Count the number previous items in by group in R [duplicate]

Answer 1

This can be approached using either specialized packages such as dplyr which has row_number(). We need to group by the variable ('alcgp') and create a new column using mutate.

library(dplyr)
df1 %>%
   group_by( alcgp) %>%
   mutate(indx= row_number())

Or using ave from base R. We group by 'alcgp' and in the FUN we can specify seq_along. I used seq_along(alcgp) as it may not work if the variable is factor class.

 df1$indx <- with(df1, ave(seq_along(alcgp), alcgp, FUN=seq_along))

Another convenient function in splitstackshape i.e. getanID

 library(splitstackshape)
 getanID(df1, 'alcgp')

Answer 2

Here's a data.table solution:

R> (data.table(Data)[,index := 1:.N, by = "agegp,alcgp"])
    agegp     alcgp    tobgp ncases ncontrols index
 1: 25-34 0-39g/day 0-9g/day      0        40     1
 2: 25-34 0-39g/day    10-19      0        10     2
 3: 25-34 0-39g/day    20-29      0         6     3
 4: 25-34 0-39g/day      30+      0         5     4
 5: 25-34     40-79 0-9g/day      0        27     1
 6: 25-34     40-79    10-19      0         7     2
 7: 25-34     40-79      30+      0         7     3
 8: 25-34     40-79    20-29      0         4     4
 9: 25-34    80-119 0-9g/day      0         2     1
10: 25-34    80-119      30+      0         2     2

---

library(data.table)
##
Data <- read.table(
  text = "   agegp     alcgp    tobgp ncases ncontrols
1  25-34 0-39g/day 0-9g/day      0        40
2  25-34 0-39g/day    10-19      0        10
3  25-34 0-39g/day    20-29      0         6
4  25-34 0-39g/day      30+      0         5
5  25-34     40-79 0-9g/day      0        27
6  25-34     40-79    10-19      0         7
8  25-34     40-79      30+      0         7
7  25-34     40-79    20-29      0         4
9  25-34    80-119 0-9g/day      0         2
11 25-34    80-119      30+      0         2",
  header = TRUE,
  stringsAsFactors = FALSE
)