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我在尝试嵌入lmer函数时遇到了问题。这是一个使用来自lexdec. 如果我lmer直接在数据框上运行，是没有问题的。例如，假设我想查看词汇决策任务中的阅读时间是否作为Trial的函数而不同。有两种类型的词刺激，“动物”（例如“狗”）和“植物”（例如“樱桃”）。我可以计算动物词的混合效果模型：

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（比如不要为每个级别的类键入相同的命令），我会收到一条错误消息。你知道为什么吗？任何建议将不胜感激！

#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
4

1 回答 1

2

我不知道问题是什么，你的代码对我来说运行得很好。（你的包是最新的吗？你运行的是什么 R 版本？你清理过你的工作空间并从头开始尝试你的代码吗？）

也就是说，这是一个很好的用例plyr::dlply。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（复制和粘贴而不修改）时，我得到了类似的输出。除了线之外，它是相同的Data:。

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
于 2014-03-20T20:54:42.787 回答