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?
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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