我有一个关于在混合glmm模型中进行事后对比的问题。基本上，我的数据是一系列具有“主题类型”（三种类型的主题），“治疗”（四种治疗）和“年”（两年）的固定因子的响应变量，具有“主题编号”和“位置'作为随机效应。
我将在下面讨论估算值，以下是按主题类型和处理方式的均值和标准差：
我的模型工作正常，如下：
library(nlme)
ctrl <- lmeControl(opt='optim')
myModel <- lme(y ~ Treatment * SubjectType * Year, random=list(SubjectNumber=~1, Location=~1), control=ctrl, method="REML", data = dframe1, na.action="na.exclude")
然后我得到以下输出，看起来都很好：
anova.lme(myModel)
numDF denDF F-value p-value
(Intercept) 1 168 3199.405 <.0001
Treatment 3 168 5.837 0.0008
SubjectType 2 168 7.502 0.0008
Year 1 168 9.410 0.0025
Treatment:SubjectType 6 168 0.933 0.4725
Treatment:Year 3 168 3.858 0.0106
SubjectType:Year 2 168 6.322 0.0023
Treatment:SubjectType:Year 6 168 1.383 0.2239
然而，我现在希望进行事后对比，看看差异在于“治疗”和“主题类型”的主要影响（即“主题类型1是否与主题类型2和3不同，或者只是2/3？”。
我做了相当多的研究并尝试了几种方法，但我似乎有点迷失，一切都给了我不同的结果，所以一些澄清将非常感激。
最初，我使用了以下代码的Tukey's，我很高兴，但我被告知，这并没有考虑到其他主要影响的变化。这是我最初使用的代码和输出，当我在问题的后面提到差异估计时，提供了lsmeans：
library(lsmeans)
lsmeans(myModel, pairwise~SubjectType, adjust="tukey")
NOTE: Results may be misleading due to involvement in interactions
$lsmeans
SubjectType lsmean SE df lower.CL upper.CL
SubjectType1 0.5531300 0.01853118 168 0.5165460 0.5897140
SubjectType2 0.6078394 0.01853118 168 0.5712554 0.6444234
SubjectType3 0.6545389 0.01853118 168 0.6179549 0.6911229
Results are averaged over the levels of: Treatment
Confidence level used: 0.95
$contrasts
contrast estimate SE df t.ratio p.value
SubjectType1 - SubjectType2 -0.05470938 0.02620704 168 -2.088 0.0955
SubjectType1 - SubjectType3 -0.10140890 0.02620704 168 -3.870 0.0005
SubjectType2 - SubjectType3 -0.04669951 0.02620704 168 -1.782 0.1788
Results are averaged over the levels of: Treatment
P value adjustment: tukey method for comparing a family of 3 estimates
lsmeans(myModel, pairwise~Treatment, adjust="tukey")
NOTE: Results may be misleading due to involvement in interactions
$lsmeans
Treatment lsmean SE df lower.CL upper.CL
Treatment1 0.6379318 0.02139796 168 0.5956883 0.6801754
Treatment2 0.5299945 0.02139796 168 0.4877510 0.5722380
Treatment3 0.6123516 0.02139796 168 0.5701081 0.6545952
Treatment4 0.6403998 0.02139796 168 0.5981563 0.6826434
Results are averaged over the levels of: SubjectType
Confidence level used: 0.95
$contrasts
contrast estimate SE df t.ratio p.value
Treatment1 - Treatment2 0.107937332 0.03026128 168 3.567 0.0026
Treatment1 - Treatment3 0.025580191 0.03026128 168 0.845 0.8327
Treatment1 - Treatment4 -0.002468028 0.03026128 168 -0.082 0.9998
Treatment2 - Treatment3 -0.082357141 0.03026128 168 -2.722 0.0358
Treatment2 - Treatment4 -0.110405360 0.03026128 168 -3.648 0.0020
Treatment3 - Treatment4 -0.028048219 0.03026128 168 -0.927 0.7905
Results are averaged over the levels of: SubjectType
P value adjustment: tukey method for comparing a family of 4 estimates
根据数据，这一点的估计看起来是正确的，而标记为重要的那些在绘制时似乎彼此不同。我还对'multcomp'包进行了类似的比较，它给出了不同的结果，并且估计的差异看起来不正确：
library("multcomp")
Contrasts2 <- glht(myModel, linfct = mcp(SubjectType = "Tukey"))
Warning message:
In mcp2matrix(model, linfct = linfct) :
covariate interactions found -- default contrast might be inappropriate
summary(Contrasts2)
Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
Fit: lme.formula(fixed = y ~ Treatment * SubjectType * Year, data = dframe1,
random = list(SubjectNumber = ~1, SubjectLocation = ~1), method = "REML",
na.action = "na.exclude", control = ctrl)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
