[R] Great difference for piecewise linear function between R and SAS
Duncan Murdoch
murdoch at stats.uwo.ca
Mon Mar 24 13:08:39 CET 2008
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 <murdoch at stats.uwo.ca> 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.
>>>
>>>
>>
>
>
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