似乎将因子变量转换为序数因子变量会将默认对比度从“contr.treatment”更改为“contr.poly”
cars.data$cyl <- factor(cars.data$cyl)
现在,考虑使用 "contr.treatment" 和 "contr.poly" 来输出
lm
的结果。
lm(mpg ~ disp + cyl.ord, data=cars.data)
Call:
lm(formula = mpg ~ disp + cyl.ord, data = cars.data)
Coefficients:
(Intercept) disp cyl.ord.L cyl.ord.Q
26.34212 -0.02731 -3.38852 1.95127
lm(mpg ~ disp + cyl, data=cars.data,
contrasts=list(cyl="contr.poly"))
Call:
lm(formula = mpg ~ disp + cyl, data = cars.data,
contrasts=list(cyl = "contr.poly"))
Coefficients:
(Intercept) disp cyl.L cyl.Q
26.34212 -0.02731 -3.38852 1.95127
对于有序因子,使用“contr.poly”作为默认对比度,我们可以从无序因子中获得相同的结果。现在,考虑无序因子。
lm(mpg ~ disp + cyl, data=cars.data)
Call:
lm(formula = mpg ~ disp + cyl, data = cars.data)
Coefficients:
(Intercept) disp cyl6 cyl8
29.53477 -0.02731 -4.78585 -4.79209
lm(mpg ~ disp + cyl.ord, data=cars.data, contrasts=list(cyl.ord="contr.treatment"))
Call:
lm(formula = mpg ~ disp + cyl.ord, data = cars.data,
contrasts=list(cyl.ord="contr.treatment"))
Coefficients:
(Intercept) disp cyl.ord6 cyl.ord8
29.53477 -0.02731 -4.78585 -4.79209
因此,无序因子变量默认使用“contr.treatment”,我们可以通过显式要求来从有序因子中获得相同的结果。
但是让我们仔细看一下在回归中使用的模型矩阵。
# Show model matrix
model.matrix(mpg ~ disp + cyl, data=cars.data)
(Intercept) disp cyl6 cyl8
Mazda RX4 1 160.0 1 0
Mazda RX4 Wag 1 160.0 1 0
Datsun 710 1 108.0 0 0
...
attr(,"assign")
[1] 0 1 2 2
attr(,"contrasts")
attr(,"contrasts")$cyl
[1] "contr.treatment"
现在,将“contr.poly”用作对比。
model.matrix(mpg ~ disp + cyl, data=cars.data, contrasts.arg=list(cyl="contr.poly"))
(Intercept) disp cyl.L cyl.Q
Mazda RX4 1 160.0 -9.073800e-17 -0.8164966
Mazda RX4 Wag 1 160.0 -9.073800e-17 -0.8164966
Datsun 710 1 108.0 -7.071068e-01 0.4082483
...
attr(,"assign")
[1] 0 1 2 2
attr(,"contrasts")
attr(,"contrasts")$cyl
[1] "contr.poly"
接下来,尝试使用cyl.ord代替cyl
model.matrix(mpg ~ disp + cyl.ord, data=cars.data)
(Intercept) disp cyl.ord.L cyl.ord.Q
Mazda RX4 1 160.0 -9.073800e-17 -0.8164966
Mazda RX4 Wag 1 160.0 -9.073800e-17 -0.8164966
Datsun 710 1 108.0 -7.071068e-01 0.4082483
...
attr(,"assign")
[1] 0 1 2 2
attr(,"contrasts")
attr(,"contrasts")$cyl.ord
[1] "contr.poly"
最后两个矩阵的条目相同,因此使用"contr.poly"似乎可以解释最初的差异。要了解有关对比的更多信息,请参阅
?contrasts
。