我正在尝试将rep
与dplyr
一起使用,但我不完全理解为什么我无法使其发挥作用。
我的数据看起来像这样。我想要的是简单地为每个dayweek
重复n
id
。
head(dt4)
id dayweek n
1 1 Friday 3
2 1 Monday 3
3 1 Saturday 3
4 1 Sunday 3
5 1 Thursday 3
6 1 Tuesday 3
我想要做的是在dplyr
流程
cbind(rep(dt4$id, dt4$n), rep(as.character(dt4$dayweek), dt4$n) )
给出了
[,1] [,2]
[1,] "1" "Friday"
[2,] "1" "Friday"
[3,] "1" "Friday"
[4,] "1" "Monday"
[5,] "1" "Monday"
[6,] "1" "Monday"
我不明白为什么这段代码不起作用
dt4 %>%
group_by(id) %>%
summarise(rep(dayweek, n))
Error: expecting a single value
有人可以帮我吗?
数据
dt4 = structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), dayweek = structure(c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 1L, 2L, 3L,
4L, 5L, 6L, 7L), .Label = c("Friday", "Monday", "Saturday", "Sunday",
"Thursday", "Tuesday", "Wedesnday"), class = "factor"), n = c(3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3)), class = "data.frame", .Names = c("id",
"dayweek", "n"), row.names = c(NA, -21L))
答案 0 :(得分:5)
data.table可以作为此类操作的有用替代方法 - 我发现这更容易阅读:
library("data.table")
dt4 <- as.data.table(dt4)
head(dt4[, rep(dayweek, n), by=id], 10)
,并提供:
id V1
1: 1 Friday
2: 1 Friday
3: 1 Friday
4: 1 Monday
5: 1 Monday
6: 1 Monday
7: 1 Saturday
8: 1 Saturday
9: 1 Saturday
10: 1 Sunday
答案 1 :(得分:3)
要获得与cbind
相同的结果,我们可以使用do
。正如@DavidArenburg所提到的,summarise
每组组合输出一个值/行,而使用mutate
我们得到的输出具有相同的行数。但是,在这里我们正在做一个可以在do
环境中完成的不同操作。代码.
表示数据集。如果我们想要提取“id”#39;来自dt4
的列,我们可以使用dt4$id
或dt4[['id']]
。将dt4
替换为.
。
library(dplyr)
dt4 %>%
group_by(id) %>%
do(data.frame(id=.$id, v1=rep(.$dayweek, .$n)))
#Source: local data frame [63 x 2]
#Groups: id
# id v1
#1 1 Friday
#2 1 Friday
#3 1 Friday
#4 1 Monday
#5 1 Monday
#6 1 Monday
#7 1 Saturday
#8 1 Saturday
#9 1 Saturday
#10 1 Sunday
#.. .. ...
或者基于@Frank评论的另一个选项是指定从rep
内slice
生成的行索引和select
我们需要保留的列。< / p>
dt4 %>%
slice(rep(1:n(),n)) %>%
select(-n)