Special grouping number for each pair

Question

There is already part of the answer to the question special-group-number-for-each-combination-of-data . In most cases, we have pairs and other data values ​​inside the data. What we want to achieve is the number of those groups if these pairs exist and their number up to the next pairs.

Since I concentrated each pair, such as c("bad","good"), I would like to group them, and for couples c('Veni',"vidi","Vici")to assign a unique number 666.

Here is sample data

names <- c(c("bad","good"),1,2,c("good","bad"),111,c("bad","J.James"),c("good","J.James"),333,c("J.James","good"),761,'Veni',"vidi","Vici")

  df <- data.frame(names)

Here is the real and overall expected result

     names  Group
1      bad    1
2     good    1
3        1    1
4        2    1
5     good    2
6      bad    2
7      111    2
8      bad    3
9  J.James    3
10    good    4
11 J.James    4
12     333    4
13 J.James    5
14    good    5
15     761    5
16    Veni    666
17    vidi    666
18    Vici    666

Answer 1

Here are two approaches that reproduce the expected OP result for a given sample dataset. `

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data.table

library(data.table)
names <- c(c("bad","good"),1,2,c("good","bad"),111,c("bad","J.James"),c("good","J.James"),333,c("J.James","good"),761,'Veni',"vidi","Vici")
exempt <- c("Veni", "vidi", "Vici")
data.table(names)[is.na(as.numeric(names)) & !names %in% exempt, 
                  grp := rep(1:.N, each = 2L, length.out = .N)][
                    names %in% exempt, grp := 666L][
                      , grp := zoo::na.locf(grp)][]

      names grp
 1:     bad   1
 2:    good   1
 3:       1   1
 4:       2   1
 5:    good   2
 6:     bad   2
 7:     111   2
 8:     bad   3
 9: J.James   3
10:    good   4
11: J.James   4
12:     333   4
13: J.James   5
14:    good   5
15:     761   5
16:    Veni 666
17:    vidi 666
18:    Vici 666

dplyr/tidyr

dplyr/tidyr:

library(dplyr)
as_tibble(names) %>% 
  mutate(grp = if_else(is.na(as.numeric(names)) & !names %in% exempt,  
                       rep(1:n(), each = 2L, length.out = n()),
                       if_else(names %in% exempt, 666L, NA_integer_))) %>% 
  tidyr::fill(grp)

# A tibble: 18 x 2
   value     grp
   <chr>   <int>
 1 bad         1
 2 good        1
 3 1           1
 4 2           1
 5 good        3
 6 bad         3
 7 111         3
 8 bad         4
 9 J.James     5
10 good        5
11 J.James     6
12 333         6
13 J.James     7
14 good        7
15 761         7
16 Veni      666
17 vidi      666
18 Vici      666