Question

I need to take a data.frame in the format of:

  id1 id2 mean start end
1   A   D    4    12  15
2   B   E    5    14  15
3   C   F    6     8  10

and generate duplicate rows based on the difference in start - end. For example, I need 3 rows for the first row, 1 for the second, and 2 for the third. The start and end fields should be in sequential order in the final data.frame. The end result for this data.frame should be:

   id1 id2 mean start end
1    A   D    4    12  13
2    A   D    4    13  14
3    A   D    4    14  15
21   B   E    5    14  15
31   C   F    6     8   9
32   C   F    6     9  10

I have written this function which works, but isn’t written in very R’esque code:

dupData <- function(df){
    diff <- abs(df$start - df$end)
    ret <- {}

    #Expand our dataframe into the appropriate number of rows.
    for (i in 1:nrow(df)){
        for (j in 1:diff[i]){
            ret <- rbind(ret, df[i,])
        } 
    }

    #If matching ID1 and ID2, generate a sequential ordering of start & end dates
    for (k in 2:nrow(ret) - 1) {
        if ( ret[k,1] == ret[k + 1, 1] & ret[k, 2] == ret[k, 2]  ){ 
            ret[k, 5] <- ret[k, 4] + 1
            ret[k + 1, 4] <- ret[k, 5]  
        }
    }
    return(ret)
}

Does anyone have suggestions on how to optimize this code? Is there a function in plyr which may be applicable?

#sample daters
df <- data.frame(id1 = c("A", "B", "C")
        , id2 = c("D", "E", "F")
        , mean = c(4,5,6)  
        , start = c(12,14,8)
        , end = c(15, 15, 10)
)

Answer 1

There’s probably a more general way to do this, but below uses rbind.fill.

cbind(df[rep(1:nrow(df), times = apply(df[,4:5], 1, diff)), 1:3],
      rbind.fill(apply(df[,4:5], 1, function(x)
                       data.frame(start = x[1]:(x[2]-1), end = (x[1]+1):x[2]))))


##     id1 id2 mean start end
## 1     A   D    4    12  13
## 1.1   A   D    4    13  14
## 1.2   A   D    4    14  15
## 2     B   E    5    14  15
## 3     C   F    6     8   9
## 3.1   C   F    6     9  10