Ralph: I looked at Henrique's solution and he does 2 things which make
it better than mine.
1) He splits based off the first two columns where I just split based on
the second. So, my split assumes that the "same rows" are next to each
other
which is an unnecessary assumption.
2) He actually checks to make sure that 1 and 2 are actually in the
third column of the resulting dataframes that split returns. I assumed
that , if a
dataframe was of length 2, then the latter would be true automatically.
So, even though mine worked for what you needed, in the spirit of
generality and minimal assumptions, it better to use Henrique's
solution. Also,
make sure you understand it because you can learn a lot from it. ( this
is also true of his solutions in general ).
On Wed, Aug 13, 2008 at 3:37 PM, Ralph S. wrote:
yes this work, very elegant thank you. I didn't get Henriques message in
my mailbox immediately for some reason -
-Ralph
___________________________________
Date: Wed, 13 Aug 2008 14:23:33 -0500
From: [EMAIL PROTECTED]
Subject: RE: [R] subsetting matrix according to columns with character
index
To: [EMAIL PROTECTED]
CC: r-help@r-project.org
sorry ralph. i meant the OR instead of the AND so that was my bad
mistake. the subset function should also work with the OR.
i think i understand better what you want now also. the approach below
for doing what you want assumes that , if there are 2 rows associated
with the
values in the first 2 columns , then they will be 1 and 2. If they are
1,1 or 2,2, then it won't work. So, henrique's solution could be better
and more general.
Assume your dataframe is called DF.
tempres<-split(DF$x,DF$y)
onlytwo<-lapply(tempres, function(.df)
if (nrow(.df) == 2) {
return(.df) } else {
return(NULL) }
)
onlytwo<-onlytwo[!sapply(onlytwo,is.null)
result<-do.call(rbind,onlytwo)
On Wed, Aug 13, 2008 at 2:45 PM, Ralph S. wrote:
I tried this - I get an empty set:
<0 rows> (or 0-length row.names)
I guess this happens because the z variable takes only one value per
row??
What works is:
DFsub<-DF[DF$z == 1 | DF$z == 2,]
but then, I do not eliminate the entries where there is only one entry
for z given an a and c combination.
Any idea what to do?
-Ralph
> Date: Wed, 13 Aug 2008 13:05:25 -0500 From: [EMAIL PROTECTED]
> Subject: RE: [R] subsetting matrix according to columns with character
> index To: [EMAIL PROTECTED]
> it must be a dataframe so, if it was DF, then, assuming i understand
> what you want then either of the following should work:
> DFsub<-DF[DF$z == 1 & DF$z == 2,]
> or
> DFsub<-subset(DF, z == 1 & z == 2 )
>
> On Wed, Aug 13, 2008 at 2:00 PM, Ralph S. wrote:
>> Hi,
>> I have a long matrix of the following form which I would like to
>> subset according to the third column:
>> [x y z]:
>> a1 c1 1 a1 c1 2 a2 c1 1 a1 c2 1 a1 c2 2 . . .
>>
>> The first two columns a characters ai and cj.
>> I would like to keep all the rows where there are two entries for z,
>> 1 and 2.
>> That is, I want: a1 c1 1 a1 c1 2 a1 c2 1 a1 c2 2 . . .
>> I try to use something like df[by(df,c(df$x,df$y),sum(z)==3),] but
>> that only gives me one line of data per x y combination.
>> Is there an easy way of coding to keep all rows for a and c
>> combinations where z has entries both 1 and 2? Many thanks,
>> Ralph
>> _________________________________________________________________
>>
>> LM_WLYIA_whichathlete_us
>> ______________________________________________ R-help@r-project.org
>> mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do
>> read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
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