Question

我在尝试在函数中嵌入lmer时遇到了问题。以下是使用lexdec数据的可重现示例。如果我直接在数据框上运行lmer，则没有问题。例如，假设我想看看词汇决策任务中的阅读时间是否与试验的函数不同。有两种类型的单词刺激，“动物”（例如“狗”）和“植物”（例如“樱桃”）。我可以为动物词计算混合效应模型：

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（比如不为每个级别类型输入相同的命令），我会收到一条错误消息。 你知道为什么吗？非常感谢任何建议！

#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

Answer 1

我不知道问题是什么，你的代码运行得很好。（您的软件包是否是最新的？您运行的是什么R版本？您是否已清理工作区并从头开始尝试代码？）

尽管如此，这是plyr::dlply的一个很好的用例。我会这样做：

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

当我运行你的代码（复制和粘贴而不做修改）时，我得到了类似的输出。除了Data:行之外，它是相同的。

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

无法在函数内运行lmer

1 个答案: