如何为lmer模型结果绘制带有标准误差的预测值? [英] How to plot predicted values with standard errors for lmer model results?
问题描述
我有一个针对四个位置和四个底物(从每个位置获取)的移植实验.我已经确定了每个人群在每个位置和底物组合中的存活率.该实验重复了三遍.
I have a transplant experiment for four locations and four substrates (taken from each location). I have determined survival for each population in each location and substrate combination. This experiment was replicated three times.
我创建了一个如下所示的lmm:
I have created a lmm as follows:
Survival.model <- lmer(Survival ~ Location + Substrate + Location:Substrate + (1|Replicate), data=Transplant.Survival,, REML = TRUE)
我想使用predict命令提取预测，例如:
I would like to use the predict command to extract predictions, for example:
Survival.pred <- predict(Survival.model)
然后提取标准错误，以便我可以将其与预测一起绘制以生成类似以下图的内容:
Then extract standard errors so that I can plot them with the predictions to generate something like the following plot:
我知道如何使用标准的glm(这是我创建示例图的方式)来执行此操作，但是不确定是否可以或应该使用lmm来执行此操作.
I know how to do this with a standard glm (which is how I created the example plot), but am not sure if I can or should do this with an lmm.
我可以做这件事吗，或者我是线性混合模型的新用户，但缺少一些基本知识?
Can I do this or am I as a new user of linear mixed models missing something fundamental?
我确实在Stack Overflow上找到了这个 post 没有帮助.
I did find this post on Stack Overflow which was not helpful.
根据RHertel的评论，也许我应该用 问题:如何绘制模型估计值和置信区间 我的lmer模型得到了结果，因此我可以获得与我相似的图 在上面创建了?
Based on a comment from RHertel, maybe I should have phrased the question: How do I plot model estimates and confidence intervals for my lmer model results so that I can get a similar plot to the one I have created above?
样本数据:
Transplant.Survival <- structure(list(Location = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("Steninge", "Molle",
"Kampinge", "Kaseberga"), class = "factor"), Substrate = structure(c(1L,
1L, 1L, 2L, 2L, 2L, 3L, 3L, 4L, 4L, 4L, 1L, 1L, 2L, 2L, 2L, 3L,
3L, 3L, 4L, 4L, 4L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L,
4L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L), .Label = c("Steninge",
"Molle", "Kampinge", "Kaseberga"), class = "factor"), Replicate = structure(c(1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 1L, 2L, 3L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("1",
"2", "3"), class = "factor"), Survival = c(1, 1, 1, 0.633333333333333,
0.966666666666667, 0.5, 0.3, 0.233333333333333, 0.433333333333333,
0.966666666666667, 0.866666666666667, 0.5, 0.6, 0.266666666666667,
0.733333333333333, 0.6, 0.3, 0.5, 0.3, 0.633333333333333, 0.9,
0.266666666666667, 0.633333333333333, 0.7, 0.633333333333333,
0.833333333333333, 0.9, 0.6, 0.166666666666667, 0.333333333333333,
0.433333333333333, 0.6, 0.9, 0.6, 0.133333333333333, 0.566666666666667,
0.633333333333333, 0.633333333333333, 0.766666666666667, 0.766666666666667,
0.0333333333333333, 0.733333333333333, 0.3, 1.03333333333333,
0.6, 1)), .Names = c("Location", "Substrate", "Replicate", "Survival"
), class = "data.frame", row.names = c(NA, -46L))
推荐答案
修复了函数/图形中的错误.
如果要使用CI绘制估计值，则可能需要查看sjPlot包中的sjp.lmer函数.
If you like to plot estimates with CI, you may want to look at the sjp.lmer function in the sjPlot package. See some example of the various plot types here.
此外，arm软件包提供了用于计算标准错误(arm::se.fixef和arm::se.ranef)
Furthermore, the arm package provides function for computing standard Errors (arm::se.fixef and arm::se.ranef)
sjp.setTheme("forestgrey") # plot theme
sjp.lmer(Survival.model, type = "fe")
将给出以下情节
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