
1       1    0      10 10.000000
2       1    1      12 10.000000
3       1    2       8 10.000000
4       2    0       7  2.333333
5       2    1       0  2.333333
6       2    2       0  2.333333

请注意,我within只是为了缩短代码而使用。这是没有的等价物within()

mydata$mean <- ave(mydata$measure, mydata$subject, FUN=mean)
mydata
  subject time measure      mean
1       1    0      10 10.000000
2       1    1      12 10.000000
3       1    2       8 10.000000
4       2    0       7  2.333333
5       2    1       0  2.333333
6       2    2       0  2.333333
于 2013-02-11T12:52:21.310 回答
9

或者与data.table包:

require(data.table)
dt <- data.table(mydata, key = "subject")
dt[, mn_measure := mean(measure), by = subject]

#   subject time measure mn_measure
# 1:       1    0      10  10.000000
# 2:       1    1      12  10.000000
# 3:       1    2       8  10.000000
# 4:       2    0       7   2.333333
# 5:       2    1       0   2.333333
# 6:       2    2       0   2.333333
于 2013-02-11T12:52:41.880 回答
6

您可以ddplyplyr包中使用:

library(plyr)
res = ddply(mydata, .(subject), mutate, mn_measure = mean(measure))
res
  subject time measure mn_measure
1       1    0      10  10.000000
2       1    1      12  10.000000
3       1    2       8  10.000000
4       2    0       7   2.333333
5       2    1       0   2.333333
6       2    2       0   2.333333
于 2013-02-11T12:50:13.043 回答