本文介绍了glmmTMB中变量之间的对比的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
作为可重现的示例,我们使用下一个无意义的示例:
As a reproducible example, let's use the next no-sense example:
> library(glmmTMB)
> summary(glmmTMB(am ~ disp + hp + (1|carb), data = mtcars))
Family: gaussian ( identity )
Formula: am ~ disp + hp + (1 | carb)
Data: mtcars
AIC BIC logLik deviance df.resid
34.1 41.5 -12.1 24.1 27
Random effects:
Conditional model:
Groups Name Variance Std.Dev.
carb (Intercept) 2.011e-11 4.485e-06
Residual 1.244e-01 3.528e-01
Number of obs: 32, groups: carb, 6
Dispersion estimate for gaussian family (sigma^2): 0.124
Conditional model:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.7559286 0.1502385 5.032 4.87e-07 ***
disp -0.0042892 0.0008355 -5.134 2.84e-07 ***
hp 0.0043626 0.0015103 2.889 0.00387 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
实际上,我的真实模型家族是nbinom2。我想在 disp
和 hp
之间进行对比测试。因此,我尝试:
Actually, my real model family is nbinom2. I want to make a contrast test between disp
and hp
. So, I try:
> glht(glmmTMB(am ~ disp + hp + (1|carb), data = mtcars), linfct = matrix(c(0,1,-1)))
Error in glht.matrix(glmmTMB(am ~ disp + hp + (1 | carb), data = mtcars), :
‘ncol(linfct)’ is not equal to ‘length(coef(model))’
如何避免此错误?
谢谢!
推荐答案
问题实际上很简单: linfct
必须是一个列等于参数个数。您指定了 matrix(c(0,1,-1))
而不指定行数或列数,因此R会默认创建一个列矩阵。添加 nrow = 1
似乎可行。
The problem is actually fairly simple: linfct
needs to be a matrix with the number of columns equal to the number of parameters. You specified matrix(c(0,1,-1))
without specifying numbers of rows or columns, so R made a column matrix by default. Adding nrow=1
seems to work.
library(glmmTMB)
library(multcomp)
m1<- glmmTMB(am ~ disp + hp + (1|carb), data = mtcars)
modelparm.glmmTMB <- function (model, coef. = function(x) fixef(x)[[component]],
vcov. = function(x) vcov(x)[[component]],
df = NULL, component="cond", ...) {
multcomp:::modelparm.default(model, coef. = coef., vcov. = vcov.,
df = df, ...)
}
glht(m1, linfct = matrix(c(0,1,-1),nrow=1))
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