当“对比只能应用于具有 2 个或更多水平的因素”时如何进行 GLM？

Question

我想在 R 中使用 glm 进行回归，但有没有办法做到这一点，因为我得到了对比错误。

mydf <- data.frame(Group=c(1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12),
                   WL=rep(c(1,0),12), 
                   New.Runner=c("N","N","N","N","N","N","Y","N","N","N","N","N","N","Y","N","N","N","Y","N","N","N","N","N","Y"), 
                   Last.Run=c(1,5,2,6,5,4,NA,3,7,2,4,9,8,NA,3,5,1,NA,6,10,7,9,2,NA))

mod <- glm(formula = WL~New.Runner+Last.Run, family = binomial, data = mydf)
#Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
# contrasts can be applied only to factors with 2 or more levels

Answer 1

使用此处定义的debug_contr_error 和debug_contr_error2 函数：How to debug “contrasts can be applied only to factors with 2 or more levels” error? 我们可以轻松定位问题：变量New.Runner 中只剩下一个级别。

info <- debug_contr_error2(WL ~ New.Runner + Last.Run, mydf)

info[c(2, 3)]
#$nlevels
#New.Runner 
#         1 
#
#$levels
#$levels$New.Runner
#[1] "N"

## the data frame that is actually used by `glm`
dat <- info$mf

不能对单个级别的因子应用对比，因为任何类型的对比都会将级别数减少1。通过1 - 1 = 0，此变量将从模型矩阵中删除。

那么，我们是否可以简单地要求不对单级因素应用对比？不可以。所有对比方法都禁止这样做：

contr.helmert(n = 1, contrasts = FALSE)
#Error in contr.helmert(n = 1, contrasts = FALSE) : 
#  not enough degrees of freedom to define contrasts

contr.poly(n = 1, contrasts = FALSE)
#Error in contr.poly(n = 1, contrasts = FALSE) : 
#  contrasts not defined for 0 degrees of freedom

contr.sum(n = 1, contrasts = FALSE)
#Error in contr.sum(n = 1, contrasts = FALSE) : 
#  not enough degrees of freedom to define contrasts

contr.treatment(n = 1, contrasts = FALSE)
#Error in contr.treatment(n = 1, contrasts = FALSE) : 
#  not enough degrees of freedom to define contrasts

contr.SAS(n = 1, contrasts = FALSE)
#Error in contr.treatment(n, base = if (is.numeric(n) && length(n) == 1L) n else length(n),  : 
#  not enough degrees of freedom to define contrasts

其实，如果你仔细想想，你会得出结论，没有对比，一个单一水平的因素只是一个全1的虚拟变量，即截距。所以，你绝对可以做到以下几点：

dat$New.Runner <- 1    ## set it to 1, as if no contrasts is applied

mod <- glm(formula = WL ~ New.Runner + Last.Run, family = binomial, data = dat)
#(Intercept)   New.Runner     Last.Run  
#     1.4582           NA      -0.2507

由于rank-deficiency，您将获得New.Runner 的NA 系数。事实上，applying contrasts is a fundamental way to avoid rank-deficiency。只是当一个因素只有一个层次时，对比的应用就成了一个悖论。

我们也来看看模型矩阵：

model.matrix(mod)
#   (Intercept) New.Runner Last.Run
#1            1          1        1
#2            1          1        5
#3            1          1        2
#4            1          1        6
#5            1          1        5
#6            1          1        4
#8            1          1        3
#9            1          1        7
#10           1          1        2
#11           1          1        4
#12           1          1        9
#13           1          1        8
#15           1          1        3
#16           1          1        5
#17           1          1        1
#19           1          1        6
#20           1          1       10
#21           1          1        7
#22           1          1        9
#23           1          1        2

(intercept) 和 New.Runner 具有相同的列，并且只能估计其中之一。如果你想估计New.Runner，就把截距去掉：

glm(formula = WL ~ 0 + New.Runner + Last.Run, family = binomial, data = dat)
#New.Runner    Last.Run  
#    1.4582     -0.2507 

确保彻底消化排名不足的问题。如果您有多个单级因子并将它们全部替换为 1，则删除单个截距仍会导致排名不足。

dat$foo.factor <- 1
glm(formula = WL ~ 0 + New.Runner + foo.factor + Last.Run, family = binomial, data = dat)
#New.Runner  foo.factor    Last.Run  
#    1.4582          NA     -0.2507