问题
Consider the following table :
DB <- data.frame(
Y =rnorm(6),
X1=c(T, T, F, T, F, F),
X2=c(T, F, T, F, T, T)
)
Y X1 X2
1 1.8376852 TRUE TRUE
2 -2.1173739 TRUE FALSE
3 1.3054450 FALSE TRUE
4 -0.3476706 TRUE FALSE
5 1.3219099 FALSE TRUE
6 0.6781750 FALSE TRUE
I'd like to explain my quantitative variable Y by two binary variables (TRUE or FALSE) without intercept.
The argument of this choice is that, in my study, we can't observe X1=FALSE
and X2=FALSE
at the same time, so it doesn't make sense to have a mean, other than 0, for this level.
With intercept
m1 <- lm(Y~X1+X2, data=DB)
summary(m1)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.9684 1.0590 -1.859 0.1600
X1TRUE 0.7358 0.9032 0.815 0.4749
X2TRUE 3.0702 0.9579 3.205 0.0491 *
Without intercept
m0 <- lm(Y~0+X1+X2, data=DB)
summary(m0)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
X1FALSE -1.9684 1.0590 -1.859 0.1600
X1TRUE -1.2325 0.5531 -2.229 0.1122
X2TRUE 3.0702 0.9579 3.205 0.0491 *
I can't explain why two coefficients are estimated for the variable X1. It seems to be equivalent to the intercept coefficient in the model with intercept.
Same results
When we display the estimation for all the combinations of variables, the two models are the same.
DisplayLevel <- function(m){
R <- outer(
unique(DB$X1),
unique(DB$X2),
function(a, b) predict(m,data.frame(X1=a, X2=b))
)
colnames(R) <- paste0('X2:', unique(DB$X2))
rownames(R) <- paste0('X1:', unique(DB$X1))
return(R)
}
DisplayLevel(m1)
X2:TRUE X2:FALSE
X1:TRUE 1.837685 -1.232522
X1:FALSE 1.101843 -1.968364
DisplayLevel(m0)
X2:TRUE X2:FALSE
X1:TRUE 1.837685 -1.232522
X1:FALSE 1.101843 -1.968364
So the two models are equivalent.
Question
My question is : can we just estimate one coefficient for the first effect ? Can we force R to assign a 0 value to the combinations X1=FALSE
and X2=FALSE
?
回答1:
Yes, we can, by
DB <- as.data.frame(data.matrix(DB))
## or you can do:
## DB$X1 <- as.integer(DB$X1)
## DB$X2 <- as.integer(DB$X2)
# Y X1 X2
# 1 -0.5059575 1 1
# 2 1.3430388 1 0
# 3 -0.2145794 0 1
# 4 -0.1795565 1 0
# 5 -0.1001907 0 1
# 6 0.7126663 0 1
## a linear model without intercept
m0 <- lm(Y ~ 0 + X1 + X2, data = DB)
DisplayLevel(m0)
# X2:1 X2:0
# X1:1 0.15967744 0.2489237
# X1:0 -0.08924625 0.0000000
I have explicitly coerced your TRUE/FALSE
binary into numeric 1/0
, so that no contrast is handled by lm()
.
The data appeared in my answer are different to yours, because you did not use set.seed(?)
before rnorm()
for reproducibility. But this is not a issue here.
来源:https://stackoverflow.com/questions/38129669/how-can-i-force-dropping-intercept-or-equivalent-in-this-linear-model