问题
The problem: I cannot remove a lower order parameter (e.g., a main effects parameter) in a model as long as the higher order parameters (i.e., interactions) remain in the model. Even when doing so, the model is refactored and the new model is not nested in the higher model.
See the following example (as I am coming from ANOVAs I use contr.sum):
d <- data.frame(A = rep(c("a1", "a2"), each = 50), B = c("b1", "b2"), value = rnorm(100))
options(contrasts=c('contr.sum','contr.poly'))
m1 <- lm(value ~ A * B, data = d)
m1
## Call:
## lm(formula = value ~ A * B, data = d)
##
## Coefficients:
## (Intercept) A1 B1 A1:B1
## -0.005645 -0.160379 -0.163848 0.035523
m2 <- update(m1, .~. - A)
m2
## Call:
## lm(formula = value ~ B + A:B, data = d)
## Coefficients:
## (Intercept) B1 Bb1:A1 Bb2:A1
## -0.005645 -0.163848 -0.124855 -0.195902
As can be seen, although I remove one parameter (A), the new model (m2) is refactored and is not nested in the bigger model (m1). If I transform my factors per hand in numerical contrast variables I can get the desired results, but how do I get it using R's factor capabilities?
The Question: How can I remove a lower order factor in R and obtain a model that really misses this parameter and is not refactored (i.e., the number of parameters in the smaller model must be lower)?
But why? I want to obtain 'Type 3' like p-values for a lmer model using the KRmodcomp function from the pbkrtest package. So this example is really just an example.
Why not CrossValidated? I have the feeling that this is really more of an R then a stats question (i.e., I know that you should never fit a model with interactions but without one of the main effects, but I still want to do it).
回答1:
Here's a sort of answer; there is no way that I know of to formulate this model directly by the formula ...
Construct data as above:
d <- data.frame(A = rep(c("a1", "a2"), each = 50),
B = c("b1", "b2"), value = rnorm(100))
options(contrasts=c('contr.sum','contr.poly'))
Confirm original finding that just subtracting the factor from the formula doesn't work:
m1 <- lm(value ~ A * B, data = d)
coef(m1)
## (Intercept) A1 B1 A1:B1
## -0.23766309 0.04651298 -0.13019317 -0.06421580
m2 <- update(m1, .~. - A)
coef(m2)
## (Intercept) B1 Bb1:A1 Bb2:A1
## -0.23766309 -0.13019317 -0.01770282 0.11072877
Formulate the new model matrix:
X0 <- model.matrix(m1)
## drop Intercept column *and* A from model matrix
X1 <- X0[,!colnames(X0) %in% "A1"]
lm.fit allows direct specification of the model matrix:
m3 <- lm.fit(x=X1,y=d$value)
coef(m3)
## (Intercept) B1 A1:B1
## -0.2376631 -0.1301932 -0.0642158
This method only works for a few special cases that allow the model matrix to be specified explicitly (e.g. lm.fit, glm.fit).
More generally:
## need to drop intercept column (or use -1 in the formula)
X1 <- X1[,!colnames(X1) %in% "(Intercept)"]
## : will confuse things -- substitute something inert
colnames(X1) <- gsub(":","_int_",colnames(X1))
newf <- reformulate(colnames(X1),response="value")
m4 <- lm(newf,data=data.frame(value=d$value,X1))
coef(m4)
## (Intercept) B1 A1_int_B1
## -0.2376631 -0.1301932 -0.0642158
This approach has the disadvantage that it won't recognize multiple input variables as stemming from the same predictor (i.e., multiple factor levels from a more-than-2-level factor).
回答2:
I think the most straightforward solution is to use model.matrix. Possibly, you could achieve what you want with some fancy footwork and custom contrasts. However, if you want "type 3 esque" p-values, You probably want it for every term in your model, in which case, I think my approach with model.matrix is convenient anyway because you can easily implicitly loop through all models dropping one column at a time. The provision of a possible approach is not an endorsement of the statistical merits of it, but I do think you formulated a clear question and seem to know it may be unsound statistically so I see no reason not to answer it.
