Thanks Frank and Greg,
This makes alot more sense to me now. I appreciate you are both very busy, but
i was wondering if i could trouble you for one last piece of advice. As my data
is a little complicated for a first effort at R let alone modelling!
The response is on a range from 1-6, which indicates extinction risk - 1 being
least concern and 6 being critical - hence using a ordinal model
The factors (6) are categorical - FRUIT TYPE - fleshy/dry
HABITAT - terrestrial,
aquatic, epiphyte
etc etc
I am asking the question - How do different combinations of factors effect
extinction risk.
Based on what you have both said i have called
> predict(model1, type="fitted")
Would this be the best way predicting the probability of falling into each
response category -
y>=2 y>=3 y>=4 y>=5 y>=6
1 0.502220616 0.410236021 0.2892270912 0.2191420568 0.1774250519
2 0.745221699 0.668501579 0.5412223837 0.4486151612 0.3847379442
3 0.720381333 0.639796647 0.5095814746 0.4174618165 0.3551631876
4 0.752321112 0.676811675 0.5505781183 0.4579680710 0.3937100283
5 0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
6 0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
7 0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
8 0.824388319 0.763956402 0.6543788296 0.5663098186 0.5008981585
9 0.526291649 0.433739868 0.3094355120 0.2360800803 0.1919312111
I have 100 species for which i have their factors and i want to predict their
response, so if i do the above and use the newdata function, and present the
probabilities as above rather than trying to classify them?
I tried polr and that "classified" each response as either 1 or 6 i.e no
2,3,4,5 - as did calling predict(model1, type="fitted.ind") which resulted in
the probabilities of being 1 or 6 far outweighing 2,3,4,5 (Below) - this may
just be that my model is not powefull enough to discrimate effectively as i
know that is incorrect ( Brier score 2.01, AUC 66.9)?
EXTINCTION=1 EXTINCTION=2 EXTINCTION=3 EXTINCTION=4 EXTINCTION=5
EXTINCTION=6
1 0.4977794 0.0919845942 0.121008930 0.070085034 0.0417170048
0.1774250519
2 0.2547783 0.0767201200 0.127279196 0.092607223 0.0638772170
0.3847379442
3 0.2796187 0.0805846862 0.130215173 0.092119658 0.0622986289
0.3551631876
4 0.2476789 0.0755094367 0.126233557 0.092610047 0.0642580427
0.3937100283
5 0.1756117 0.0604319173 0.109577572 0.088069011 0.0654116601
0.5008981585
6 0.1756117 0.0604319173 0.109577572 0.088069011 0.0654116601
0.5008981585
7 0.1756117 0.0604319173 0.109577572 0.088069011 0.0654116601
0.5008981585
8 0.1756117 0.0604319173 0.109577572 0.088069011 0.0654116601
0.5008981585
9 0.4737084 0.0925517814 0.124304356 0.073355432 0.0441488692
0.1919312111
10 0.2489307 0.0757263892 0.126424896 0.092614323 0.0641934484
0.3921102030
Thanks very much for any advice given,
John
10 0.751069260 0.675342871 0.5489179746 0.4563036514 0.3921102030
On 1 Oct 2010, at 23:13, Frank Harrell wrote:
Well put Greg. The job of the statistician is to produce good estimates
(probabilities in this case). Those cannot be translated into action
without subject-specific utility functions. Classification during the
analysis or publication stage is not necessary.
Frank
-----
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
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