Amazing. Thanks everybody for the help. I have about 12,000 rows of data
with up to 50 reccurrences, but it seems to work like a charm.
Best,
Kai
On Sun, Feb 12, 2012 at 8:11 AM, Petr Savicky <savi...@cs.cas.cz> wrote:
> On Sat, Feb 11, 2012 at 04:05:25PM -0500, David Winsemius wrote:
> >
> > On Feb 11, 2012, at 1:17 PM, Kai Mx wrote:
> >
> > >Hi everybody,
> > >I have a large dataframe similar to this one:
> > >knames <-c('ab', 'aa', 'ac', 'ad', 'ab', 'ac', 'aa', 'ad','ae', 'af')
> > >kdate <- as.Date( c('20111001', '20111102', '20101001', '20100315',
> > >'20101201', '20110105', '20101001', '20110504', '20110603',
> > >'20110201'),
> > >format="%Y%m%d")
> > >kdata <- data.frame (knames, kdate)
> >
> > > ave(unclass(kdate), knames, FUN=order )
> > [1] 2 2 1 1 1 2 1 2 1 1
> >
> >
> > That was actually not using the dataframe values but you could also do
> > this:
> >
> > > kdata$ord <- with(kdata, ave(unclass(kdate), knames, FUN=order ))
> > > kdata
> > knames kdate ord
> > 1 ab 2011-10-01 2
> > 2 aa 2011-11-02 2
> > 3 ac 2010-10-01 1
> > 4 ad 2010-03-15 1
> > 5 ab 2010-12-01 1
> > 6 ac 2011-01-05 2
> > 7 aa 2010-10-01 1
> > 8 ad 2011-05-04 2
> > 9 ae 2011-06-03 1
> > 10 af 2011-02-01 1
>
> Hi.
>
> This is a good solution, if there are at most two occurrences
> of each name. If there are more occurrences, then function "order"
> should be replaced by "rank". Replacing name "aa" at row 2 by "ab",
> we get
>
> knames <-c('ab', 'ab', 'ac', 'ad', 'ab', 'ac', 'aa', 'ad','ae', 'af')
> kdate <- as.Date( c('20111001', '20111102', '20101001', '20100315',
> '20101201', '20110105', '20101001', '20110504', '20110603', '20110201'),
> format="%Y%m%d")
> kdata <- data.frame (knames, kdate)
>
> kdata$ord <- with(kdata, ave(unclass(kdate), knames, FUN=order))
> kdata$rank <- with(kdata, ave(unclass(kdate), knames, FUN=rank))
> kdata
>
> knames kdate ord rank
> 1 ab 2011-10-01 3 2
> 2 ab 2011-11-02 1 3
> 3 ac 2010-10-01 1 1
> 4 ad 2010-03-15 1 1
> 5 ab 2010-12-01 2 1
> 6 ac 2011-01-05 2 2
> 7 aa 2010-10-01 1 1
> 8 ad 2011-05-04 2 2
> 9 ae 2011-06-03 1 1
> 10 af 2011-02-01 1 1
>
> The names "ab" occur in the order row 5, row 1, row 2, so
> row 1 should get index 2, row 2 index 3.
>
> If some of the dates repeat, then rank() by default computes
> the average index. In this case, the following function f()
> may be used
>
> knames <-c('ab', 'ab', 'ac', 'ad', 'ab', 'ac', 'aa', 'ad','ae', 'af')
> kdate <- as.Date( c('20111001', '20111001', '20101001', '20100315',
> '20101201', '20110105', '20101001', '20110504', '20110603', '20110201'),
> format="%Y%m%d")
> kdata <- data.frame (knames, kdate)
>
> kdata$rank <- with(kdata, ave(unclass(kdate), knames, FUN=rank))
> f <- function(x) rank(x, ties.method="first")
> kdata$f <- with(kdata, ave(unclass(kdate), knames, FUN=f))
> kdata
>
> knames kdate rank f
> 1 ab 2011-10-01 2.5 2
> 2 ab 2011-10-01 2.5 3
> 3 ac 2010-10-01 1.0 1
> 4 ad 2010-03-15 1.0 1
> 5 ab 2010-12-01 1.0 1
> 6 ac 2011-01-05 2.0 2
> 7 aa 2010-10-01 1.0 1
> 8 ad 2011-05-04 2.0 2
> 9 ae 2011-06-03 1.0 1
> 10 af 2011-02-01 1.0 1
>
> Hope this helps.
>
> Petr Savicky.
>
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