subset() of a vector in R

Question

I\'ve written the following function based on subset(), which I find handy:

ss <- function (x, subset, ...) 
{
    r <- eval(substitute(subset         


        

Answer 1

Thanks for sharing Ken.

You could use:

x <- myDataFrame$MyVariableName; x[x > 100 & x < 180] 

Yours may require less typing but the code is less generalizable to others if you're sharing code. I have a few time saver functions like that myself but use them sparingly because they may be slowing down your code (extra steps) and requires you to also include that code for that function when ever you share the file with someone else.

Compare writing length. Almost the same length:

ss(mtcars$hp, 100 < . & . < 180)
x <- mtcars$hp; x[x > 100 & x < 180] 

Compare time on 1000 replications.

library(rbenchmark)
benchmark(
       tyler = x[x > 100 & x < 180],
       ken = ss(mtcars$hp, 100 <. & . < 180),
 replications=1000)

   test replications elapsed relative user.self sys.self user.child sys.child
2   ken         1000    0.56 18.66667      0.36     0.03         NA        NA
1 tyler         1000    0.03  1.00000      0.03     0.00         NA        NA

So I guess it depends on if you need speed and/or sharability vs convenience. If it's just for you on a small data set I'd say it's valuable.

EDIT: NEW BENCHMARKING

> benchmark(
+     tyler = {x <- mtcars$hp; x[x > 100 & x < 180]}, 
+     ken = ss(mtcars$hp, 100 <. & . < 180), 
+     ken2 = ss2(mtcars$hp, 100 <. & . < 180),
+     joran = with(mtcars,hp[hp>100 & hp< 180 ]), 
+  replications=10000)

   test replications elapsed  relative user.self sys.self user.child sys.child
4 joran        10000    0.83  2.677419      0.69     0.00         NA        NA
2   ken        10000    3.79 12.225806      3.45     0.02         NA        NA
3  ken2        10000    0.67  2.161290      0.35     0.00         NA        NA
1 tyler        10000    0.31  1.000000      0.20     0.00         NA        NA