Question

我是R的新手，我现在已经阅读了这些论坛（寻求R的帮助），但这是我第一次发帖。在谷歌搜索每个错误后，我仍然无法弄清楚并修复我的错误。

我正在尝试运行具有不相等样本大小的单向重复测量ANOVA。这是我的数据的玩具版本和我正在使用的代码。（如果重要的话，我的真实数据有12个箱子，每个箱子里的值最多为14到20个。）

## the data: average probability for a subject, given reaction time bin
bin1=c(0.37,0.00,0.00,0.16,0.00,0.00,0.08,0.06)
bin2=c(0.33,0.21,0.000,1.00,0.00,0.00,0.00,0.00,0.09,0.10,0.04)
bin3=c(0.07,0.41,0.07,0.00,0.10,0.00,0.30,0.25,0.08,0.15,0.32,0.18)

## creating the data frame

# dependent variable column
probability=c(bin1,bin2,bin3)

# condition column
bin=c(rep("bin1",8),rep("bin2",11),rep("bin3",12))

# subject column (in the order that will match them up with their respective
# values in the dependent variable column)
subject=c("S2","S3","S5","S7","S8","S9","S11","S12","S1","S2","S3","S4","S7",
  "S9","S10","S11","S12","S13","S14","S1","S2","S3","S5","S7","S8","S9","S10",
  "S11","S12","S13","S14")

# putting together the data frame
dataFrame=data.frame(cbind(probability,bin,subject))

## one-way repeated measures anova
test=aov(probability~bin+Error(subject/bin),data=dataFrame)

这些是我得到的错误：

Error in qr.qty(qr.e, resp) : 
  invalid to change the storage mode of a factor
In addition: Warning messages:
1: In model.response(mf, "numeric") :
  using type = "numeric" with a factor response will be ignored
2: In Ops.factor(y, z$residuals) : - not meaningful for factors
3: In aov(probability ~ bin + Error(subject/bin), data = dataFrame) :
  Error() model is singular

对于复杂性感到抱歉（假设它很复杂;对我而言）。谢谢你的时间。

Answer 1

对于不平衡的重复测量设计，它可能是最容易的使用lme（来自nlme包）：

## this should be the same as the data you constructed above, just 
## a slightly more compact way to do it.
datList <- list(
   bin1=c(0.37,0.00,0.00,0.16,0.00,0.00,0.08,0.06),
   bin2=c(0.33,0.21,0.000,1.00,0.00,0.00,0.00,0.00,0.09,0.10,0.04),
   bin3=c(0.07,0.41,0.07,0.00,0.10,0.00,0.30,0.25,0.08,0.15,0.32,0.18))
subject=c("S2","S3","S5","S7","S8","S9","S11","S12",
          "S1","S2","S3","S4","S7","S9","S10","S11","S12","S13","S14",
          "S1","S2","S3","S5","S7","S8","S9","S10","S11","S12","S13","S14")
d <- data.frame(probability=do.call(c,datList),
                bin=paste0("bin",rep(1:3,sapply(datList,length))),
                subject)

library(nlme)
m1 <- lme(probability~bin,random=~1|subject/bin,data=d)
summary(m1)

唯一真正的问题是解释的某些方面等。与经典的平方和分解方法相差甚远（例如，对方差分量进行显着性检验相当棘手）。 Pinheiro和Bates（Springer，2000）强烈推荐阅读，如果你的话朝着这个方向前进。

模拟/构建一些平衡数据并做到这一点可能是个好主意使用aov()和lme()进行分析，查看输出，并确保你可以看到通信的位置/知道发生了什么。

单向重复测量具有不平衡数据的ANOVA

1 个答案: