Question

这是一个玩具数据框。

self.next_track

我想将前三列分为Player，将后三列分为self.player = audiotools.player.Player( audio_output, replay_gain, self.next_track) # ^ no call parentheses，以便我得到以下内容

>library(tidyverse)
>a

id  e0  e1  e2  ee
1   0    1   2   3
1   0    1   2   3  
1   0    1   2   3
2   6    7   8   9
2   6    7   8   9
2   6    7   8   9

Answer 1

txt <- "id  e0  e1  e2  ee
1   0    1   2   3
1   0    1   2   3  
1   0    1   2   3
2   6    7   8   9
2   6    7   8   9
2   6    7   8   9"
a <- read.table(text = txt, header = TRUE)

使用dplyr：

library(dplyr)
a2 <- distinct(a)
bind_rows(
  select(a2, id, start = e0, end = e1),
  select(a2, id, start = e1, end = e2),
  select(a2, id, start = e2, end = ee)
)
#   id start end
# 1  1     0   1
# 2  2     6   7
# 3  1     1   2
# 4  2     7   8
# 5  1     2   3
# 6  2     8   9

以基地R：

do.call("rbind.data.frame",
        list(
          setNames(a2[,c("id","e0","e1")], c("id", "start", "end")),
          setNames(a2[,c("id","e1","e2")], c("id", "start", "end")),
          setNames(a2[,c("id","e2","ee")], c("id", "start", "end"))
        ))

修改：根据评论，如果可以安全地假设每个id的行数与e - 列减去1的行数完全相同，那么你可以这样做：

nc <- 3
a %>%
  group_by(id) %>%
  mutate(
    n = (row_number() - 1) %% nc + 1,
    start = recode(n, e0, e1, e2),
    end = recode(n, e1, e2, ee)
  ) %>%
  ungroup() %>%
  select(id, start, end)
# # A tibble: 6 × 3
#      id start   end
#   <int> <int> <int>
# 1     1     0     1
# 2     1     1     2
# 3     1     2     3
# 4     2     6     7
# 5     2     7     8
# 6     2     8     9

实际上，即使没有正确的行数，这也有效，但如果你不这样做，这可能是不正确的结果。

Answer 2

基本R选项是，

data.frame(ID = a$id, start = unique(c(t(a[2:4]))), end = unique(c(t(a[3:5]))))
#  ID start end
#1  1     0   1
#2  1     1   2
#3  1     2   3
#4  2     6   7
#5  2     7   8
#6  2     8   9

部分聚集4列，使2列dplyr

2 个答案: