R sorts a vector on its own accord
In response to Dwin's comment on Dirk's answer: the data are always putty in your hands. "This is R. There is no if. Only how." -- Simon Blomberg
You can add 0
like so:
df.sorted <- gsub("(walker)([[:digit:]]{1}_)", "\\10\\2", df.sorted)
If you needed to add 00
, you do it like this:
df.sorted <- gsub("(walker)([[:digit:]]{1}_)", "\\10\\2", df.sorted)
df.sorted <- gsub("(walker)([[:digit:]]{2}_)", "\\10\\2", df.sorted)
...and so on.
construct it as an ordered factor:
> df.new <- ordered(df.sorted,levels=df.sorted)
> order(df.new)
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
EDIT :
After @DWins comment, I want to add that it is even not nessecary to make it an ordered factor, just a factor is enough if you give the right order of levels :
> df.new2 <- factor(df.sorted,levels=df.sorted)
> order(df.new)
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
The difference will be noticeable when you use those factors in a regression analysis, they can be treated differently. The advantage of ordered factors is that they let you use comparison operators as < and >. This makes life sometimes a lot easier.
> df.new2[5] < df.new2[10]
[1] NA
Warning message:
In Ops.factor(df.new[5], df.new[10]) : < not meaningful for factors
> df.new[5] < df.new[10]
[1] TRUE
Use the mixedsort
(or) mixedorder
functions in package gtools:
require(gtools)
mixedorder(df.sorted)
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
[28] 28 29 30 31 32 33 34 35 36 37 38 39
Isn't this simply the same thing you get with all lexicographic shorts (as e.g. ls
on directories) where walker10_foo sorts
higher than walker1_foo
?
The easiest way around, in my book, is to use a consistent number of digits, i.e. I would change to binned_walker01_1.grd
and so on inserting a 0 for the one-digit counts.