Using for loop instead of matrix index in [r] - r

Question

I have been meaning to rewrite the following code with for loop in R:
x <- sample(1:100, 10)
x[x %% 2 == 0]
#[1] 6 26 72 62 32 86 100
which extracts those elements in vector x that are even number only.
I have come up with the following result in the end, but I believe there are more simple ways of coding this.
x <- sample(1:100, 10)
output <- integer(0)
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {
output[i] <- x[i]
output <- output[!is.na(output)]
}
}
output
#[1] 6 26 72 62 32 86 100
I would be grateful if you could help me with this.

Answer 1

You can skip the NA removes when adding the new hit to the end of output using length.
output <- integer(0)
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {
output[length(output) + 1] <- x[i]
}
}
output
Where length(output) gives you the current length of output. By adding 1 you place the new element at the end of output.
Or as #Roland (thanks!) commented using c
output <- integer(0)
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {
output <- c(output, x[i])
}
}
output
c combines output and x[i] to a new vector.
Or preallocate output with x and mark the non hits with NA
output <- x
for(i in seq_along(x)) {
if(x[i] %% 2 != 0) {
output[i] <- NA
}
}
output <- output[!is.na(output)]
output
Benchmarks:
fun <- alist(Vectorized = x[x %% 2 == 0]
, Question = {output <- integer(0)
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {output[i] <- x[i]; output <- output[!is.na(output)]}
}
output}
, NaOutside = {output <- integer(0)
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {output[i] <- x[i]}
}
output <- output[!is.na(output)]
output}
, Append = {output <- integer(0)
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {output[length(output) + 1] <- x[i]}
}
output}
, Append2 = {output <- integer(0); j <- 1
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {output[j] <- x[i]; j <- j + 1}
}
, C = {output <- integer(0)
for(i in seq_along(x)) {
if(x[i] %% 2 == 0) {
output <- c(output, x[i])
}
}
output}
, Preallocate = {output <- x
for(i in seq_along(x)) {
if(x[i] %% 2 != 0) {
output[i] <- NA
}
}
output <- output[!is.na(output)]
output}
)
library(microbenchmark)
set.seed(42)
x <- sample(1:100, 10)
microbenchmark(list = fun, control=list(order="block"))
#Unit: nanoseconds
# expr min lq mean median uq max neval
# Vectorized 644 655 781.54 666.5 721 8559 100
# Question 4377648 4419010 4682029.95 4492628.5 4579952 7512162 100
# NaOutside 3443488 3558401 3751542.62 3597273.0 3723662 5146015 100
# Append 3932586 4070234 4287628.11 4129849.0 4209361 6036142 100
# Append2 3966245 4094766 4360989.39 4147847.5 4312868 5899000 100
# C 3464081 3566902 3806531.77 3618758.5 3743058 6528224 100
# Preallocate 3162424 3263220 3435591.92 3290938.0 3374547 4823017 100
set.seed(42)
x <- sample(1:1e5, 1e4)
microbenchmark(list = fun, control=list(order="block"))
#Unit: microseconds
# expr min lq mean median uq max neval
# Vectorized 226.224 276.271 277.0027 278.993 284.4515 345.527 100
# Question 125550.120 126392.848 129287.7202 126812.309 128426.3655 157958.571 100
# NaOutside 6911.053 7020.831 7497.4403 7109.891 8158.7580 8779.448 100
# Append 7843.988 7982.987 8582.6769 8129.988 9287.5760 10775.894 100
# Append2 7647.340 7783.334 8347.7824 7954.683 9007.4500 10325.973 100
# C 27976.747 29776.632 29997.5407 30024.121 30250.9590 51630.868 100
# Preallocate 6119.198 6232.228 6679.9407 6367.618 7290.1015 8331.277 100