Dear Murdoch,
Thanks very much for your rapid response. It helps me greatly.
If i want to write the model formula according to the estimets from R, Is
it correct for the below formula? I'm not very sure about it, and i also
hope that you can recommend a good book on this area. I want to learn it
systematically. Thanks a million.
Logit*p*=12.10+5.815*x+6.654*elevation-6.755*elevation2
-1.291*distance1-10.348*distance2-3.53*distance3
-6.828*y1-4.426*y2-11.216*y3
#R code and the estimate results--Knots for distance are 16.13 and 24,
respectively, and Knots for y are -0.4357 and -0.3202
m.glm<-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13,24))
+bs(y,degree=1,knots=c(-0.4357,-0.3202
)),family=binomial(logit),data=point)
summary(m.glm)
Estimate Std. Error z
value Pr(>|z|)
(Intercept) 12.104 3.138
3.857 0.000115 ***
x 5.815 1.987
2.926 0.003433 **
poly(elevation, 2)1 6.654 4.457
1.493 0.135444
poly(elevation, 2)2 -6.755 3.441 -
1.963 0.049645 *
bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291 1.139 -
1.133 0.257220
bs(distance, degree = 1, knots = c(16.13, 24))2 -10.348 2.025 -
5.110 3.22e-07 ***
bs(distance, degree = 1, knots = c(16.13, 24))3 -3.530 3.752 -
0.941 0.346850
bs(y, degree = 1, knots = c(-0.4357, -0.3202))1 -6.828 1.836 -
3.719 0.000200 ***
bs(y, degree = 1, knots = c(-0.4357, -0.3202))2 -4.426 1.614 -
2.742 0.006105 **
bs(y, degree = 1, knots = c(-0.4357, -0.3202))3 -11.216 2.861 -
3.920 8.86e-05 ***
Thanks again.
On Mon, Mar 24, 2008 at 8:08 PM, Duncan Murdoch <[EMAIL PROTECTED]>
wrote:
> On 24/03/2008 7:06 AM, zhijie zhang wrote:
> > Dear Murdoch,
> > "Compare the predictions, not the coefficients.", this is the key
> point.
> > I have checked their predicted probability and their results are the
> > same.
> > What do you mean by "You're using different bases"? Could you please
> give
> > me a little more info on it, so that i can go to find some materials?
>
> There are many ways to represent a piecewise linear function. R and
> your SAS code have both chosen linear combinations of basis functions,
> but you have used different basis functions. R uses the B-spline basis.
> You have what is sometimes called the truncated power basis, but maybe
> not commonly in the linear context. I don't know if it has a name here.
>
> Duncan Murdoch
>
> > Thanks a lot.
> >
> >
> > On 3/24/08, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> >> On 24/03/2008 5:23 AM, zhijie zhang wrote:
> >>> Dear Rusers,
> >>> I am now using R and SAS to fit the piecewise linear functions, and
> >> what
> >>> surprised me is that they have a great differrent result. See below.
> >> You're using different bases. Compare the predictions, not the
> >> coefficients.
> >>
> >> To see the bases that R uses, do this:
> >>
> >> matplot(distance, bs(distance,degree=1,knots=c(16.13,24)))
> >> matplot(y, bs(y,degree=1,knots=c(-0.4357,-0.3202)))
> >>
> >> Duncan Murdoch
> >>
> >>> #R code--Knots for distance are 16.13 and 24, respectively, and Knots
> >> for y
> >>> are -0.4357 and -0.3202
> >>> m.glm<-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13
> >> ,24))
> >>> +bs(y,degree=1,knots=c(-0.4357,-0.3202
> >>> )),family=binomial(logit),data=point)
> >>> summary(m.glm)
> >>>
> >>> Coefficients:
> >>> Estimate Std.
> >> Error z
> >>> value Pr(>|z|)
> >>> (Intercept) 12.104
> >> 3.138
> >>> 3.857 0.000115 ***
> >>> x 5.815
> 1.987
> >>> 2.926 0.003433 **
> >>> poly(elevation, 2)1 6.654
> 4.457
> >>> 1.493 0.135444
> >>> poly(elevation, 2)2 -6.755
> >> 3.441 -
> >>> 1.963 0.049645 *
> >>> bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291
> >> 1.139 -
> >>> 1.133 0.257220
> >>> bs(distance, degree = 1, knots = c(16.13, 24))2 -10.348
> >> 2.025 -
> >>> 5.110 3.22e-07 ***
> >>> bs(distance, degree = 1, knots = c(16.13, 24))3 -3.530
> 3.752
> >> -
> >>> 0.941 0.346850
> >>> bs(y, degree = 1, knots = c(-0.4357, -0.3202))1 -6.828
> 1.836
> >> -
> >>> 3.719 0.000200 ***
> >>> bs(y, degree = 1, knots = c(-0.4357, -0.3202))2 -4.426
> 1.614
> >> -
> >>> 2.742 0.006105 **
> >>> bs(y, degree = 1, knots = c(-0.4357, -0.3202))3 -11.216
> >> 2.861 -
> >>> 3.920 8.86e-05 ***
> >>>
> >>> #SAS codes
> >>> data b;
> >>> set a;
> >>> if distance > 16.13 then d1=1; else d1=0;
> >>> distance2=d1*(distance - 16.13);
> >>> if distance > 24 then d2=1; else d2=0;
> >>> distance3=d2*(distance - 24);
> >>> if y>-0.4357 then dd1=1; else dd1=0;
> >>> y2=dd1*(y+0.4357);
> >>> if y>-0.3202 then dd2=1; else dd2=0;
> >>> y3=dd2*(y+0.3202);
> >>> run;
> >>>
> >>> proc logistic descending data=b;
> >>> model mark =x elevation elevation*elevation distance distance2
> >> distance3 y
> >>> y2 y3;
> >>> run;
> >>>
> >>>
> >>> The LOGISTIC Procedure Analysis of Maximum Likelihood
> >>> Estimates
> >>>
> >>> Standard
> Wald
> >>> Parameter DF Estimate Error
> >>> Chi-Square Pr > ChiSq
> >>>
> >>> Intercept 1 -2.6148 2.1445
> >>> 1.4867
> >>> 0.2227
> >>> x 1 5.8146 1.9872
> >>> 8.5615
> >>> 0.0034
> >>> elevation 1 0.4457 0.1506
> >>> 8.7545
> >>> 0.0031
> >>> elevation*elevation 1 -0.0279 0.0142
> >>> 3.8533
> >>> 0.0496
> >>> distance 1 -0.1091 0.0963
> >>> 1.2836
> >>> 0.2572
> >>> distance2 1 -1.0418 0.2668
> >>> 15.2424
> >>> <.0001
> >>> distance3 1 2.8633 0.7555
> >>> 14.3625
> >>> 0.0002
> >>> y 1 -16.2032 4.3568
> >>> 13.8314
> >>> 0.0002
> >>> y2 1 36.9974 10.3219
> >>> 12.8476
> >>> 0.0003
> >>> y3 1 -58.4938 14.0279
> >>> 17.3875
> >>> <.0001
> >>> Q: What is the problem? which one is correct for the piecewise linear
> >>> function?
> >>> Thanks very much.
> >>>
> >>>
> >>
> >
> >
>
>
--
With Kind Regards,
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Zhi Jie,Zhang ,PHD
Tel:+86-21-54237149
Dept. of Epidemiology,School of Public Health,Fudan University
Address:No. 138 Yi Xue Yuan Road,Shanghai,China
Postcode:200032
Email:[EMAIL PROTECTED]
Website: www.statABC.com
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