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Message-ID: <OFAE1707D9.DF900E7E-ONC1257920.004C000F-C1257920.004C3762@precheza.cz>
Date: 2011-10-05T13:54:52Z
From: PIKAL Petr
Subject: How to subset() from data frame using specific rows
In-Reply-To: <alpine.LNX.2.00.1110050554020.25250@salmo.appl-ecosys.com>

Hi

> 
> On Wed, 5 Oct 2011, Petr PIKAL wrote:
> 
> > Hm. I seldom use such approach. In your original request you said you 
want
> > split your data to smaller data frames based on sites
> 
> Petr,
> 
>    I need the additional information in the database, too.

But you do not loose them, your data frame is cut according to sites 
variable and put into a list

see

> iris.spl<- split(iris, iris$Species)
> str(iris.spl)
List of 3
 $ setosa    :'data.frame':     50 obs. of  5 variables:
  ..$ Sepal.Length: num [1:50] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
  ..$ Sepal.Width : num [1:50] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
 
> 
> > From what we know it is difficult to say if there is some common 
feature
> > in site variable. If it is organised like
> > XY-N
> > you can simply make new variable from first two letters
> 
>    Unfortunately, the site designations are not so uniform. As I went 
through
> the process of re-doing the data I discovered this lack of consistency
> resulting in duplicate records because one site had been designated XX-n 
and
> XXn. Had to clean those up, too.
> 
> > sites <- substr(chemdata$site,1,2)

Which would not matter if the first two letters designates required 
grouping variable I called sites

Regards
Petr

> >
> > then you can split your data frame according to sites
> >
> > chem.spl <- split(chemdata, sites)
> >
> > and do anything with your splitted data frames organised in list
> 
>    First thing this morning I'm upgrading to 2.13.2 and hoping that this
> fixes an issue that just showed up yesterday afternoon: not being able 
to
> access function help pages. For example, I tried ?subset and ?split 
because
> I thought the latter is really what I want, yet R told me no help was 
found.
> Strange; it was there a week ago.
> 
> Thanks,
> 
> Rich
> 
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