R sorts a vector on its own accord

问题:

df.sorted <- c("binned_walker1_1.grd", "binned_walker1_2.grd", "binned_walker1_3.grd",
    "binned_walker1_4.grd", "binned_walker1_5.grd", "binned_walker1_6.grd",
    "binned_walker2_1.grd", "binned_walker2_2.grd", "binned_walker3_1.grd",
    "binned_walker3_2.grd", "binned_walker3_3.grd", "binned_walker3_4.grd",
    "binned_walker3_5.grd", "binned_walker4_1.grd", "binned_walker4_2.grd",
    "binned_walker4_3.grd", "binned_walker4_4.grd", "binned_walker4_5.grd",
    "binned_walker5_1.grd", "binned_walker5_2.grd", "binned_walker5_3.grd",
    "binned_walker5_4.grd", "binned_walker5_5.grd", "binned_walker5_6.grd",
    "binned_walker6_1.grd", "binned_walker7_1.grd", "binned_walker7_2.grd",
    "binned_walker7_3.grd", "binned_walker7_4.grd", "binned_walker7_5.grd",
    "binned_walker8_1.grd", "binned_walker8_2.grd", "binned_walker9_1.grd",
    "binned_walker9_2.grd", "binned_walker9_3.grd", "binned_walker9_4.grd",
    "binned_walker10_1.grd", "binned_walker10_2.grd", "binned_walker10_3.grd")

One would expect that order of this vector would be 1:length(df.sorted), but that appears not to be the case. It looks like R internally sorts the vector according to its logic but tries really hard to display it the way it was created (and is seen in the output).

order(df.sorted)
 [1] 37 38 39  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22
[26] 23 24 25 26 27 28 29 30 31 32 33 34 35 36

Is there a way to "reset" the ordering to 1:length(df.sorted)? That way, ordering, and the output of the vector would be in sync.

回答1:

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

回答2:

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

回答3:

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.

回答4:

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.