无法在函数中运行lmer [英] Cannot run lmer from within a function
问题描述
我在尝试将lmer嵌入到函数中时遇到问题.这是使用来自lexdec的数据的可重现示例.如果直接在数据帧上运行lmer，则没有问题.例如，假设我想查看词汇决策任务中的阅读时间是否随 Trial 的变化而变化.有两种类型的单词刺激，动物"(例如狗")和植物"(例如樱桃").我可以为动物单词计算一个混合效果模型:
I am running into a problem trying to embed lmer within a function. Here is a reproducible example using data from lexdec. If I run lmer on the data frame directly, there is no problem. For example, say that I want to see whether reading times in a lexical decision task differed as a function of Trial. There were 2 types of word stimuli, "animal" (e.g. "dog") and "plant" (e.g. "cherry"). I can compute a mixed-effects model for animal words:
library(languageR) #load lexdec data
library(lme4) #load lmer()
s <- summary(lmer(RT ~ Trial + (1|Subject) + (1|Word), data = lexdec[lexdec$Class== "animal", ]))
s #this works well
但是，如果我将lmer模型嵌入到函数中(例如，对于每个级别的类都不要键入相同的命令)，则会收到错误消息. 你知道为什么吗?任何建议将不胜感激！
However, if I embed the lmer model inside a function (say to not type the same command for each level of class) I get an error message. Do you know why? Any suggestions will be much appreciated!
#lmer() is now embedded in a function
compute.lmer <- function(df,class) {
m <- lmer(RT ~ Trial + (1|Subject) + (1|Word),data = df[df$Class== class, ])
m <- summary(m)
return(m)
}
#Now I can use this function to iterate over the 2 levels of the **Class** factor
for (c in levels(lexdec$Class)){
s <- compute.lmer(lexdec,c)
print(c)
print(s)
}
#But this gives an error message
Error in `colnames<-`(`*tmp*`, value = c("Estimate", "Std. Error", "df", :
length of 'dimnames' [2] not equal to array extent
推荐答案
我不知道问题出在哪里，您的代码对我来说运行得很好. (您的软件包是最新的吗?您正在运行哪个R版本?您是否清理过工作区并从头开始尝试了代码?)
I don't know what the problem is, your code runs just fine for me. (Are your packages up to date? What R version are you running? Have you cleaned your workspace and tried your code from scratch?)
也就是说，这是plyr::dlply的一个很好的用例.我会这样:
That said, this is a great use case for plyr::dlply. I would do it like this:
library(languageR)
library(lme4)
library(plyr)
stats <- dlply(lexdec,
.variables = c("Class"),
.fun=function(x) return(summary(lmer(RT ~ Trial + (1 | Subject) +
(1 | Word), data = x))))
names(stats) <- levels(lexdec$Class)
然后产生
> stats[["plant"]]
Linear mixed model fit by REML ['lmerMod']
Formula: RT ~ Trial + (1 | Subject) + (1 | Word)
Data: x
REML criterion at convergence: -389.5
Scaled residuals:
Min 1Q Median 3Q Max
-2.2647 -0.6082 -0.1155 0.4502 6.0593
Random effects:
Groups Name Variance Std.Dev.
Word (Intercept) 0.003718 0.06097
Subject (Intercept) 0.023293 0.15262
Residual 0.028697 0.16940
Number of obs: 735, groups: Word, 35; Subject, 21
Fixed effects:
Estimate Std. Error t value
(Intercept) 6.3999245 0.0382700 167.23
Trial -0.0001702 0.0001357 -1.25
Correlation of Fixed Effects:
(Intr)
Trial -0.379
当我运行您的代码时(未经修改就复制粘贴)，我得到类似的输出.除了Data:行之外，其他均相同.
When I run your code (copied and pasted without modification), I get similar output. It's identical except for the Data: line.
stats = list()
compute.lmer <- function(df,class) {
m <- lmer(RT ~ Trial + (1|Subject) + (1|Word),data = df[df$Class== class, ])
m <- summary(m)
return(m)
}
for (c in levels(lexdec$Class)){
s <- compute.lmer(lexdec,c)
stats[[c]] <- s
}
> stats[["plant"]]
Linear mixed model fit by REML ['lmerMod']
Formula: RT ~ Trial + (1 | Subject) + (1 | Word)
Data: df[df$Class == class, ]
REML criterion at convergence: -389.5
Scaled residuals:
Min 1Q Median 3Q Max
-2.2647 -0.6082 -0.1155 0.4502 6.0593
Random effects:
Groups Name Variance Std.Dev.
Word (Intercept) 0.003718 0.06097
Subject (Intercept) 0.023293 0.15262
Residual 0.028697 0.16940
Number of obs: 735, groups: Word, 35; Subject, 21
Fixed effects:
Estimate Std. Error t value
(Intercept) 6.3999245 0.0382700 167.23
Trial -0.0001702 0.0001357 -1.25
Correlation of Fixed Effects:
(Intr)
Trial -0.379
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