Convert integers to decimal values - r

Question

I have a set of integer data between 1:10000. I need to bring them in range 0:1.
For example, converting
12 --> 0.12
123 --> 0.123
1234 --> 0.1234
etc. (note that I don't want to scale the values).
Any suggestions how to do this on all the data at once?

Answer 1

I would simply do
x <- c(2, 14, 128, 1940, 140, 20000)
x/10^nchar(x)
## [1] 0.200 0.140 0.128 0.194 0.140 0.200
But a much faster approach (which avoids to character conversion) offered by #Frank
x/10^ceiling(log10(x))
Benchmark
library(microbenchmark)
set.seed(123)
x <- sample(1e8, 1e6)
microbenchmark(
david = x/10^nchar(x),
davidfrank = x/10^ceiling(log10(x)),
richard1 = as.numeric(paste0(".", x)),
richard2 = as.numeric(sprintf(".%d", x))
)
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# david 691.0513 822.6482 1052.2473 956.5541 1153.4779 2391.7856 100 b
# davidfrank 130.0522 164.3227 255.8397 197.3158 339.3224 576.2255 100 a
# richard1 1130.5160 1429.8314 1972.2624 1689.8454 2473.6409 4791.0558 100 c
# richard2 712.8357 926.8013 1181.5349 1103.1661 1315.4459 2753.6795 100 b

Answer 2

The non-mathy way would be to add the decimal with paste() then coerce back to numeric.
x <- c(2, 14, 128, 1940, 140, 20000)
as.numeric(paste0(".", x))
# [1] 0.200 0.140 0.128 0.194 0.140 0.200
Update 1: There was some interest about the timings of the two originally posted methods. According to the following benchmarks, they seem to be about the same.
library(microbenchmark)
x <- 1:1e5
microbenchmark(
david = { david <- x/10^nchar(x) },
richard = { richard <- as.numeric(paste0(".", x)) }
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# david 88.94391 89.18379 89.70962 89.40736 89.71012 99.68126 100
# richard 87.89776 88.17234 89.38383 88.44439 88.77052 105.06066 100
identical(richard, david)
# [1] TRUE
Update 2: I have also remembered that sprintf() is often faster than paste0(). We can also use the following.
as.numeric(sprintf(".%d", x))
Now using the same x from above, and only comparing these two choices, we have a good improvement in the timing of sprintf() versus paste(), as shown below.
microbenchmark(
paste0 = as.numeric(paste0(".", x)),
sprintf = as.numeric(sprintf(".%d", x))
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# paste0 87.89413 88.41606 90.25795 88.82484 89.65674 107.8080 100
# sprintf 61.16524 61.23328 62.26202 61.29192 61.48316 79.1202 100