Hi All, I have conducted the following survival analysis which appears to be OK (thanks BRipley for solving my earlier problem).> surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset,dist="weibull", scale = 1)> summary(surv.mod1)Call: survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, dist = "weibull", scale = 1) Value Std. Error z p (Intercept) 7.36 0.259 28.42 1.27e-177 randgrpc -0.59 0.156 -3.80 1.47e-04 Scale fixed at 1 Weibull distribution Loglik(model)= -1268.6 Loglik(intercept only)= -1276 Chisq= 14.72 on 1 degrees of freedom, p= 0.00012 Number of Newton-Raphson Iterations: 5 n= 400> version_ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 1.1 year 2005 month 06 day 20 language R>I emailed this output to a colleague and received an email requesting for the 2 groups (randgrpc in the code) the 'Weibull lambda and p values' in the analysis. I checked the mailings for direction but to no avail. Could someone please provide direction as how to extract the Weibull "lambda" and "p" for both randgrpc values? Many thanks S. ???? ?"? ???? ???? http://mail.nana.co.il [[alternative HTML version deleted]]
Firstly, I assume that your variable is a numeric one. For seperat values p and lambda for diferent categories, you should convert it to factor. However, this has no effect in your case, since you have only 2 categories. You can have only one p and lambda for a variable with only 2 values. The model can only evaluate the diference, which is what you got, assuming that your groups are coded in such a way, that the difference between the codes is 1. Best, Ales Ziberna ----- Original Message ----- From: "Stephen" <szlevine at nana.co.il> To: <r-help at stat.math.ethz.ch> Sent: Thursday, November 24, 2005 10:13 AM Subject: [R] Survreg Weibull lambda and p> Hi All, > > > > I have conducted the following survival analysis which appears to be OK > > (thanks BRipley for solving my earlier problem). > > > >> surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, > dist="weibull", scale = 1) > >> summary(surv.mod1) > > > > Call: > > survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, > > dist = "weibull", scale = 1) > > Value Std. Error z p > > (Intercept) 7.36 0.259 28.42 1.27e-177 > > randgrpc -0.59 0.156 -3.80 1.47e-04 > > > > Scale fixed at 1 > > > > Weibull distribution > > Loglik(model)= -1268.6 Loglik(intercept only)= -1276 > > Chisq= 14.72 on 1 degrees of freedom, p= 0.00012 > > Number of Newton-Raphson Iterations: 5 > > n= 400 > > > >> version > > _ > > platform i386-pc-mingw32 > > arch i386 > > os mingw32 > > system i386, mingw32 > > status > > major 2 > > minor 1.1 > > year 2005 > > month 06 > > day 20 > > language R > >> > > I emailed this output to a colleague and received an email requesting > for the 2 groups (randgrpc in the code) > > the 'Weibull lambda and p values' in the analysis. I checked the > mailings for direction but to no avail. > > Could someone please provide direction as how to extract the Weibull > "lambda" and "p" for both randgrpc values? > > Many thanks > > S. > > > ???? ?"? ???? ???? > http://mail.nana.co.il > > [[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
Hi Ales, Thanks I had tried converting randgrpc previously to a factor - still didn't get the effect for the different groups. Weird - do you think not getting different effects for each group might have to do with tidying up tmp files and the like? I have had tmp file issues with well known commercial software. Thanks Regards S. ________________________________ From: Ales Ziberna [mailto:aleszib@gmail.com] Sent: Thu 24/11/2005 14:12 To: Stephen; r-help@stat.math.ethz.ch Subject: Re: [R] Survreg Weibull lambda and p Firstly, I assume that your variable is a numeric one. For seperat values p and lambda for diferent categories, you should convert it to factor. However, this has no effect in your case, since you have only 2 categories. You can have only