For-loop vs while loop in R

Because 1 is numeric, but not integer (i.e. it's a floating point number), and 1:6000 is numeric and integer.

> print(class(1))[1] "numeric"> print(class(1:60000))[1] "integer"

60000 squared is 3.6 billion, which is NOT representable in signed 32-bit integer, hence you get an overflow error:

> as.integer(60000)*as.integer(60000)[1] NAWarning message:In as.integer(60000) * as.integer(60000) : NAs produced by integer overflow

3.6 billion is easily representable in floating point, however:

> as.single(60000)*as.single(60000)[1] 3.6e+09

To fix your for code, convert to a floating point representation:

function (N){    for(i in as.single(1:N)) {        y <- i*i    }}

The variable in the for loop is an integer sequence, and so eventually you do this:

> y=as.integer(60000)*as.integer(60000)Warning message:In as.integer(60000) * as.integer(60000) : NAs produced by integer overflow

whereas in the while loop you are creating a floating point number.

Its also the reason these things are different:

> seq(0,2,1)[1] 0 1 2> seq(0,2)[1] 0 1 2

Don't believe me?

> identical(seq(0,2),seq(0,2,1))[1] FALSE

because:

> is.integer(seq(0,2))[1] TRUE> is.integer(seq(0,2,1))[1] FALSE

And about timing:

fn1 <- function (N) {    for(i in as.numeric(1:N)) { y <- i*i }}fn2 <- function (N) {    i=1    while (i <= N) {        y <- i*i        i <- i + 1    }}system.time(fn1(60000))# user  system elapsed # 0.06    0.00    0.07 system.time(fn2(60000))# user  system elapsed # 0.12    0.00    0.13

And now we know that for-loop is faster than while-loop. You cannot ignore warnings during timing.