Re: [R] CRAN and Multiple Linear Regression

From: Jorge Ivan Velez <jorgeivanvelez_at_gmail.com>
Date: Tue, 13 May 2008 20:20:03 -0400

Hi Thomas,

Perhaps:

 test = read.table("test.dat", header=T, dec=',')  x <- test[,1:21]
 y <- test[, "Y"]
 mymodel <- lm(y ~ x)

 # Coefficients and more information
 summary(mymodel)

 # Plots for the residuals
 plotspar(mfrow=c(2,2))
 plot(mymodel)

See also ?lm

HTH, Jorge

On Tue, May 13, 2008 at 6:54 PM, Anja und Th. Sponsel <anjaspp_at_netscape.net> wrote:

> I just have read the guide and I can do some small steps with cran but
> I still have no clue...
>
> I have data like this:
> X1 X2 X3 ... X21 Y
> 1 0 0 0 ... 18 -0,07254
> 2 1 0 0 ... 6 -0,14921
> 3 0 1 0 ... 12 -0,04165
> 4 0 0 0 ... 8 0,08359
> 5 ... ... ... ... ... ...
> 120 0 0 1 1 8 0,07928
>
> My script is:
> > require(stats); require(graphics)
> > test = read.table("test.dat", header=T)
> > x <- test[, 1:21]
> > y <- test[, "Y"]
> > lm(y ~ x)
>
> The error may be an invalid type (list) for variable x (see below):
> Fehler in model.frame.default(formula = y ~ x, drop.unused.levels =
> TRUE) :
> ungültiger Typ (list) für die Variable 'x'
>
> Then I tried:
> ...
> > lm(data=test)
> Then I get a lot of coefficients, but I'm not sure. That cant be the
> result.
>
> Then I tried:
> > summary(lm(data = wetter))
>
> Call:
> lm(data = wetter)
>
> Residuals:
> ALL 120 residuals are 0: no residual degrees of freedom!
>
> Coefficients: (139 not defined because of singularities)
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) 0.044747 NA NA NA
> X2 0.660275 NA NA NA
> X3 -0.097629 NA NA NA
> X4 0.647851 NA NA NA
> X5 -3.305631 NA NA NA
> ... 0.249788 NA NA NA
>
> I am confused...
>
> Best regards,
> Thomas
>
> ______________________________________________
> R-help_at_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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