将 ID 分配给连续的组列 r

Question

我想在 data.frame 中生成一个列，用于计算组的连续 ID（虚拟 df 中的 s 列）

dummy_df = data.frame(s = c("a", "a", "b","b", "b", "c","c", "a", "a", "c", "c","a","a"),
                  desired_output= c(1,1,1,1,1,1,1,2,2,2,2,3,3))
dummy_df$rleid_output = rleid(dummy_df$s)
dummy_df

   s desired_output rleid_output
1  a              1            1
2  a              1            1
3  b              1            2
4  b              1            2
5  b              1            2
6  c              1            3
7  c              1            3
8  a              2            4
9  a              2            4
10 c              2            5
11 c              2            5
12 a              3            6
13 a              3            6

我会说它类似于 rleid() 所做的，但在看到新组时重新开始计数。但是，我找不到以如此直接的方式做到这一点的方法。谢谢

Answer 1

您可以尝试将 tidyverse 与基础 R rle 函数结合使用

library(tidyverse)
rle(dummy_df$s) %>% 
  with(., data.frame(a=.$length, b=.$value)) %>% 
  group_by(b) %>% 
  mutate(n = 1:n()) %>%
  with(., rep(n, times=a)) %>% 
  bind_cols(dummy_df, res=.)
   s desired_output res
1  a              1   1
2  a              1   1
3  b              1   1
4  b              1   1
5  b              1   1
6  c              1   1
7  c              1   1
8  a              2   2
9  a              2   2
10 c              2   2
11 c              2   2
12 a              3   3
13 a              3   3

Answer 2

你可以这样做：

dummy_df$out <- with(rle(dummy_df$s), rep(ave(lengths, values, FUN = seq_along), lengths))

结果：

   s desired_output out
1  a              1   1
2  a              1   1
3  b              1   1
4  b              1   1
5  b              1   1
6  c              1   1
7  c              1   1
8  a              2   2
9  a              2   2
10 c              2   2
11 c              2   2
12 a              3   3
13 a              3   3

Answer 3

如果您愿意使用data.table（rleid是软件包的一部分），您可以分两步完成：

library(data.table)
dummy_df = data.frame(s = c("a", "a", "b", "b", "b", "c", "c", "a", "a", "c", "c", "a", "a"))
# cast data.frame to data.table
setDT(dummy_df)
# create auxiliary variable
dummy_df[, rleid_output := rleid(s)]
# obtain desired output
dummy_df[, desired_output := rleid(rleid_output), by = "s"]
# end result
dummy_df
#>     s rleid_output desired_output
#>  1: a            1              1
#>  2: a            1              1
#>  3: b            2              1
#>  4: b            2              1
#>  5: b            2              1
#>  6: c            3              1
#>  7: c            3              1
#>  8: a            4              2
#>  9: a            4              2
#> 10: c            5              2
#> 11: c            5              2
#> 12: a            6              3
#> 13: a            6              3

^{由 reprex package (v0.3.0)}

于 2020-10-16 创建

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