[R] Help with lm with a nested effect
Timothy Clough
tclough at purdue.edu
Fri Sep 18 03:22:03 CEST 2009
Dear All,
I am running an ANOVA model with three factors: FEATURE (3 levels),
GROUP (5 levels), and PATIENT (2 levels), where PATIENT is nested
within GROUP.
fit <- lm(ABUNDANCE ~ FEATURE + GROUP + FEATURE:GROUP + GROUP/PATIENT,
example)
However, the design is not balanced: PATIENT1 in GROUP1 is missing,
and when I run summary(fit), I receive the following output.
Call:
lm(formula = ABUNDANCE ~ FEATURE + GROUP + FEATURE:GROUP + GROUP/
PATIENT,
data = example)
Residuals:
Min 1Q Median 3Q Max
-3.569e-01 -1.048e-01 1.479e-17 1.048e-01 3.569e-01
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 13.4542954 0.2790175 48.220 3.78e-11 ***
FEATURE4227 -0.9393679 0.3417252 -2.749 0.0251 *
FEATURE6374 0.1260711 0.3417252 0.369 0.7218
GROUP1 0.0020501 0.4411653 0.005 0.9964
GROUP2 0.0992019 0.3945903 0.251 0.8078
GROUP3 0.0823760 0.3945903 0.209 0.8399
GROUP4 -0.0062425 0.3945903 -0.016 0.9878
FEATURE4227:GROUP1 -0.2178760 0.5918854 -0.368 0.7223
FEATURE6374:GROUP1 -0.1109533 0.5918854 -0.187 0.8560
FEATURE4227:GROUP2 0.0001341 0.4832724 0.000278 0.9998
FEATURE6374:GROUP2 0.2720569 0.4832724 0.563 0.5889
FEATURE4227:GROUP3 0.3848577 0.4832724 0.796 0.4488
FEATURE6374:GROUP3 0.1829045 0.4832724 0.378 0.7149
FEATURE4227:GROUP4 -0.7113272 0.4832724 -1.472 0.1793
FEATURE6374:GROUP4 -0.1151362 0.4832724 -0.238 0.8177
GROUP0:PATIENT2 0.1264295 0.2790175 0.453 0.6625
GROUP1:PATIENT2 NA NA NA NA
GROUP2:PATIENT2 -0.2636043 0.2790175 -0.945 0.3724
GROUP3:PATIENT2 0.0429295 0.2790175 0.154 0.8815
GROUP4:PATIENT2 0.5981179 0.2790175 2.144 0.0644 .
---
As you can see, the estimate for GROUP1:PATIENT2 is NA because of the
missing values from PATIENT1 in GROUP1.
This becomes a problem when I want to estimate contrasts, as the
estimable function gives me an error.
Is there a way to get around this? When running the same model in
SAS, instead of an NA there is simply a 0 for that level, and
contrasts have no problem being estimated.
Sincerely,
Tim
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