SubjectType2 - SubjectType1 == 0 0.26395 0.16575 1.593 0.249
SubjectType3 - SubjectType1 == 0 0.31642 0.16575 1.909 0.136
SubjectType3 - SubjectType2 == 0 0.05247 0.16575 0.317 0.946
(Adjusted p values reported -- single-step method)
ContrastsTreatment <- glht(myModel, linfct = mcp(Treatment = "Tukey"))
Warning message:
In mcp2matrix(model, linfct = linfct) :
covariate interactions found -- default contrast might be inappropriate
summary(ContrastsTreatment)
Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
Fit: lme.formula(fixed = y ~ Treatment * SubjectType * Year, data = dframe1,
random = list(SubjectNumber = ~1, SubjectLocation = ~1), method = "REML",
na.action = "na.exclude", control = ctrl)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
Treatment2 - Treatment1 == 0 -0.49321 0.16575 -2.976 0.01542 *
Treatment3 - Treatment1 == 0 0.08369 0.16575 0.505 0.95794
Treatment4 - Treatment1 == 0 -0.02263 0.16575 -0.137 0.99909
Treatment3 - Treatment2 == 0 0.57690 0.16575 3.481 0.00294 **
Treatment4 - Treatment2 == 0 0.47059 0.16575 2.839 0.02373 *
Treatment4 - Treatment3 == 0 -0.10631 0.16575 -0.641 0.91856
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Adjusted p values reported -- single-step method)
然后我尝试用glht做一个对比矩阵，这又给出了相当不同的结果，并且基于数据估计看起来不正确：
contrast.matrix.SubjectType <- rbind(
`SubjectType1 vs. Soanta` = c(1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
`SubjectType1 vs. SubjectType3` = c(1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
`SubjectType2 vs. SubjectType3` = c(0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0))
compsSubjectType <- glht(myModel, contrast.matrix.SubjectType)
summary(compsSubjectType)
Simultaneous Tests for General Linear Hypotheses
Fit: lme.formula(fixed = y ~ Treatment * SubjectType * Year, data = dframe1,
random = list(SubjectNumber = ~1, SubjectLocation = ~1), method = "REML",
na.action = "na.exclude", control = ctrl)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
SubjectType1 vs. Soanta == 0 0.33426 0.26207 1.275 0.401
SubjectType1 vs. SubjectType3 == 0 0.28180 0.26207 1.075 0.522
SubjectType2 vs. SubjectType3 == 0 -0.05247 0.16575 -0.317 0.945
(Adjusted p values reported -- single-step method)
contrast.matrix.Treatment <- rbind(
`Treatment1 vs. Treatment 1` = c(1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
`Treatment1 vs. Treatment 2` = c(1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
`Treatment1 vs. Treatment4` = c(1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
`Treatment 1 vs. Treatment 2` = c(0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
`Treatment 1 vs. Treatment4` = c(0, 1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
`Treatment 2 vs. Treatment4` = c(0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0))
compsTreatment <- glht(myModel, contrast.matrix.Treatment)
summary(compsTreatment)
Simultaneous Tests for General Linear Hypotheses
Fit: lme.formula(fixed = y ~ Treatment * SubjectType * Year, data = dframe1,
random = list(SubjectNumber = ~1, SubjectLocation = ~1), method = "REML",
na.action = "na.exclude", control = ctrl)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
Treatment1 vs. Treatment 1 == 0 1.0914 0.2621 4.165 < 0.001 ***
Treatment1 vs. Treatment 2 == 0 0.5145 0.2621 1.963 0.19465
Treatment1 vs. Treatment4 == 0 0.6208 0.2621 2.369 0.07935 .
