R: scan vectors once instead of 4 times? - r

Question

Suppose I have two equal length logical vectors.
Computing the confusion matrix the easy way:
c(sum(actual == 1 & predicted == 1),
sum(actual == 0 & predicted == 1),
sum(actual == 1 & predicted == 0),
sum(actual == 0 & predicted == 0))
requires scanning the vectors 4 times.
Is it possible to do that in a single pass?
PS. I tried table(2*actual+predicted) and table(actual,predicted) but both are obviously much slower.
PPS. Speed is not my main consideration here, I am more interested in understanding the language.

Answer 1

You could try using data.table
library(data.table)
DT <- data.table(actual, predicted)
setkey(DT, actual, predicted)[,.N, .(actual, predicted)]$N
data
set.seed(24)
actual <- sample(0:1, 10 , replace=TRUE)
predicted <- sample(0:1, 10, replace=TRUE)
Benchmarks
Using data.table_1.9.5 and dplyr_0.4.0
library(microbenchmark)
set.seed(245)
actual <- sample(0:1, 1e6 , replace=TRUE)
predicted <- sample(0:1, 1e6, replace=TRUE)
f1 <- function(){
DT <- data.table(actual, predicted)
setkey(DT, actual, predicted)[,.N, .(actual, predicted)]$N}
f2 <- function(){table(actual, predicted)}
f3 <- function() {data_frame(actual, predicted) %>%
group_by(actual, predicted) %>%
summarise(n())}
microbenchmark(f1(), f2(), f3(), unit='relative', times=20L)
#Unit: relative
# expr min lq mean median uq max neval cld
#f1() 1.000000 1.000000 1.000000 1.00000 1.000000 1.000000 20 a
#f2() 20.818410 22.378995 22.321816 22.56931 22.140855 22.984667 20 b
#f3() 1.262047 1.248396 1.436559 1.21237 1.220109 2.504662 20 a
Including the count from dplyr and tabulate also in the benchmarks on a slightly bigger dataset
set.seed(498)
actual <- sample(0:1, 1e7 , replace=TRUE)
predicted <- sample(0:1, 1e7, replace=TRUE)
f4 <- function() {data_frame(actual, predicted) %>%
count(actual, predicted)}
f5 <- function(){tabulate(4-actual-2*predicted, 4)}
Update
Including another data.table solution (provided by #Arun) also in the benchmarks
f6 <- function() {setDT(list(actual, predicted))[,.N, keyby=.(V1,V2)]$N}
microbenchmark(f1(), f3(), f4(), f5(), f6(), unit='relative', times=20L)
#Unit: relative
#expr min lq mean median uq max neval cld
#f1() 2.003088 1.974501 2.020091 2.015193 2.080961 1.924808 20 c
#f3() 2.488526 2.486019 2.450749 2.464082 2.481432 2.141309 20 d
#f4() 2.388386 2.423604 2.430581 2.459973 2.531792 2.191576 20 d
#f5() 1.034442 1.125585 1.192534 1.217337 1.239453 1.294920 20 b
#f6() 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 20 a

Answer 2

Like this:
tabulate(4 - actual - 2*predicted, 4)
(tabulate here is much faster than table because it knows the output will be a vector of length 4).

Answer 3

There is table which computes a cross tabulation and should give similar results if actual and predicted contain only zeros and ones:
table(actual, predicted)
Internally, this works by pasteing the vectors -- horribly inefficient. It seems that the coercion to character also happens when tabulating only one value, and this might be the very reason for the bad performance also of table(actual*2 + predicted).