所以我是R新手，正在尝试GLMM并进行事后分析...帮助！我已经在6个光照水平下收集了9个豆娘的二进制数据，1 =对光电鼓的运动响应，0 =无响应。我的数据以“ Animal_ID，light_intensity，response”为标题导入了R。每种光强度（3.36-0.61）重复的动物ID（1-9）（见下文）
使用以下代码（lme4程序包），我执行了GLMM并发现光照水平对响应有重大影响：
d = data.frame(id = data[,1], var = data$Light_Intensity, Response = data$Response)
model <- glmer(Response~var+(1|id),family="binomial",data=d)
summary(model)
返回
Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) [glmerMod]
Family: binomial ( logit )
Formula: Response ~ var + (1 | Animal_ID)
Data: d
AIC BIC logLik deviance df.resid
66 72 -30 60 51
Scaled residuals:
Min 1Q Median 3Q Max
-3.7704 -0.6050 0.3276 0.5195 1.2463
Random effects:
Groups Name Variance Std.Dev.
Animal_ID (Intercept) 1.645 1.283
Number of obs: 54, groups: Animal_ID, 9
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.7406 1.0507 -1.657 0.0976 .
var 1.1114 0.4339 2.561 0.0104 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Correlation of Fixed Effects:
(Intr)
var -0.846
然后运行：
m1 <- update(model, ~.-var)
anova(model, m1, test = 'Chisq')
返回
Data: d
Models:
m1: Response ~ (1 | Animal_ID)
model: Response ~ var + (1 | Animal_ID)
Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq)
m1 2 72.555 76.533 -34.278 68.555
model 3 66.017 71.983 -30.008 60.017 8.5388 1 0.003477 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
我已经安装了multcomp和lsmeans软件包，以尝试执行Tukey post hoc来查看区别在哪里，但是在这两者上都遇到了困难。
运行：
summary(glht(m1,linfct=mcp("Animal_ID"="Tukey")))
返回： “ mcp2matrix（model，linfct = linfct）中的错误： 在“ linfct”中指定了变量“ Animal_ID”，但在“模型”中找不到！ “
运行：
lsmeans(model,pairwise~Animal_ID,adjust="tukey")
返回： “ lsmeans.character.ref.grid（object = new（“ ref.grid”，model.info = list（： 参考网格中没有名为Animal_ID的变量”
我知道我在这里可能很愚蠢，但是任何帮助将不胜感激。我的困惑是滚雪球。
此外，有人对我如何最好地可视化结果（以及如何做到这一点）有任何建议吗？
非常感谢您！
更新：
新代码-
Light <- c("3.36","3.36","3.36","3.36","3.36","3.36","3.36","3.36","3.36","2.98","2.98","2.98","2.98","2.98","2.98","2.98","2.98","2.98","2.73","2.73","2.73","2.73","2.73","2.73","2.73","2.73","2.73","2.15","2.15","2.15","2.15","2.15","2.15","2.15","2.15","2.15","1.72","1.72","1.72","1.72","1.72","1.72","1.72","1.72","1.72","0.61","0.61","0.61","0.61","0.61","0.61","0.61","0.61","0.61")
Subject <- c("1","2","3","4","5","6","7","8","9","1","2","3","4","5","6","7","8","9","1","2","3","4","5","6","7","8","9","1","2","3","4","5","6","7","8","9","1","2","3","4","5","6","7","8","9","1","2","3","4","5","6","7","8","9")
Value <- c("1","0","1","0","1","1","1","0","1","1","0","1","1","1","1","1","1","1","0","1","1","1","1","1","1","0","1","0","0","1","1","1","1","1","1","1","0","0","0","1","0","0","1","0","1","0","0","0","1","1","0","1","0","0")
data <- data.frame(Light, Subject, Value)
library(lme4)
model <- glmer(Value~Light+(1|Subject),family="binomial",data=data)
summary(model)
返回：
Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) [
glmerMod]
Family: binomial ( logit )
Formula: Value ~ Light + (1 | Subject)
Data: data
AIC BIC logLik deviance df.resid
67.5 81.4 -26.7 53.5 47
Scaled residuals:
Min 1Q Median 3Q Max
-2.6564 -0.4884 0.2193 0.3836 1.2418
Random effects:
Groups Name Variance Std.Dev.
Subject (Intercept) 2.687 1.639
Number of obs: 54, groups: Subject, 9
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.070e+00 1.053e+00 -1.016 0.3096
Light1.72 -7.934e-06 1.227e+00 0.000 1.0000
Light2.15 2.931e+00 1.438e+00 2.038 0.0416 *
Light2.73 2.931e+00 1.438e+00 2.038 0.0416 *
Light2.98 4.049e+00 1.699e+00 2.383 0.0172 *
Light3.36 2.111e+00 1.308e+00 1.613 0.1067
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Correlation of Fixed Effects:
(Intr) Lg1.72 Lg2.15 Lg2.73 Lg2.98
Light1.72 -0.582
Light2.15 -0.595 0.426
Light2.73 -0.595 0.426 0.555
Light2.98 -0.534 0.361 0.523 0.523
Light3.36 -0.623 0.469 0.553 0.553 0.508
然后运行：
m1 <- update(model, ~.-Light)
anova(model, m1, test= 'Chisq')
返回：
Data: data
Models:
m1: Value ~ (1 | Subject)
model: Value ~ Light + (1 | Subject)
Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq)
m1 2 72.555 76.533 -34.278 68.555
model 7 67.470 81.393 -26.735 53.470 15.086 5 0.01 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
最后，运行：
library(lsmeans)
lsmeans(model,list(pairwise~Light),adjust="tukey")
返回（现在可以正常工作了！）：
$`lsmeans of Light`
Light lsmean SE df asymp.LCL asymp.UCL
0.61 -1.070208 1.053277 NA -3.1345922 0.9941771
1.72 -1.070216 1.053277 NA -3.1345997 0.9941687
2.15 1.860339 1.172361 NA -0.4374459 4.1581244
2.73 1.860332 1.172360 NA -0.4374511 4.1581149
2.98 2.978658 1.443987 NA 0.1484964 5.8088196
3.36 1.040537 1.050317 NA -1.0180467 3.0991215
Results are given on the logit (not the response) scale.
Confidence level used: 0.95
$`pairwise differences of contrast`
contrast estimate SE df z.ratio p.value
0.61 - 1.72 7.933829e-06 1.226607 NA 0.000 1.0000
0.61 - 2.15 -2.930547e+00 1.438239 NA -2.038 0.3209
0.61 - 2.73 -2.930539e+00 1.438237 NA -2.038 0.3209
0.61 - 2.98 -4.048866e+00 1.699175 NA -2.383 0.1622
0.61 - 3.36 -2.110745e+00 1.308395 NA -1.613 0.5897
1.72 - 2.15 -2.930555e+00 1.438239 NA -2.038 0.3209
1.72 - 2.73 -2.930547e+00 1.438238 NA -2.038 0.3209
1.72 - 2.98 -4.048874e+00 1.699175 NA -2.383 0.1622
1.72 - 3.36 -2.110753e+00 1.308395 NA -1.613 0.5897
2.15 - 2.73 7.347728e-06 1.357365 NA 0.000 1.0000
2.15 - 2.98 -1.118319e+00 1.548539 NA -0.722 0.9793
2.15 - 3.36 8.198019e-01 1.302947 NA 0.629 0.9889
2.73 - 2.98 -1.118326e+00 1.548538 NA -0.722 0.9793
2.73 - 3.36 8.197945e-01 1.302947 NA 0.629 0.9889
2.98 - 3.36 1.938121e+00 1.529202 NA 1.267 0.8029
结果以对数优势比（而非响应）量表给出。 P值调整：用于比较6个估算值的族的tukey方法
答案 0 :(得分：1)
您的模型将Animal_ID指定为随机效果。 glht和lsmeans函数仅适用于固定效果比较。