43
data = c(1,2,3,4)
names = c("foo", "bar", "baz", "yak")
d = data.frame(data, row.names=names)

This returns:

    data
foo    1
bar    2
baz    3
yak    4

Now, I'd like to sort this dataframe by the column, without losing the row names attached. So, my result would be:

    data
yak    4
baz    3
bar    2
foo    1

I've already tried the following:

  • d[order(-d$data),], which of course only gives me the one-dimensional list.

  • arrange(d, desc(data)) from the plyr package, which drops the row names.

  • Find out the order with o = order(-d$data), then reconstruct the frame with

    data.frame(d[o,], row.names=rownames(d)[o])
    

    … which still leaves me with the wrong column name.

Is there any sane way to do this?

slhck
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1 Answers1

49

You were close with your first attempt, just forgot about using drop = FALSE:

> d[order(-d$data), , drop = FALSE]
    data
yak    4
baz    3
bar    2
foo    1
flodel
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  • Ah, interesting. I somehow didn't find out about this only after reading the keyword "drop" here, which now makes me think whether R dropping dimensions is a good idea or not. I'll read up on that. – slhck Oct 31 '12 at 11:14
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    I think everyone has already asked this question. The consensus of the R gods was that this default was not so good, but that it cannot be changed without obsoleting most existing code. Will be corrected in Q. – Dieter Menne Oct 31 '12 at 17:06
  • When I use this solution I filter out all of my rows but one. How did this happen? – Kory Jul 29 '15 at 14:45