Equivalent code to regress(), using fitglm()

2 次查看（过去 30 天）
the cyclist
the cyclist 2014-9-3
评论： the cyclist 2014-9-4
Given the same input data, I would expect
b = regress(log(Y),[ones(size(X,1),1) X]);
and
mdl = fitglm(X,Y,'Link','log');
B = mdl.Coefficients.Estimate;
to give the same coefficients.
Below is some sample code where they do not. Am I missing something simple?
NTRIALS = 31;
dateString = {'01-04-1993','01-07-1995','01-03-1996','01-03-1997', ...
'01-02-1998','01-07-1998','01-11-1998','01-08-1999', ...
'01-09-1999','01-09-2002','01-01-2006','01-03-2006', ...
'01-02-2008','01-05-2008','01-10-2008','01-01-2010', ...
'01-02-2010','01-03-2010','01-10-2011','01-11-2011', ...
'01-01-2012','01-04-2012','01-03-2012','01-04-2012', ...
'01-09-2012','01-09-2012','01-01-2012','01-01-2012', ...
'01-04-2013','01-06-2013','01-02-2014'...
}';
ARR = [1.27 0.84 0.90 1.26 ...
1.61 0.50 1.28 0.98 ...
1.08 1.24 0.61 0.67 ...
0.70 0.44 0.41 0.33 ...
0.40 0.90 0.54 0.80 ...
0.28 1.30 0.39 1.20 ...
0.40 0.36 0.501 0.40 ...
1.04 0.505 0.34 ]';
N = [372 251 301 150 ...
40 127 560 168 ...
293 49 1651 942 ...
104 267 257 1326 ...
1272 29 1088 218 ...
66 165 1106 120 ...
1417 1234 1169 1083 ...
80 1404 1331 ]';
dateNumber = datenum(dateString,'dd-mm-yyyy');
dateVector = datevec(dateNumber);
year = dateVector(:,1);
[b,bint,r,rint,stats] = regress(log(ARR),[ones(NTRIALS,1) year]);
mdl = fitglm(year,ARR,'Link','log');
B = mdl.Coefficients.Estimate;
figure
hold on
h = line(datenum({'1993-01-01','2014-01-01'}),exp(1).^[b(1)+b(2)*1994 b(1)+b(2)*2014]);
hg = line(datenum({'1993-01-01','2014-01-01'}),exp(1).^[B(1)+B(2)*1994 B(1)+B(2)*2014]);
set(h,'Color','r')
set(hg,'Color','g')
semilogy(dateNumber,ARR,'o')
axis tight
set(gca,'YLim',[0.1 2],'YTick',[0.2 0.4 0.6 0.8 1.0 1.5 2])
set(gca,'XTick',1994:2:2014)
datetick('x')

请先登录，再回答此问题。

采纳的回答

Tom Lane
Tom Lane 2014-9-4
I believe these are not the same. For the fitglm model we are saying that Y has a normal distribution whose log(mean) has a linear relationship with the predictors. Y could take on negative values in that case. For the regress model we are saying that the log of Y has a normal distribution whose mean has a linear relationship with the predictors. Y would have to be positive in that case.
  1 个评论
the cyclist
the cyclist 2014-9-4
Proving once again that I am not a statistician, but just play one on TV. Thanks for the clear explanation.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Resampling Techniques 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by