one p and lambda for a variable with only 2 values. The model can only evaluate the diference, which is what you got, assuming that your groups are coded in such a way, that the difference between the codes is 1. Best, Ales Ziberna ----- Original Message ----- From: "Stephen" <szlevine@nana.co.il> To: <r-help@stat.math.ethz.ch> Sent: Thursday, November 24, 2005 10:13 AM Subject: [R] Survreg Weibull lambda and p> Hi All, > > > > I have conducted the following survival analysis which appears to beOK> > (thanks BRipley for solving my earlier problem). > > > >> surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, > dist="weibull", scale = 1) > >> summary(surv.mod1) > > > > Call: > > survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, > > dist = "weibull", scale = 1) > > Value Std. Error z p > > (Intercept) 7.36 0.259 28.42 1.27e-177 > > randgrpc -0.59 0.156 -3.80 1.47e-04 > > > > Scale fixed at 1 > > > > Weibull distribution > > Loglik(model)= -1268.6 Loglik(intercept only)= -1276 > > Chisq= 14.72 on 1 degrees of freedom, p= 0.00012 > > Number of Newton-Raphson Iterations: 5 > > n= 400 > > > >> version > > _ > > platform i386-pc-mingw32 > > arch i386 > > os mingw32 > > system i386, mingw32 > > status > > major 2 > > minor 1.1 > > year 2005 > > month 06 > > day 20 > > language R > >> > > I emailed this output to a colleague and received an email requesting > for the 2 groups (randgrpc in the code) > > the 'Weibull lambda and p values' in the analysis. I checked the > mailings for direction but to no avail. > > Could someone please provide direction as how to extract the Weibull > "lambda" and "p" for both randgrpc values? > > Many thanks > > S. > > > ???? ?"? ???? ???? > http://mail.nana.co.il > > [[alternative HTML version deleted]] > > ______________________________________________ > 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???? ?"? ???? ???? http://mail.nana.co.il [[alternative HTML version deleted]]
Hi Ales, Sorry Mis-read your posting - I blindly saw 'no effect' and was thrown - so its like a dummy regression you loose 1 category I can conclude that p of randgrp is 1.47e-04 or 0.000147 which is a significant difference between the groups? Sorry this is the first time I have done this. Thanks S ________________________________ From: Ales Ziberna [mailto:aleszib@gmail.com] Sent: Thu 24/11/2005 14:12 To: Stephen; r-help@stat.math.ethz.ch Subject: Re: [R] Survreg Weibull lambda and p Firstly, I assume that your variable is a numeric one. For seperat values p and lambda for diferent categories, you should convert it to factor. However, this has no effect in your case, since you have only 2 categories. You can have only one p and lambda for a variable with only 2 values. The model can only evaluate the diference, which is what you got, assuming that your groups are coded in such a way, that the difference between the codes is 1. Best, Ales Ziberna ----- Original Message ----- From: "Stephen" <szlevine@nana.co.il> To: <r-help@stat.math.ethz.ch> Sent: Thursday, November 24, 2005 10:13 AM Subject: [R] Survreg Weibull lambda and p> Hi All, > > > > I have conducted the following survival analysis which appears to beOK> > (thanks BRipley for solving my earlier problem). > > > >> surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, > dist="weibull", scale = 1) > >> summary(surv.mod1) > > > > Call: > > survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, > > dist = "weibull", scale = 1) > > Value Std. Error z p > > (Intercept) 7.36 0.259 28.42 1.27e-177 > > randgrpc -0.59 0.156 -3.80 1.47e-04 > > > > Scale fixed at 1 > > > > Weibull distribution > > Loglik(model)= -1268.6 Loglik(intercept only)= -1276 > > Chisq= 14.72 on 1 degrees of freedom, p= 0.00012 > > Number of Newton-Raphson Iterations: 5 > > n= 400 > > > >> version > > _ > > platform i386-pc-mingw32 > > arch i386 > > os mingw32 > > system i386, mingw32 > > status > > major 2 > > minor 1.1 > > year 2005 > > month 06 > > day 20 > > language R > >> > > I emailed this output to a colleague and received an email requesting > for the 2 groups (randgrpc in the code) > > the 'Weibull lambda and p values' in the analysis. I checked the > mailings for direction but to no avail. > > Could someone please provide direction as how to extract the Weibull > "lambda" and "p" for both randgrpc values? > > Many thanks > > S. > > > ???? ?"