Question

我有这样的数据集：

df=data.frame(subject= c(rep(1, 3), rep(2, 2),rep(3,4)), visit=c(1:3,1:2,1:4),time=c('2003-03-07 6:34','2003-03-07 7:33','2003-03-07 8:15','2003-03-15 6:42','2003-03-15 7:42','2003-03-16 6:20','2003-03-16 6:40','2003-03-16 7:38','2003-03-16 8:42')) 

  subject visit            time
1       1     1 2003-03-07 6:34
2       1     2 2003-03-07 7:33
3       1     3 2003-03-07 8:15
4       2     1 2003-03-15 6:42
5       2     2 2003-03-15 7:42
6       3     1 2003-03-16 6:20
7       3     2 2003-03-16 6:40
8       3     3 2003-03-16 7:38
9       3     4 2003-03-16 8:42

我希望创建一个列，使其包含每次访问时每个人的基准时间，预期输出应如下所示：

df1=data.frame(subject= c(rep(1, 3), rep(2, 2),rep(3,4)), visit=c(1:3,1:2,1:4),time=c('2003-03-07 6:34','2003-03-07 6:34','2003-03-07 6:34','2003-03-15 6:42','2003-03-15 6:42','2003-03-16 6:20','2003-03-16 6:20','2003-03-16 6:20','2003-03-16 6:20')) 

  subject visit            time
1       1     1 2003-03-07 6:34
2       1     2 2003-03-07 6:34
3       1     3 2003-03-07 6:34
4       2     1 2003-03-15 6:42
5       2     2 2003-03-15 6:42
6       3     1 2003-03-16 6:20
7       3     2 2003-03-16 6:20
8       3     3 2003-03-16 6:20
9       3     4 2003-03-16 6:20

有没有人知道如何实现这个目标？

Answer 1

选项1（假设排序顺序）：

do.call(rbind, lapply(split(df, df$subject), function(x) cbind(x,time2 = with(x, x$time[1]))))

选项2（一个稍微更强大的解决方案，确定哪个是第一次访问）：

do.call(rbind, lapply(split(df, df$subject), function(x) cbind(x,time2 = with(x, x$time[which(x$visit==1)]))))

选项3（转换为POSIXct并使用min）：

do.call(rbind, lapply(split(df, df$subject), function(x) cbind(x,time2 = min(as.POSIXct(x$time)))))

选项4（可能最快/最简单）：

within(df, time2 <- ave(as.POSIXct(time), subject, FUN = min))

选项5（再次假定排序顺序）：

within(df, time2 <- ave(time, subject, FUN = function(x) head(x, 1)))

所有这些都会给你：

    subject visit            time           time2
1.1       1     1 2003-03-07 6:34 2003-03-07 6:34
1.2       1     2 2003-03-07 7:33 2003-03-07 6:34
1.3       1     3 2003-03-07 8:15 2003-03-07 6:34
2.4       2     1 2003-03-15 6:42 2003-03-15 6:42
2.5       2     2 2003-03-15 7:42 2003-03-15 6:42
3.6       3     1 2003-03-16 6:20 2003-03-16 6:20
3.7       3     2 2003-03-16 6:40 2003-03-16 6:20
3.8       3     3 2003-03-16 7:38 2003-03-16 6:20
3.9       3     4 2003-03-16 8:42 2003-03-16 6:20

Answer 2

data.table方法

library(data.table)
setDT(df)[, time2 := min(as.POSIXct(time)), by = subject]

dplyr方法

library(dplyr)
df %>%
  group_by(subject) %>%
  mutate(time = min(as.POSIXct(time)))

Answer 3

您可以使用dplyr。

require(dplyr)

df %>%
  group_by(subject) %>%
  summarize(time2 = time[1]) %>%
  left_join(df, by = "subject")

这里是结果数据框：

  subject           time2 visit            time
1       1 2003-03-07 6:34     1 2003-03-07 6:34
2       1 2003-03-07 6:34     2 2003-03-07 7:33
3       1 2003-03-07 6:34     3 2003-03-07 8:15
4       2 2003-03-15 6:42     1 2003-03-15 6:42
5       2 2003-03-15 6:42     2 2003-03-15 7:42
6       3 2003-03-16 6:20     1 2003-03-16 6:20
7       3 2003-03-16 6:20     2 2003-03-16 6:40
8       3 2003-03-16 6:20     3 2003-03-16 7:38
9       3 2003-03-16 6:20     4 2003-03-16 8:42

创建一个包含每个人的基准时间的列

3 个答案: