當“對比度只能應用於兩個或兩個以上級別的因子”時，如何執行GLM？

Question

我想使用glm在R中進行回歸，但是有一種方法，因為我得到了對比度誤差。

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

使用此處定義的debug_contr_error和debug_contr_error2函數： 如何調試“對比度只能應用於具有兩個或多個級別的因數”錯誤？ 我們可以輕松地找到問題所在：變量New.Runner只剩下一個級別。

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

不能將單個級別的因數應用於對比度，因為任何種類的對比都會使級別數減少1 。 通過1 - 1 = 0該變量將從模型矩陣中刪除。

那么，我們可以簡單地要求不對單個級別的因素應用任何對比嗎？ 否。所有對比方法均禁止這樣做：

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

實際上，如果仔細考慮，您將得出結論， 沒有對比，具有單個水平的因子只是所有1的虛擬變量，即intercept 。 因此，您絕對可以執行以下操作：

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

由於等級不足，您將獲得New.Runner的NA系數。 實際上， 應用對比是避免等級不足的一種基本方法 。 只是當一個因素只有一個層次時，對比的應用就變成了一個悖論。

我們還來看看模型矩陣：

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，並且只能估計其中之一。 如果要估算New.Runner ，則刪除截距：

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

確保您徹底消化了排名不足的問題。 如果您有一個以上的單層因子，並將它們全部替換為1，則丟棄單個截距仍會導致秩不足。

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