## initial data
set.seed(10)
d <- data.frame(
A = rep(c("a1", "a2"), each = 50),
B = c("b1", "b2"),
value = rnorm(100))
options(contrasts=c('contr.sum','contr.poly'))
## create design matrix
X <- model.matrix(~ A * B, data = d)
## fit models dropping one effect at a time
## change from 1:ncol(X) to 2:ncol(X)
## to avoid a no intercept model
m <- lapply(1:ncol(X), function(i) {
lm(value ~ 0 + X[, -i], data = d)
})
## fit (and store) the full model
m$full <- lm(value ~ 0 + X, data = d)
## fit the full model in usual way to compare
## full and regular should be equivalent
m$regular <- lm(value ~ A * B, data = d)
## extract and view coefficients
lapply(m, coef)
This results in this final output:
[[1]]
X[, -i]A1 X[, -i]B1 X[, -i]A1:B1
-0.2047465 -0.1330705 0.1133502
[[2]]
X[, -i](Intercept) X[, -i]B1 X[, -i]A1:B1
-0.1365489 -0.1330705 0.1133502
[[3]]
X[, -i](Intercept) X[, -i]A1 X[, -i]A1:B1
-0.1365489 -0.2047465 0.1133502
[[4]]
X[, -i](Intercept) X[, -i]A1 X[, -i]B1
-0.1365489 -0.2047465 -0.1330705
$full
X(Intercept) XA1 XB1 XA1:B1
-0.1365489 -0.2047465 -0.1330705 0.1133502
$regular
(Intercept) A1 B1 A1:B1
-0.1365489 -0.2047465 -0.1330705 0.1133502
That is nice so far for models using lm. You mentioned this is ultimately for lmer(), so here is an example using mixed models. I believe it may become more complex if you have more than a random intercept (i.e., effects need to be dropped from the fixed and random parts of the model).
## mixed example
require(lme4)
## data is a bit trickier
set.seed(10)
mixed <- data.frame(
ID = factor(ID <- rep(seq_along(n <- sample(3:8, 60, TRUE)), n)),
A = sample(c("a1", "a2"), length(ID), TRUE),
B = sample(c("b1", "b2"), length(ID), TRUE),
value = rnorm(length(ID), 3) + rep(rnorm(length(n)), n))
## model matrix as before
X <- model.matrix(~ A * B, data = mixed)
## as before but allowing a random intercept by ID
## becomes trickier if you need to add/drop random effects too
## and I do not show an example of this
mm <- lapply(1:ncol(X), function(i) {
lmer(value ~ 0 + X[, -i] + (1 | ID), data = mixed)
})
## full model
mm$full <- lmer(value ~ 0 + X + (1 | ID), data = mixed)
## full model regular way
mm$regular <- lmer(value ~ A * B + (1 | ID), data = mixed)
## view all the fixed effects
lapply(mm, fixef)
Which gives us...
[[1]]
X[, -i]A1 X[, -i]B1 X[, -i]A1:B1
0.009202554 0.028834041 0.054651770
[[2]]
X[, -i](Intercept) X[, -i]B1 X[, -i]A1:B1
2.83379928 0.03007969 0.05992235
[[3]]
X[, -i](Intercept) X[, -i]A1 X[, -i]A1:B1
2.83317191 0.02058800 0.05862495
[[4]]
X[, -i](Intercept) X[, -i]A1 X[, -i]B1
2.83680235 0.01738798 0.02482256
$full
X(Intercept) XA1 XB1 XA1:B1
2.83440919 0.01947658 0.02928676 0.06057778
$regular
(Intercept) A1 B1 A1:B1
2.83440919 0.01947658 0.02928676 0.06057778
来源：https://stackoverflow.com/questions/11335923/how-to-remove-a-lower-order-parameter-in-a-model-when-the-higher-order-parameter