首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >Emmeans函数-参考网格中没有变量

Emmeans函数-参考网格中没有变量
EN

Stack Overflow用户
提问于 2021-10-13 19:16:45
回答 2查看 357关注 0票数 0

我试图在一个较大的数据集上运行emmeans函数,但它不起作用。以下是我的数据:

代码语言:javascript
复制
structure(list(Date = structure(c(16578, 16578, 16578, 16578, 
16578, 16578), class = "Date"), Time = c(7, 7, 7, 9, 11, 11), 
    Turtle = c("R3L12", "R3L12", "R3L12", "R3L12", "R3L12", "R3L12"
    ), Tex = c(11.891, 12.008, 12.055, 13.219, 18.727, 18.992
    ), m.Tb = c(12.477, 12.54, 12.54, 12.978, 16.362, 16.612), 
    m.HR = c(7.56457, 6.66759, 17.51107, 9.72277, 19.44553, 13.07674
    ), season = c("beginning", "beginning", "beginning", "beginning", 
    "beginning", "beginning"), year = c(2015L, 2015L, 2015L, 
    2015L, 2015L, 2015L), Mass = c(360L, 360L, 360L, 360L, 360L, 
    360L)), row.names = c(NA, 6L), class = "data.frame") 

模型代码:model1 <- lmer(m.HR ~ season + (1|Time) + (1|Date) + (1|Turtle), turtledata)

code表示代码:

代码语言:javascript
复制
model1.emmeans <- emmeans(model1, "Turtle")

以下是我得到的错误:

代码语言:javascript
复制
To enable adjustments, add the argument 'pbkrtest.limit = 20608' (or larger)
[or, globally, 'set emm_options(pbkrtest.limit = 20608)' or larger];
but be warned that this may result in large computation time and memory use.
Note: D.f. calculations have been disabled because the number of observations exceeds 3000.
To enable adjustments, add the argument 'lmerTest.limit = 20608' (or larger)
[or, globally, 'set emm_options(lmerTest.limit = 20608)' or larger];
but be warned that this may result in large computation time and memory use.
Error in emmeans(model1, "Turtle") : 
  No variable named Turtle in the reference grid

我不知道为什么它会说没有Turtle,因为它是我数据集中的一个字符变量。

基本上,我只希望emmeans运行,但我也担心它不会运行,因为整个数据集有20,000行长。

EN

回答 2

Stack Overflow用户

发布于 2021-10-13 22:13:55

Function emmeans测试固定效果(被操纵的东西),而不是随机效果(由于设计而发生的东西)。下面的示例显示了这一点,并提供了一种创建最小可重现示例的方法:

代码语言:javascript
复制
library(emmeans)
library(lme4)

# some artificial data
set.seed(143)
foo <- data.frame(
  m.HR   <- rnorm(100, mean=c(rep(c(5, 6), 25), rep(c(8, 9), 25))),
  season <- rep(c("a", "b"), each=50),
  Turtle <- rep(c("T1", "T2"), 50)
)

# simplified model with one fixed and one raqndom effect
model1 <- lmer(m.HR ~ season +  (1|Turtle), foo)

(model1.emmeans <- emmeans(model1, "Turtle"))
# --> error as Turtle is a random effect

(model1.emmeans <- emmeans(model1, "season"))
# --> works as season is a fixed effect

#season emmean    SE   df lower.CL upper.CL
#a        5.73 0.535 1.07  -0.0567     11.5
#b        8.61 0.535 1.07   2.8254     14.4
#
#Degrees-of-freedom method: kenward-roger 
#Confidence level used: 0.95 

关于随机和固定的更多讨论可以在Cross Validated中找到。

票数 2
EN

Stack Overflow用户

发布于 2021-10-13 23:34:13

您不一定能让emmeans直接做您想做的事情,但某种合理的计算是可能的。

最简单的事情是用季节平均值得到每只海龟的平均预测值:

代码语言:javascript
复制
ref_grid <- with(turtledata, 
   expand.grid(season=unique(season), turtle=unique(Turtle)))
pp <- predict(model1, newdata = ref_grid)
aggregate(pp, by=ref_grid$turtle, FUN=mean)

置信区间更难。

票数 1
EN
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/69561158

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档