R, dplyr: cumulative version of n_distinct

Question

I have a dataframe as follows. It is ordered by column time.

Input -

df = data.frame(time = 1:20,
            grp = sort(rep(1:5,4)),
             


        

Answer 1

Here's another solution using data.table that's pretty quick.

Generic Function

cum_n_distinct <- function(x, na.include = TRUE){
  # Given a vector x, returns a corresponding vector y
  # where the ith element of y gives the number of unique
  # elements observed up to and including index i
  # if na.include = TRUE (default) NA is counted as an 
  # additional unique element, otherwise it's essentially ignored

  temp <- data.table(x, idx = seq_along(x))
  firsts <- temp[temp[, .I[1L], by = x]$V1]
  if(na.include == FALSE) firsts <- firsts[!is.na(x)]
  y <- rep(0, times = length(x))
  y[firsts$idx] <- 1
  y <- cumsum(y)

  return(y)
}

Example Use

cum_n_distinct(c(5,10,10,15,5))  # 1 2 2 3 3
cum_n_distinct(c(5,NA,10,15,5))  # 1 2 3 4 4
cum_n_distinct(c(5,NA,10,15,5), na.include = FALSE)  # 1 1 2 3 3

Solution To Your Question

d_out = df %>%
  arrange(time) %>%
  group_by(grp) %>%
  mutate(var2 = cum_n_distinct(var1))