[R] add one variable to a data frame

[Bert Gunter]

Sarah et. al.:

As a matter of aesthetics (i.e. my personal ocd-ness) I prefer using the
public API of an object, i.e. *not* to makes use of the representation of a
factor as essentially an integer vector with labels, but rather to use its
documented behavior. (Feel free to ignore this remark!)

Anyway,

>cumsum(!duplicated(dat1$B))
 [1] 1 1 1 2 2 3 3 3 3 3 4 4

will do it.

This is very efficient (almost certainly of no concern here, btw). But the
price for this efficiency is that it depends completely on the data beig
grouped in order as the OP showed. It will fail if this is not the case.
If, for example, the data appeared as:

> set.seed(1234)
> ix <- sample(1:12)

> dat1[ix,]
    N      B
2   2 29_log
7   7  1_log
11 11  3_cat
6   6  1_log
10 10  1_log
5   5 27_cat
1   1 29_log
12 12  3_cat
3   3 29_log
8   8  1_log
4   4 27_cat
9   9  1_log

then Don's solution will still work. The above doesn't.

So this emphasizes the importance of precisely and completely specifying
the nature of your data. Hence: which is it? -- all the groups appearing
together or possibly mixed up?

But I have another question: why do this at all? The new column adds no new
information -- I believe that anything you want to do with the integer
codes can be done in R with the original factor representation (and just as
efficiently, as Sarah's "aesthetically displeasing to Bert" suggestion
makes clear). Note: counterexample welcome! So as AFAICS, there is no need
for this at all.

Cheers,
Bert





Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

On Fri, May 11, 2018 at 11:39 AM, Sarah Goslee <sarah.goslee using gmail.com>
wrote:

> Hi,
>
> Here's one way to approach it, using the coercion of factor to numeric.
>
> Note that I changed your data.frame() statement to avoid coercing
> strings to factors, just to make it simpler to set the levels.
>
> dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log",
> "27_cat", "27_cat", "1_log", "1_log", "1_log", "1_log", "1_log",
> "3_cat", "3_cat"), stringsAsFactors=FALSE)
>
>
> dat1$C1 <- as.numeric(factor(dat1$B, levels=unique(dat1$B)))
>
> And here's a way using rle()
>
> dat1$C2 <- rep(seq_len(length(unique(dat1$B))),
> times=rle(as.vector(dat1$B))$lengths)
>
> (That second will work even if B is a factor.)
>
> > dat1
>     N      B C1 C2
> 1   1 29_log  1  1
> 2   2 29_log  1  1
> 3   3 29_log  1  1
> 4   4 27_cat  2  2
> 5   5 27_cat  2  2
> 6   6  1_log  3  3
> 7   7  1_log  3  3
> 8   8  1_log  3  3
> 9   9  1_log  3  3
> 10 10  1_log  3  3
> 11 11  3_cat  4  4
> 12 12  3_cat  4  4
>
>
> Sarah
>
> On Fri, May 11, 2018 at 1:52 PM, Ding, Yuan Chun <ycding using coh.org> wrote:
> > Hi All,
> >
> > I have a data frame dat1:
> > dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log",
> "27_cat", "27_cat",
> >
> "1_log", "1_log", "1_log", "1_log", "1_log",
> >
>  "3_cat", "3_cat"))
> >
> > Then I need to add one column or variable to reflect uniqueness of B
> variable in sequential order as below.
> > dat1$C <-c(1,1,1,2,2,3,3,3,3,3,4,4)
> >
> > I only show 12 rows, my real data frame has over 1000 rows, I can not
> manually to add column C.
> >
> > It should be easy, but I can not figure out. Can you help me?
> >
> > Thanks,
> >
> > Ding
> >
> --
> Sarah Goslee
> http://www.functionaldiversity.org
>
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