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我是 R 新手，我已经阅读这些论坛（寻求 R 帮助）有一段时间了，但这是我第一次发帖。在这里搜索每个错误后，我仍然无法弄清楚并修复我的错误。

我正在尝试以不相等的样本量运行单向重复测量方差分析。这是我的数据和我正在使用的代码的玩具版本。（如果重要的话，我的真实数据有 12 个 bin，每个 bin 最多有 14 到 20 个值。）

## 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

对复杂性感到抱歉（假设它很复杂；对我来说）。感谢您的时间。

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1 回答 1

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For an unbalanced repeated-measures design, it might be easiest to use lme (from the nlme package):

## 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)

The only real problem is that some aspects of the interpretation etc. are pretty far from the classical sum-of-squares-decomposition approach (e.g. it's fairly tricky to do significance tests of variance components). Pinheiro and Bates (Springer, 2000) is highly recommended reading if you're going to head in this direction.

It might be a good idea to simulate/make up some balanced data and do the analysis with both aov() and lme(), look at the output, and make sure you can see where the correspondences are/know what's going on.

于 2013-07-16T20:34:42.003 回答