Hello, I would like to fit a linear regression and when I use summary(), I got the following result: Call: lm(formula = weight ~ group - 1) Residuals: Min 1Q Median 3Q Max -1.0710 -0.4938 0.0685 0.2462 1.3690 Coefficients: Estimate Std. Error t value Pr(>|t|) groupCtl 5.0320 0.2202 22.85 9.55e-15 *** groupTrt 4.6610 0.2202 21.16 3.62e-14 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.6964 on 18 degrees of freedom Multiple R-Squared: 0.9818, Adjusted R-squared: 0.9798 F-statistic: 485.1 on 2 and 18 DF, p-value: < 2.2e-16 In fact, I do not need them all. Is there a way of exacting part of the infomation, like the Coefficient or Multiple R-Squared? -- View this message in context: http://www.nabble.com/Linear-Regression-tf4554258.html#a12996725 Sent from the R help mailing list archive at Nabble.com.
look at ?summary.lm(), and specifically at the `Value' section, e.g., try this: lmFit <- lm(weight ~ group - 1) summ.lmFit <- summary(lmFit) summ.lmFit$coefficients summ.lmFit$r.squared 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://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm ----- Original Message ----- From: "livia" <yn19832 at msn.com> To: <r-help at r-project.org> Sent: Tuesday, October 02, 2007 1:04 PM Subject: [R] Linear Regression> > Hello, > > I would like to fit a linear regression and when I use summary(), I > got the > following result: > > Call: > lm(formula = weight ~ group - 1) > > Residuals: > Min 1Q Median 3Q Max > -1.0710 -0.4938 0.0685 0.2462 1.3690 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > groupCtl 5.0320 0.2202 22.85 9.55e-15 *** > groupTrt 4.6610 0.2202 21.16 3.62e-14 *** > --- > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > Residual standard error: 0.6964 on 18 degrees of freedom > Multiple R-Squared: 0.9818, Adjusted R-squared: 0.9798 > F-statistic: 485.1 on 2 and 18 DF, p-value: < 2.2e-16 > > In fact, I do not need them all. Is there a way of exacting part of > the > infomation, like the Coefficient or Multiple R-Squared? > -- > View this message in context: > http://www.nabble.com/Linear-Regression-tf4554258.html#a12996725 > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > 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. >Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
Livia, Try the following: fit1<-summary(lm(weight~group-1) summary(fit1) names(fit1) #The code above wil fit the model and print the results. #The statement names(fit1) will give you the components of #the summary such as #coefficients, R.squared adj.R.squared, etc. #You can then access the components, viz. coeffs<-summary(fit1)$coefficients #or RSq<-summary(fit1)$r.squared John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing)>>> livia <yn19832 at msn.com> 10/2/2007 7:04 AM >>>Hello, I would like to fit a linear regression and when I use summary(), I got the following result: Call: lm(formula = weight ~ group - 1) Residuals: Min 1Q Median 3Q Max -1.0710 -0.4938 0.0685 0.2462 1.3690 Coefficients: Estimate Std. Error t value Pr(>|t|) groupCtl 5.0320 0.2202 22.85 9.55e-15 *** groupTrt 4.6610 0.2202 21.16 3.62e-14 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.6964 on 18 degrees of freedom Multiple R-Squared: 0.9818, Adjusted R-squared: 0.9798 F-statistic: 485.1 on 2 and 18 DF, p-value: < 2.2e-16 In fact, I do not need them all. Is there a way of exacting part of the infomation, like the Coefficient or Multiple R-Squared? -- View this message in context: http://www.nabble.com/Linear-Regression-tf4554258.html#a12996725 Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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. Confidentiality Statement: This email message, including any attachments, is for the so...{{dropped}}
Hello, I am performing a multiple linear regression, is there a way of get the mean and standard error of the response variable? Could anyone give me some advice? Many thanks. -- View this message in context: http://www.nabble.com/Linear-Regression-tf4574635.html#a13057768 Sent from the R help mailing list archive at Nabble.com.
livia wrote:> > Hello, > > I am performing a multiple linear regression, is there a way of get the > mean and standard error of the response variable? > > Could anyone give me some advice? Many thanks. >You may need to give us more information/express your request more clearly. Two answers I can think of (if your model is LM1 = lm(z~x+y)) are mean(z) sd(z)/sqrt(length(z)) or predict(LM1,se.fit=TRUE) Ben Bolker -- View this message in context: http://www.nabble.com/Linear-Regression-tf4574635.html#a13058756 Sent from the R help mailing list archive at Nabble.com.
Thank you very much for your reply and I did not make myself clear. My model is like lm(y ~ x1+x2+x3,data), I would like to see E(y) and the standard error. Ben Bolker wrote:> > > > livia wrote: >> >> Hello, >> >> I am performing a multiple linear regression, is there a way of get the >> mean and standard error of the response variable? >> >> Could anyone give me some advice? Many thanks. >> > > You may need to give us more information/express your request more > clearly. Two > answers I can think of (if your model is LM1 = lm(z~x+y)) are > > mean(z) > sd(z)/sqrt(length(z)) > > or > > predict(LM1,se.fit=TRUE) > > Ben Bolker >-- View this message in context: http://www.nabble.com/Linear-Regression-tf4574635.html#a13059177 Sent from the R help mailing list archive at Nabble.com.