R, dplyr: cumulative version of n_distinct - r

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)),
var1 = rep(c('A','B'),10)
)
head(df,10)
time grp var1
1 1 1 A
2 2 1 B
3 3 1 A
4 4 1 B
5 5 2 A
6 6 2 B
7 7 2 A
8 8 2 B
9 9 3 A
10 10 3 B
I want to create another variable var2 which computes no of distinct var1 values so far i.e. until that point in time for each group grp . This is a little different from what I'd get if I were to use n_distinct.
Expected output -
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
I want to create a function say cum_n_distinct for this and use it as -
d_out = df %>%
arrange(time) %>%
group_by(grp) %>%
mutate(var2 = cum_n_distinct(var1))

Answer 1

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

Answer 2

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))

Answer 3

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

Answer 4

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))