在 [r] 中使用 for 循环代替矩阵索引

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