R-like way to add a % column to a data frame - r

Question

I want to add a column that shows me the percentage part of that row compared to the some of the column. (sorry for my bad mathematical english here).
> trees['Heigth_%'] <- round((100 / sum(trees$Height) * trees$Height), digits=2)
> head(trees)
Girth Height Volume Heigth_%
1 8.3 70 10.3 2.97
2 8.6 65 10.3 2.76
3 8.8 63 10.2 2.67
4 10.5 72 16.4 3.06
5 10.7 81 18.8 3.44
6 10.8 83 19.7 3.52
This work.
But the question is if this is a good and R-like way?
e.g. Is sum() called for each row? Or is R intelligent enough here?

Answer 1

To answer you question if sum is called for every row or is R intelligent enough, you can use trace:
df = data.frame(a = 1:10, b = 21:30)
df['b_%'] = round((100 / sum(df$b) * df$b), digits=2)
trace('sum')
round((100 / sum(df$b) * df$b), digits=2)
untrace('sum')
Which shows only one call to the sum function. Afterwards, R recognizes that the lengths of trees$Height and sum(trees$Height) differ and tries to replicate the shorter one until is has the same length as the bigger one.

Answer 2

Converting first to data.table and then using prop.table is a bit faster then f1-f3 (IF you neglect the conversion to a data.table, as this is usually only done once for all subsequent commands).
# Get data
data(trees)
# Load package & convert to data.table
library(data.table)
trees <- as.data.table(trees)
# data.table way to create the new variable
f4 <- function(trees) {
trees[, Heigth_percentage:=round(prop.table(Height)*100,2)]
}
Here are the bechmark results:
# > microbenchmark(r1 <- f1(trees),
# + r2 <- f2(trees),
# + r3 <- f3(trees),
# + r3 <- f4(trees),
# + times = 10000)
# Unit: microseconds
# expr min lq mean median uq max neval cld
# r1 <- f1(trees) 616.616 666.290 730.8883 683.122 708.164 8390.818 10000 b
# r2 <- f2(trees) 617.437 666.701 730.3211 683.533 709.191 8100.574 10000 b
# r3 <- f3(trees) 596.500 655.616 721.1057 672.243 697.080 55048.757 10000 b
# r4 <- f4(trees) 551.342 612.922 680.7581 633.037 665.059 54672.712 10000 a

Answer 3

To begin, Vandenman's answer is much more adequate and precise. What follows is not really worth an answer, but as usual - not readable as a comment.
I have added prop.table() and data.table() (see majom's answer) approaches to the timings. With 40k rows data.table() is a bit closer to the rest, but still slower (~3 ms to ~3.7 ms), with 400k rows it starts to be comparable, and with 4M rows it is finally faster than the rest:
library(microbenchmark)
trees <- data.frame(Height = runif(400000, 9, 11),
Heigth_PCT = numeric(4000000))
trees_dt <- as.data.table(trees)
f1 <- function(trees) {
trees$Heigth_PCT <- round((100 / sum(trees$Height) * trees$Height), digits = 2)
return(trees)
}
f2 <- function(trees) {
sum_trees <- sum(trees$Height)
trees$Heigth_PCT <- round((100 / sum_trees * trees$Height), digits = 2)
return(trees)
}
f3 <- function(trees) {
trees$Heigth_PCT <- round(prop.table(trees$Height)*100, digits = 2)
return(trees)
}
f4 <- function(trees_dt) {
trees_dt[, Heigth_PCT := round(prop.table(Height)*100, 2)]
}
# Time both functions
microbenchmark(r1 <- f1(trees),
r2 <- f2(trees),
r3 <- f3(trees),
r4 <- f4(trees_dt),
times = 100)
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# r1 <- f1(trees) 296.4452 309.3853 373.5945 318.7987 400.0373 639.8556 100 a
# r2 <- f2(trees) 296.3453 310.6638 381.4048 323.0655 474.9295 682.2172 100 a
# r3 <- f3(trees) 304.3185 317.0654 383.9600 328.5494 395.6238 783.2435 100 a
# r4 <- f4(trees_dt) 304.3327 315.4685 361.9526 325.8711 366.1153 722.7629 100 a
sapply(list(r2, r3, as.data.frame(r4)), identical, r1)
# [1] TRUE TRUE TRUE
Edit: prop.table() added.
Edit 2: data.table() added.