我在R中有一个这样的列表:
cat1
cat7
cat10
cat4
frog
dino11
dino12
dino15
rabbit
我需要制作一个新的数据框,如下所示:
cat1 frog
cat7 frog
cat10 frog
cat4 frog
dino11 rabbit
dino12 rabbit
dino15 rabbit
想法?谢谢!
答案 0 :(得分:2)
我们根据'v1'中数字的不出现情况创建分组变量,以lag
,创建一个新列'v2'作为'v1'的last
元素,删除每个组的last
行和感兴趣的select
列
library(tidyverse)
df %>%
group_by(grp = lag(cumsum(grepl("^[^0-9]+$", v1)), default = 0)) %>%
mutate(v2 = last(v1)) %>%
slice(-n()) %>%
ungroup %>%
select(-grp)
# A tibble: 7 x 2
# v1 v2
# <chr> <chr>
#1 cat1 frog
#2 cat7 frog
#3 cat10 frog
#4 cat4 frog
#5 dino11 rabbit
#6 dino12 rabbit
#7 dino15 rabbit
df <- structure(list(v1 = c("cat1", "cat7", "cat10", "cat4", "frog",
"dino11", "dino12", "dino15", "rabbit")), .Names = "v1",
class = "data.frame", row.names = c(NA, -9L))
答案 1 :(得分:2)
类似于@akrun的答案,但带有data.table:
library(data.table)
setDT(df)
df[, .(
anum = v1[-.N],
a = v1[.N]
), by=.(g = cumsum(!(shift(v1) %like% "\\d")))]
g anum a
1: 1 cat1 frog
2: 1 cat7 frog
3: 1 cat10 frog
4: 1 cat4 frog
5: 2 dino11 rabbit
6: 2 dino12 rabbit
7: 2 dino15 rabbit
答案 2 :(得分:2)
仅对于基数R,您可以使用grepl
和rle
来实现。
where <- grepl("[[:digit:]]", x)
r <- rle(where)
A <- x[where]
B <- rep.int(x[!where], times = r$lengths[r$values])
data.frame(A, B)
# A B
#1 cat1 frog
#2 cat7 frog
#3 cat10 frog
#4 cat4 frog
#5 dino11 rabbit
#6 dino12 rabbit
#7 dino15 rabbit
数据。
x <- scan(what = character(), text = "
cat1
cat7
cat10
cat4
frog
dino11
dino12
dino15
rabbit
")