问题描述
这是我的数据:
subject arm treat bline change
'subject1' 'L' N 6.3597 4.9281
'subject1' 'R' T 10.3499 1.8915
'subject3' 'L' N 12.4108 -0.9008
'subject3' 'R' T 13.2422 -0.7357
'subject4' 'L' T 8.7383 2.756
'subject4' 'R' N 10.8257 -0.531
'subject5' 'L' N 7.1766 2.0536
'subject5' 'R' T 8.1369 1.9841
'subject6' 'L' T 10.3978 9.0743
'subject6' 'R' N 11.3184 3.381
'subject8' 'L' T 10.7251 2.9658
'subject8' 'R' N 10.9818 2.9908
'subject9' 'L' T 7.3745 2.9143
'subject9' 'R' N 9.4863 -3.0847
'subject10' 'L' T 11.8132 -2.1629
'subject10' 'R' N 9.5287 0.1401
'subject11' 'L' T 8.2977 6.2219
'subject11' 'R' N 9.3691 0.7408
'subject12' 'L' T 12.6003 -0.7645
'subject12' 'R' N 11.7329 0.0342
'subject13' 'L' N 9.4918 2.0716
'subject13' 'R' T 9.6205 1.5705
'subject14' 'L' T 9.3945 4.6176
'subject14' 'R' N 11.0176 1.445
'subject16' 'L' T 8.0221 1.4751
'subject16' 'R' N 9.8307 -2.3697
当我使用 treat
和 arm
作为因子拟合混合模型时:
m <- lmer(change ~ bline + treat + arm + (1|subject),data=change1)
ls_means(m,which = NULL,level=0.95,ddf="Kenward-Roger")
ls_means
语句不返回结果。任何人都可以帮助解决问题吗?
解决方法
我们需要更多信息。这是一个看起来不错的可重现示例:
# Standard
logical.vector = c(TRUE,TRUE,FALSE,TRUE)
# Test Patterns
# Case 1 (non-match): should be FALSE
!any(logical.vector != c(TRUE,TRUE))
# Case 2 (non-match): should be FALSE
!any(logical.vector != c(FALSE,TRUE))
# Case 3 (non-match): should be FALSE
!any(logical.vector != c(TRUE,TRUE))
# Case 4 (non-match): should be FALSE
!any(logical.vector != c(FALSE,FALSE))
# Case 5 (match): should be TRUE
!any(logical.vector != c(TRUE,TRUE))
此时我最好的猜测是您在模型拟合方面遇到了一些问题。 set.seed(101)
library(lme4)
library(lmerTest)
dd <- expand.grid(subject=factor(1:40),arm=c("L","R"))
## replicate N/T in random order for each subject
dd$treat <- c(replicate(40,sample(c("N","T"))))
dd$bline <- rnorm(nrow(dd))
dd$change <- simulate(~bline+treat+arm+(1|subject),newdata=dd,newparams=list(beta=rep(1,4),theta=1,sigma=1))[[1]]
m <- lmer(change ~ bline + treat + arm + (1|subject),data=dd)
ls_means(m,which = NULL,level=0.95,ddf="Kenward-Roger")
## Least Squares Means table:
##
## Estimate Std. Error df t value lower upper Pr(>|t|)
## armL 1.37494 0.22716 55.6 6.0527 0.91981 1.83007 1.275e-07 ***
## armR 2.54956 0.22716 55.6 11.2235 2.09443 3.00469 6.490e-16 ***
有时可能是不透明/吞咽警告或错误消息。你有没有收到任何你忽略告诉我们的警告?如果您使用 lmerTest
重新运行模型(即使用 lme4::lmer(...)
中的基本版本,而不是 lme4
中的增强版本),您是否看到任何警告?
我也看到空结果:
> ls_means(m,ddf="Kenward-Roger")
Least Squares Means table:
Estimate Std. Error df t value lower upper Pr(>|t|)
Confidence level: 95%
Degrees of freedom method: Kenward-Roger
但是,emmeans 包运行良好。您可以使用 emmeans()
或 lsmeans()
- 后者只是重新标记 emmeans()
结果。 “估计边际均值”是一个更通用的术语。
> library(emmeans)
> lsmeans(m,"treat")
treat lsmean SE df lower.CL upper.CL
N 0.996 0.72 15 -0.539 2.53
T 2.290 0.72 15 0.755 3.82
Results are averaged over the levels of: arm
Degrees-of-freedom method: kenward-roger
Confidence level used: 0.95
> lsmeans(m,"arm")
arm lsmean SE df lower.CL upper.CL
L 1.97 0.737 15.6 0.403 3.53
R 1.32 0.737 15.6 -0.248 2.88
Results are averaged over the levels of: treat
Degrees-of-freedom method: kenward-roger
Confidence level used: 0.95
我怀疑 lmerTest::ls_means()
不支持“字符”类的预测器。如果您将 treat
和 arm
更改为因数,它可能会起作用。