r - How to do a GLM when "contrasts can be applied only to factors with 2 or more levels"?

Question

I want to do a regression in R using glm, but is there a way to do it since I get the contrasts error.

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

Using the debug_contr_error and debug_contr_error2 function defined here: How to debug “contrasts can be applied only to factors with 2 or more levels” error? we can easily locate the problem: only a single level is left in variable New.Runner.

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

A factor of single level can not be applied contrasts to, since any kind of contrasts would reduce the number of levels by 1. By 1 - 1 = 0 this variable would be dropped from the model matrix.

Well then, can we simply require that no contrasts be applied to a single-level factor? No. All contrasts methods forbid this:

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

Actually, if you think it carefully, you will conclude that without contrasts, a factor with a single level is just a dummy variable of all 1, i.e., the intercept. So, you can definitely do the following:

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

You get an NA coefficient for New.Runner due to rank-deficiency. In fact, applying contrasts is a fundamental way to avoid rank-deficiency. It is just that when a factor has only one level, application of contrasts becomes a paradox.

Let's also have a look at the model matrix:

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

The (intercept) and New.Runner have identical columns and only one of them can be estimated. If you want to estimate New.Runner, drop the intercept:

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

Make sure you digest the rank-deficiency issue thoroughly. If you have more than one single-level factors and you replace all of them by 1, dropping a single intercept still results in rank-deficiency.

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