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？
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？
提问于 2018-05-12 01:20:54
我想用glm在R中做一个回归，但是有没有办法这样做，因为我得到了对比度错误。
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回答 1
Stack Overflow用户
回答已采纳
发布于 2018-07-29 02:11:15
使用这里定义的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?，我们可以很容易地定位问题: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$mf单一水平的因子不能应用于对比，因为任何一种对比都会减少1的水平数。通过1 - 1 = 0，这个变量将从模型矩阵中删除。
那么，我们能不能简单地要求没有对比度应用于单级因子？不是的。所有的对比方法都禁止这样做：
contr.helmert(n = 1, contrasts = FALSE)
#Error in contr.helmert(n = 1, contrasts = FALSE) :
# not enough degrees of freedom to define contrasts
contr.poly(n = 1, contrasts = FALSE)
#Error in contr.poly(n = 1, contrasts = FALSE) :
# contrasts not defined for 0 degrees of freedom
contr.sum(n = 1, contrasts = FALSE)
#Error in contr.sum(n = 1, contrasts = FALSE) :
# not enough degrees of freedom to define contrasts
contr.treatment(n = 1, contrasts = FALSE)
#Error in contr.treatment(n = 1, contrasts = FALSE) :
# not enough degrees of freedom to define contrasts
contr.SAS(n = 1, contrasts = FALSE)
#Error in contr.treatment(n, base = if (is.numeric(n) && length(n) == 1L) n else length(n), :
# not enough degrees of freedom to define contrasts实际上，如果你仔细考虑，你会得出结论，没有对比的因子，只有一个水平的因子只是全1的一个虚拟变量，即截距。所以，你绝对可以做到以下几点：
dat$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，你可以得到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水平时，对比的应用就成了一种悖论。
让我们来看看模型矩阵：
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，并且只能估计其中的一个。如果你想估计New.Runner，去掉截取：
glm(formula = WL ~ 0 + New.Runner + Last.Run, family = binomial, data = dat)
#New.Runner Last.Run
# 1.4582 -0.2507 确保你彻底消化了等级不足的问题。如果您有多个单级因子，并且将所有因子都替换为1，则丢弃单个截取仍然会导致秩不足。
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 页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50297260
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