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 López Serna" <[EMAIL PROTECTED]> To: <r-help@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 López Serna - ---- Instituto de Ingeniería del Conocimiento (http://www.iic.uam.es) Universidad Autónoma 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@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 ______________________________________________ R-help@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