Extracting indices for data frame rows that have MAX value for named field

Question

I have a data frame that is rather large and I need a good way (explained bellow) to extract indices for rows that have maximum values for a given field, within a certain set of labels. To explain this a bit better, here is an example 10 row data frame:

      value label
1  5.531637     D
2  5.826498     A
3  8.866210     A
4  1.387978     C
5  8.128505     C
6  7.391311     B
7  1.829392     A
8  4.373273     D
9  7.380244     A
10 6.157304     D

To generate:

structure(list(value = c(5.531637, 5.826498, 8.86621, 1.387978, 8.128505, 
7.391311, 1.829392, 4.373273, 7.380244, 6.157304), 
label = c("D", "A", "A", "C", "C", "B", "A", "D", "A", "D")), 
.Names = c("value", "label"), class = "data.frame", row.names = c(NA, -10L))

If I want to know what the index is for rows that have the maximum value per label, I currently use the following code:

idx <- sapply(split(1:nrow(d), d$label), function(x) {
  x[which.max(d[x,"value"])]
})

Generating this answer:

A  B  C  D 
3  6  5 10

I have also played around with ddply but have yet to find a better way to do this. By "better" in this case I mean faster (ddply is pretty slow and what I currently use is not far behind) as well as more elegant since the above solution seems way to wordy too me.

Answer 1

First of all: you can get the speed up using:

idx <- sapply(split(seq_len(nrow(d)), d$label), function(x) {
      x[which.max(d$value[x])]})

For a 100k data.frame, on my machine it is 5x faster than d[x,"value"] version.

For a large data.frame and many labels you could use a similar method that I posted in earlier question:

dd <- d[i<-order(d$label, d$value),] # dd is sorted by label and value
ind <- c(dd$label[-1] != dd$label[-n], TRUE)
idx <- setNames(seq_len(nrow(d))[i][ind], dd$label[ind])

---

edit: A more efficient solution with the use of a trick from Martin Morgan answer:

v <- d$label[i<-order(d$value)] # we need only label, and with Martin
                                # trick sorting over label is not needed
ind <- !duplicated(v, fromLast=TRUE) # it finds last (max) occurrence of label
idx <- setNames(seq_len(nrow(d))[i][ind], v[ind])

NOTE: order of final vector is different.

It depends on your actual data structure but you should gain a nice speed-up:

Timings:

# NOTE: different machine, so timing differ from previous
set.seed(6025051)
n <- 100000; k <- 20000
d <- data.frame(value=rnorm(n), 
    label=sample(paste("A",seq_len(k),sep="_"), n, replace=TRUE))

system.time(
    idx_1 <- sapply(split(1:nrow(d), d$label), function(x) {
        x[which.max(d[x,"value"])]})
)
# user  system elapsed 
# 1.30    0.02    1.31 
system.time(
    idx_1b <- sapply(split(seq_len(nrow(d)), d$label), function(x) {
        x[which.max(d$value[x])]})
)
# user  system elapsed 
# 0.23    0.00    0.23
all.equal(idx_1, idx_1b)
# [1] TRUE
system.time({
    dd <- d[i<-order(d$label, d$value),]
    ind <- c(dd$label[-1] != dd$label[-n], TRUE)
    idx_2 <- setNames(seq_len(nrow(d))[i][ind],dd$label[ind])
})
# user  system elapsed 
# 0.19    0.00    0.19 
all.equal(idx_1, idx_2)
# [1] TRUE

---

new solution

system.time({
    v <- d$label[i<-order(d$value)]
            ind <- !duplicated(v, fromLast=TRUE)
            idx_3 <- setNames(seq_len(nrow(d))[i][ind], v[ind])
})
# user  system elapsed 
# 0.05    0.00    0.04 
all.equal(sort(idx_1), sort(idx_3))
# [1] TRUE