[R] Help with data.frame and lapply

[Dimitris Rizopoulos]

you could use something like:

d <- data.frame(x1 = rnorm(100), x2 = rnorm(100), x3 = rnorm(100), cls 
= rnorm(100))
lapply(d[-match("cls", names(d))], function(x, y) lm(y ~ x), y = 
d$cls)


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm



----- Original Message ----- 
From: "Juan Daniel López Serna" <juanda.lopez at iic.uam.es>
To: <r-help at stat.math.ethz.ch>
Sent: Friday, December 16, 2005 9:33 AM
Subject: [R] Help with data.frame and lapply


-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

Hello:
I'm having problems with this line of code:

X.lm <- lapply(names(d), function(x) lm(d["cls"] ~ d[x], data=d))

d[x] is what is giving trouble here, but I don't know exactly how to
solve it. What I'm trying to do is to create a linear model from each
column of the data frame 'd' to apply ANOVA later.
Thanks very much in advance. Regards:


Juan Daniel López Serna

- ----
Instituto de Ingeniería del Conocimiento
(http://www.iic.uam.es)
Universidad Autónoma de Madrid
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.2.6 (GNU/Linux)
Comment: Using GnuPG with Fedora - http://enigmail.mozdev.org

iD8DBQFDonvRXHsVbn2qIYMRAqi8AJ0X6zOAevAGzMczQ+ahHlVJnUK4ZQCeIDi6
PPB3baK8JNOa3eoIgbmVCdM=
=WKlt
-----END PGP SIGNATURE-----

______________________________________________
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


Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm