Count visits (cumsum) per ID while ignoring NA's and 0's

Question

I have the following df:

df <- data.frame(ID = c(1,1,2,2,2,3,3,3,3),
                 Attendance = c(1, 1, NA, 1,1, NA, 1, NA, 1 ))

And I want this one:

df <- data.frame(ID = c(1,1,2,2,2,3,3,3,3),
                 Attendance = c(1, 1, NA, 1,1, NA, 1, NA, 1),
                 Visit = c(1,2,0,1,2,0,1,0,2))

How can I count every time (cumsum) an ID appears , in 'Visit' column, based on 'Attendance' column value while ignoring NA's or 0's?

I have tried something with ave function like this one, but unsuccessfully:

df$Visit <- ifelse(!is.na(df$ID), (ave(df$ID, df$ID, FUN=cumsum))/df$ID, 0)

I have achieved the result by creating an auxiliar df with:

aux <- df[complete.cases(df$Attendance),] 

Counting the visits with Ave function and then merging , but I'm sure there exists an easiest way

Answer 1

We can use data.table . convert the 'data.frame' to 'data.table' ( setDT(df) ), grouped by 'ID', specify the i as a logical vector which is TRUE for non-NA elements in 'Attendance', assign ( := ) the 'rowid' of 'Attendance' as the 'Visit' column. Then, replace the NA in 'Visit' to 0

library(data.table)
setDT(df)[!is.na(Attendance), Visit := rowidv(Attendance), 
                   ID][is.na(Visit), Visit := 0]
df
#   ID Attendance Visit
#1:  1          1     1
#2:  1          1     2
#3:  2         NA     0
#4:  2          1     1
#5:  2          1     2
#6:  3         NA     0
#7:  3          1     1
#8:  3         NA     0
#9:  3          1     2

---

Or if we are using ave , then create an index for non-NA rows, and then use ave on those rows

i1 <- !is.na(df$Attendance)
df$Visit <- 0
df$Visit[i1] <- with(df[i1, ], ave(Attendance, ID, FUN = cumsum))

Answer 2

library(dplyr)
df %>%
    group_by(ID) %>%
    mutate(Visit = if_else(is.na(Attendance), 0, cumsum(if_else(is.na(Attendance), 0, 1))))