Using R's plyr package to reorder groups within a dataframe - r

Question

I have a data reorganization task that I think could be handled by R's plyr package. I have a dataframe with numeric data organized in groups. Within each group I need to have the data sorted largest to smallest.
The data looks like this (code to generate below)
group value
2 b 0.1408790
6 b 1.1450040 #2nd b is smaller than 1st
1 c 5.7433568
3 c 2.2109819
4 d 0.5384659
5 d 4.5382979
What I would like is this.
group value
b 1.1450040 #1st b is largest
b 0.1408790
c 5.7433568
c 2.2109819
d 4.5382979
d 0.5384659
So, what I need plyr to do is go through each group & apply something like order on the numeric data, reorganize by order, save the reordered subset of data, & put it all back together at the end.
I can process this "by hand" with a list & some loops, but it takes a long long time. Can this be done by plyr in a couple of lines?
Example data
df.sz <- 6;groups <-c("a","b","c","d")
df <- data.frame(group = sample(groups,df.sz,replace = TRUE),
value = runif(df.sz,0,10),stringsAsFactors = FALSE)
df <- df[order(df$group),] #order by group letter
The inefficient approach using loops:
My current approach is to separate the dataframe df into a list by groups, apply order to each element of the list, and overwrite the original list element with the reordered element. I then use a loop to re-assemble the dataframe. (As a learning exercise, I'd interested also in how to make this code more efficient. In particular, what would be the most efficient way using base R functions to turn a list into a dataframe?)
Vector of the unique groups in the dataframe
groups.u <- unique(df$group)
Create empty list
my.list <- as.list(groups.u); names(my.list) <- groups.u
Break up df by $group into list
for(i in 1:length(groups.u)){
i.working <- which(df$group == groups.u[i])
my.list[[i]] <- df[i.working, ]
}
Sort elements within list using order
for(i in 1:length(my.list)){
order.x <- order(my.list[[i]]$value,na.last = TRUE, decreasing = TRUE)
my.list[[i]] <- my.list[[i]][order.x, ]
}
Finally rebuild df from the list. 1st, make seed for loop
new.df <- my.list[[1]][1,];; new.df[1,] <- NA
for(i in 1:length(my.list)){
new.df <- rbind(new.df,my.list[[i]])
}
Remove seed
new.df <- new.df[-1,]

Answer 1

You could use dplyr which is a newer version of plyr that focuses on data frames:
library(dplyr)
arrange(df, group, desc(value))

Answer 2

It's virtually sacrilegious to include a "data.table" response in a question tagged "plyr" or "dplyr", but your comment indicates you're looking for fast compact code.
In "data.table", you could use setorder, like this:
setorder(setDT(df), group, -value)
That command does two things:
It converts your data.frame to a data.table without copying.
It sorts your columns by reference (again, no copying).
You mention "> 50k rows". That's actually not very large, and even base R should be able to handle it well. In terms of "dplyr" and "data.table", you're looking at measurements in the milliseconds. That could make a difference as your input datasets become larger.
set.seed(1)
df.sz <- 50000
groups <- c(letters, LETTERS)
df <- data.frame(
group = sample(groups, df.sz, replace = TRUE),
value = runif(df.sz,0,10), stringsAsFactors = FALSE)
library(data.table)
library(dplyr)
library(microbenchmark)
dt1 <- function() as.data.table(df)[order(group, -value)]
dt2 <- function() setorder(as.data.table(df), group, -value)[]
dp1 <- function() arrange(df, group, desc(value))
microbenchmark(dt1(), dt2(), dp1())
# Unit: milliseconds
# expr min lq mean median uq max neval
# dt1() 5.749002 5.981274 7.725225 6.270664 8.831899 67.402052 100
# dt2() 4.956020 5.096143 5.750724 5.229124 5.663545 8.620155 100
# dp1() 37.305364 37.779725 39.837303 38.169298 40.589519 96.809736 100