R：按一列分组，然后在其他任何列中返回值大于0的第一行，然后返回此行之后的所有行

Question

I'm new to R programming and hope someone could help me with the situation below:

I have a dataframe shown in the picture (Original Dataframe), I would like to return the first record grouped by the [ID] column that has a value >= 1 in any of the four columns (A, B, C, or D) and all the records after based off the [Date] column (the desired dataframe should look like the Output Dataframe shown in the picture). Basically, remove all the records highlighted in yellow. I would appreciate greatly if you can provide the R code to achieve this.

structure(list(ID = c(101L, 101L, 101L, 101L, 101L, 101L, 103L, 
103L, 103L, 103L), Date = c(43338L, 43306L, 43232L, 43268L, 43183L, 
43144L, 43310L, 43246L, 43264L, 43209L), A = c(0L, 0L, 0L, 0L, 
0L, 0L, 0L, 1L, 0L, 0L), B = c(0L, 2L, 0L, 0L, 0L, 0L, 0L, 1L, 
0L, 0L), C = c(0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), D = c(0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("ID", "Date", 
"A", "B", "C", "D"), row.names = c(NA, -10L), class = c("data.table", 
"data.frame"))

Answer 1

Here is a solution,

    ID       Date A B C D
1  101 26.08.2018 0 0 0 0
2  101 25.07.2018 0 2 0 0
3  101 12.05.2018 0 0 1 0
4  101 17.06.2018 0 0 0 0
5  101 24.03.2018 0 0 0 0
6  101 13.02.2018 0 0 0 0
7  103 29.07.2018 0 0 0 0
8  103 26.05.2018 1 1 0 0
9  103 13.06.2018 0 0 0 0
10 103 19.04.2018 0 0 0 0


data$Check <- rowSums(data[3:6]) 

data$Date <- as.Date(data$Date , "%d.%m.%Y")


data <- data[order(data$ID,data$Date),]


id <- unique(data$ID)

for(i in 1:length(id)) {

    data_sample <- data[data$ID == id[i],]

    data_sample <- data_sample[ min(which(data_sample$Check>0 )):nrow(data_sample),]

    if(i==1) {

        final <- data_sample


    } else {

        final <- rbind(final,data_sample)

    }

}

final <- final[,-7]

   ID       Date A B C D
3 101 2018-05-12 0 0 1 0
4 101 2018-06-17 0 0 0 0
2 101 2018-07-25 0 2 0 0
1 101 2018-08-26 0 0 0 0
8 103 2018-05-26 1 1 0 0
9 103 2018-06-13 0 0 0 0
7 103 2018-07-29 0 0 0 0

Answer 2

Here's a tidyverse solution. The filter condition deserves some explanation:

1. first, we sort by ID and Date and group_by ID
2. Then, for each ID (since we're grouped by ID) we apply the filter condition:
   1. Test, for each row, whether any of the variables are > 0
   2. Get the row number for all rows (in the group) where this is the case
   3. Find the lowest one (since rows are sorted by Date, this will be the earliest)
   4. Get the value of Date for that row.
   5. Then filter rows where Date is >= than this.

Since we're still grouping by ID , all these calculations will happen separately for each group:

df %>%
    arrange(ID, Date) %>%
    group_by(ID) %>%
    filter(Date >= Date[min(which(A > 0 | B > 0 | C > 0 | D > 0))])

# A tibble: 7 x 6
# Groups:   ID [2]
     ID  Date     A     B     C     D
  <int> <int> <int> <int> <int> <int>
1   101 43232     0     0     1     0
2   101 43268     0     0     0     0
3   101 43306     0     2     0     0
4   101 43338     0     0     0     0
5   103 43246     1     1     0     0
6   103 43264     0     0     0     0
7   103 43310     0     0     0     0