I have a data set in the following format: x<-data.frame(id=c(‘a’,’b’,’c’),’2005-01-15’=c(100,225,425), ’2005-02-23’=c(1100,2325,4525)) > x id X2005.01.15 X2005.02.23 1 a 100 1100 2 b 225 2325 3 c 425 4525 I want: id a b c X2005.01.15 100 225 425 X2005.02.23 1100 2325 4525 Any Suggestions? __________________________________________________ [[alternative HTML version deleted]]
t c wrote:> I have a data set in the following format: > > x<-data.frame(id=c(?a?,?b?,?c?),?2005-01-15?=c(100,225,425), ?2005-02-23?=c(1100,2325,4525)) > > > x > id X2005.01.15 X2005.02.23 > 1 a 100 1100 > 2 b 225 2325 > 3 c 425 4525 > > > I want: > id > a > b > c > X2005.01.15 > 100 > 225 > 425 > X2005.02.23 > 1100 > 2325 > 4525 > > > Any Suggestions?I do not get your point, since subject line and body of your message are telling different stories. Do you want to have a list of vectors? You certainly do not want to transpose a data frame with both factors (or character) and numeric values in it..... Uwe Ligges> > __________________________________________________ > > > > [[alternative HTML version deleted]] > > > > ------------------------------------------------------------------------ > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Not sure what you want from the explanation you gave but to transpose try that: x <- as.matrix(x) x.trans <- t(x) One method that's may also of use is help.search("transpose") David On Dec 20, 2005, at 17:26, t c wrote:> I have a data set in the following format: > > x<-data.frame(id=c(?a?,?b?,?c?),?2005-01-15?=c(100,225,425), > ?2005-02-23?=c(1100,2325,4525)) > >> x > id X2005.01.15 X2005.02.23 > 1 a 100 1100 > 2 b 225 2325 > 3 c 425 4525 > > > I want: > id > a > b > c > X2005.01.15 > 100 > 225 > 425 > X2005.02.23 > 1100 > 2325 > 4525 > > > Any Suggestions? > > > __________________________________________________ > > > > [[alternative HTML version deleted]] > > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html >-----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.1 (Darwin) iD8DBQFDqUsI7EoGVUIQyhERArLdAKCX7FniqohYs646riJopkqs6/rboQCcDBUK 4n9JS+hzHOwLZLc6HKQWJcM=U6e6 -----END PGP SIGNATURE-----
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Not sure what you want from the explanation you gave but to transpose try that: x <- as.matrix(x) x.trans <- t(x) One method that's may also of use is help.search("transpose") David On Dec 20, 2005, at 17:26, t c wrote:> I have a data set in the following format: > > x<-data.frame(id=c(?a?,?b?,?c?),?2005-01-15?=c(100,225,425), > ?2005-02-23?=c(1100,2325,4525)) > >> x > id X2005.01.15 X2005.02.23 > 1 a 100 1100 > 2 b 225 2325 > 3 c 425 4525 > > > I want: > id > a > b > c > X2005.01.15 > 100 > 225 > 425 > X2005.02.23 > 1100 > 2325 > 4525 > > > Any Suggestions? > > > __________________________________________________ > > > > [[alternative HTML version deleted]] > > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html >-----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.1 (Darwin) iD8DBQFDqUtZ7EoGVUIQyhERAohmAJ9ASkIETkTvBP7nT8Tr739H7gwWjwCgiBji LpWmylZvZ6al91b02rrAtGg=a98J -----END PGP SIGNATURE-----
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