Treatment 1 vs. Treatment 2 == 0 -0.5769 0.1657 -3.481 0.00268 **
Treatment 1 vs. Treatment4 == 0 -0.4706 0.1657 -2.839 0.02195 *
Treatment 2 vs. Treatment4 == 0 0.1063 0.1657 0.641 0.91584
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Adjusted p values reported -- single-step method)
更改修正显然会影响结果，但同样，估计看起来不正确：
summary(glht(myModel, contrast.matrix.SubjectType), test = adjusted("none"))
Simultaneous Tests for General Linear Hypotheses
Fit: lme.formula(fixed = y ~ Treatment * SubjectType * Year, data = dframe1,
random = list(SubjectNumber = ~1, SubjectLocation = ~1), method = "REML",
na.action = "na.exclude", control = ctrl)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
SubjectType1 vs. Soanta == 0 0.33426 0.26207 1.275 0.202
SubjectType1 vs. SubjectType3 == 0 0.28180 0.26207 1.075 0.282
SubjectType2 vs. SubjectType3 == 0 -0.05247 0.16575 -0.317 0.752
(Adjusted p values reported -- none method)
summary(glht(myModel, contrast.matrix.Treatment), test = adjusted("none"))
Simultaneous Tests for General Linear Hypotheses
Fit: lme.formula(fixed = y ~ Treatment * SubjectType * Year, data = dframe1,
random = list(SubjectNumber = ~1, SubjectLocation = ~1), method = "REML",
na.action = "na.exclude", control = ctrl)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
Treatment1 vs. Treatment 1 == 0 1.0914 0.2621 4.165 3.12e-05 ***
Treatment1 vs. Treatment 2 == 0 0.5145 0.2621 1.963 0.04961 *
Treatment1 vs. Treatment4 == 0 0.6208 0.2621 2.369 0.01784 *
Treatment 1 vs. Treatment 2 == 0 -0.5769 0.1657 -3.481 0.00050 ***
Treatment 1 vs. Treatment4 == 0 -0.4706 0.1657 -2.839 0.00452 **
Treatment 2 vs. Treatment4 == 0 0.1063 0.1657 0.641 0.52125
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Adjusted p values reported -- none method)
最后，模型摘要中的模型比较似乎与数据显示的内容相符，但是再次提供不同的p值，估计似乎与数据不匹配：
summary(myModel)
Linear mixed-effects model fit by REML
Data: dframe1
AIC BIC logLik
0.9785978 104.0406 26.5107
Random effects:
Formula: ~1 | SubjectNumber
(Intercept)
StdDev: 0.0001623009
Formula: ~1 | SubjectLocation %in% SubjectNumber
(Intercept) Residual
StdDev: 0.0001623009 0.2029986
Fixed effects: y ~ Treatment * SubjectType * Year
Value Std.Error DF t-value p-value
(Intercept) 0.5982163 0.11720132 168 5.104177 0.0000
TreatmentTreatment2 -0.4932126 0.16574769 168 -2.975683 0.0034
TreatmentTreatment3 0.0836861 0.16574769 168 0.504900 0.6143
TreatmentTreatment4 -0.0226273 0.16574769 168 -0.136517 0.8916
SubjectTypeSubjectType2 0.2639539 0.16574769 168 1.592504 0.1132
SubjectTypeSubjectType3 0.3164198 0.16574769 168 1.909045 0.0580
Year 0.0082773 0.07412461 168 0.111667 0.9112
TreatmentTreatment2:SubjectTypeSubjectType2 0.0807378 0.23440263 168 0.344441 0.7309
TreatmentTreatment3:SubjectTypeSubjectType2 -0.0983580 0.23440263 168 -0.419611 0.6753
TreatmentTreatment4:SubjectTypeSubjectType2 0.1477391 0.23440263 168 0.630279 0.5294
TreatmentTreatment2:SubjectTypeSubjectType3 0.3533741 0.23440263 168 1.507552 0.1335
TreatmentTreatment3:SubjectTypeSubjectType3 -0.2917698 0.23440263 168 -1.244738 0.2150
TreatmentTreatment4:SubjectTypeSubjectType3 0.0481905 0.23440263 168 0.205589 0.8374
TreatmentTreatment2:Year 0.2465353 0.10482803 168 2.351807 0.0198
TreatmentTreatment3:Year -0.0791173 0.10482803 168 -0.754734 0.4515
TreatmentTreatment4:Year -0.0326548 0.10482803 168 -0.311508 0.7558
SubjectTypeSubjectType2:Year -0.1651260 0.10482803 168 -1.575208 0.1171
SubjectTypeSubjectType3:Year -0.1671907 0.10482803 168 -1.594904 0.1126
TreatmentTreatment2:SubjectTypeSubjectType2:Year -0.0516816 0.14824922 168 -0.348613 0.7278
TreatmentTreatment3:SubjectTypeSubjectType2:Year 0.0718031 0.14824922 168 0.484341 0.6288
TreatmentTreatment4:SubjectTypeSubjectType2:Year -0.0043490 0.14824922 168 -0.029336 0.9766
TreatmentTreatment2:SubjectTypeSubjectType3:Year -0.2067818 0.14824922 168 -1.394826 0.1649
TreatmentTreatment3:SubjectTypeSubjectType3:Year 0.2071016 0.14824922 168 1.396983 0.1643
TreatmentTreatment4:SubjectTypeSubjectType3:Year 0.0218841 0.14824922 168 0.147617 0.8828
所以我的整体问题是什么最适合用于确定治疗/受试者类型之间的比较？为什么两个Tukey的方法提供不同的结果？对不起这个冗长的问题，非常感谢你的时间和建议。如果可以用相对简单的术语提供任何答案，我将不胜感激！谢谢！
答案 0 :(得分：0)
首先，我建议您查看emmeans Vignettes（以前是 lsmeans ）。您可以遵循一些很好的示例，以更好地了解软件包的功能以及结果的含义。我相信这些小插图是了解软件包工作原理的最佳资源之一。至于为什么您的结果因方法而异。我没有足够的统计知识来尝试回答这个问题。
第二，对于我的模型，我遵循StatsExchange this question中提供的说明来完成特定的成对比较。