r - How to add row index to a data frame, based on combination of factors [duplicate]

Answer 1

Is this what you are trying to achieve ?

library(ggplot2)
df <- data.frame(
  Dim1 = c("A","A","A","A","A","A","B","B"),
  Dim2 = c(100,100,100,100,200,200,100,200),
  Value = sample(1:10, 8)
)
df$index <- c(1,2,3,4,1,2,1,1)

ggplot(df,aes(x=index,y=Value))+geom_point()+facet_wrap(Dim1~Dim2)

The output is as follows:

Answer 2

Here you go, using data.table:

library(data.table)
df <- data.table(
    Dim1 = c("A","A","A","A","A","A","B","B"),
    Dim2 = c(100,100,100,100,200,200,100,200),
    Value = sample(1:10, 8)
        )

df[, index := seq_len(.N), by = list(Dim1, Dim2)]

Answer 3

This is probably going to look like cheating since I am passing a vector into a function which I then totally ignore except to get its length:

 df$Index <- ave( 1:nrow(df), df$Dim1, factor( df$Dim2), FUN=function(x) 1:length(x) )

The ave function returns a vector of the same length as its first argument but computed within categories defined by all of the factors between the first argument and the argument named FUN. (I often forget to put the "FUN=" in for my function and get a cryptic error message along the lines of unique() applies only to vectors, since it was trying to determine how many unique values an anonymous function possesses and it fails.

There's actually another even more compact way of expressing function(x) 1:length(x) using the seq_along function whch is probably safer since it would fail properly if passed a vector of length zero whereas the anonymous function form would fail improperly by returning 1:0 instead of numeric(0):

ave( 1:nrow(df), df$Dim1, factor( df$Dim2), FUN=seq_along )