? ???? ???? > http://mail.nana.co.il > > [[alternative HTML version deleted]] > > ______________________________________________ > 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???? ?"? ???? ???? http://mail.nana.co.il [[alternative HTML version deleted]]
Thanks Ales, Best wishes Stephen. ________________________________ From: Ales Ziberna [mailto:aleszib@gmail.com] Sent: Fri 25/11/2005 10:05 To: Stephen; r-help@stat.math.ethz.ch Subject: Re: [R] Survreg Weibull lambda and p Yes, that is correct! Best, Ales Ziberna ----- Original Message ----- From: Stephen <mailto:szlevine@nana.co.il> To: Ales Ziberna <mailto:aleszib@gmail.com> ; r-help@stat.math.ethz.ch Sent: Thursday, November 24, 2005 4:17 PM Subject: RE: [R] Survreg Weibull lambda and p Hi Ales, Sorry Mis-read your posting - I blindly saw 'no effect' and was thrown - so its like a dummy regression you loose 1 category I can conclude that p of randgrp is 1.47e-04 or 0.000147 which is a significant difference between the groups? Sorry this is the first time I have done this. Thanks S ________________________________ From: Ales Ziberna [mailto:aleszib@gmail.com] Sent: Thu 24/11/2005 14:12 To: Stephen; r-help@stat.math.ethz.ch Subject: Re: [R] Survreg Weibull lambda and p Firstly, I assume that your variable is a numeric one. For seperat values p and lambda for diferent categories, you should convert it to factor. However, this has no effect in your case, since you have only 2 categories. You can have only one p and lambda for a variable with only 2 values. The model can only evaluate the diference, which is what you got, assuming that your groups are coded in such a way, that the difference between the codes is 1. Best, Ales Ziberna ----- Original Message ----- From: "Stephen" <szlevine@nana.co.il> To: <r-help@stat.math.ethz.ch> Sent: Thursday, November 24, 2005 10:13 AM Subject: [R] Survreg Weibull lambda and p > Hi All, > > > > I have conducted the following survival analysis which appears to be OK > > (thanks BRipley for solving my earlier problem). > > > >> surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, > dist="weibull", scale = 1) > >> summary(surv.mod1) > > > > Call: > > survreg(formula = Surv(timep1, relall6) ~ randgrpc, data Dataset, > > dist = "weibull", scale = 1) > > Value Std. Error z p > > (Intercept) 7.36 0.259 28.42 1.27e-177 > > randgrpc -0.59 0.156 -3.80 1.47e-04 > > > > Scale fixed at 1 > > > > Weibull distribution > > Loglik(model)= -1268.6 Loglik(intercept only)= -1276 > > Chisq= 14.72 on 1 degrees of freedom, p= 0.00012 > > Number of Newton-Raphson Iterations: 5 > > n= 400 > > > >> version > > _ > > platform i386-pc-mingw32 > > arch i386 > > os mingw32 > > system i386, mingw32 > > status > > major 2 > > minor 1.1 > > year 2005 > > month 06 > > day 20 > > language R > >> > > I emailed this output to a colleague and received an email requesting > for the 2 groups (randgrpc in the code) > > the 'Weibull lambda and p values' in the analysis. I checked the > mailings for direction but to no avail. > > Could someone please provide direction as how to extract the Weibull > "lambda" and "p" for both randgrpc values? > > Many thanks > > S. > > > ???? ?"? ???? ???? > http://mail.nana.co.il > > [[alternative HTML version deleted]] > > ______________________________________________ > 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 ???? ?"? ???? ???? http://mail.nana.co.il ???? ?"? ???? ???? http://mail.nana.co.il [[alternative HTML version deleted]]