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Stephen

2005-Nov-24 09:13 UTC

[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]]

Ales Ziberna

2005-Nov-24 12:12 UTC

head link

[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 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

Stephen

2005-Nov-24 14:44 UTC

head link

[R] Survreg Weibull lambda and p

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 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

	[[alternative HTML version deleted]]

Stephen

2005-Nov-24 15:17 UTC

head link

[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

	[[alternative HTML version deleted]]

Stephen

2005-Nov-25 11:52 UTC

head link

[R] Survreg Weibull lambda and p

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]]

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