R, dplyr: cumulative version of n_distinct

Assuming stuff is ordered by time already, first define a cumulative distinct function:

dist_cum <- function(var)
  sapply(seq_along(var), function(x) length(unique(head(var, x))))

Then a base solution that uses ave to create groups (note, assumes var1 is factor), and then applies our function to each group:

transform(df, var2=ave(as.integer(var1), grp, FUN=dist_cum))

A data.table solution, basically doing the same thing:

library(data.table)
(data.table(df)[, var2:=dist_cum(var1), by=grp])

And dplyr, again, same thing:

library(dplyr)
df %>% group_by(grp) %>% mutate(var2=dist_cum(var1))

Try:

Update

With your new dataset, an approach in base R

  df$var2 <-  unlist(lapply(split(df, df$grp),
              function(x) {x$var2 <-0
               indx <- match(unique(x$var1), x$var1)
               x$var2[indx] <- 1
               cumsum(x$var2) }))

  head(df,7)
  #   time grp var1 var2
  # 1    1   1    A    1
  # 2    2   1    B    2
  # 3    3   1    A    2
  # 4    4   1    B    2
  # 5    5   2    A    1
  # 6    6   2    B    2
  # 7    7   2    A    2

A dplyr solution inspired from @akrun's answer -

Ths logic is basically to set 1st occurrence of each unique values of var1 to 1 and rest to 0 for each group grp and then apply cumsum on it -

df = df %>%
  arrange(time) %>%
  group_by(grp,var1) %>%
  mutate(var_temp = ifelse(row_number()==1,1,0)) %>%
  group_by(grp) %>%
  mutate(var2 = cumsum(var_temp)) %>%
  select(-var_temp)

head(df,10)

Source: local data frame [10 x 4]
Groups: grp

   time grp var1 var2
1     1   1    A    1
2     2   1    B    2
3     3   1    A    2
4     4   1    B    2
5     5   2    A    1
6     6   2    B    2
7     7   2    A    2
8     8   2    B    2
9     9   3    A    1
10   10   3    B    2