1

与我问的上一个问题相关（ggplot2 如何获得 2 个直方图，y 值 = 一个计数 / 两个计数的总和），我尝试编写一个函数，该函数将 data.frame 作为响应的输入几个参与者在几个条件下的时间（RT）和准确度（正确），并输出一个“汇总”data.frame，其中的数据像直方图一样聚合。这里的特殊性是我不想获得每个箱中响应的绝对数量，而是相对计数。

我所说的相对计数是对于直方图的每个 bin，该值对应于： 

relative_correct   = ncorrect / sum(ncorrect+nincorrect)
relative_incorrect = nincorrect / sum(ncorrect+nincorrect)

结果实际上接近于密度图，只是它不是每条曲线的总和等于 1，而是正确和不正确曲线的总和。

这是创建示例数据的代码：

# CREATE EXAMPLE DATA
subjectname <- factor(rep(c("obs1","obs2"),each=50))
Visibility  <- factor(rep(rep(c("cond1","cond2"),each=25),2)) 
RT          <- rnorm(100,300,50)
correct     <- sample(c(rep(0,25),rep(1,75)),100)
my.data <- data.frame(subjectname,Visibility,RT,correct)

首先，我需要定义一个稍后在 ddply 中使用的函数

histRTcounts <- function(df) {out = hist(df$RT, breaks=seq(5, 800, by=10), plot=FALSE)
                          out = out$counts}

然后是 main 函数（有 2 个小问题阻止它在函数内部工作，请参见带有 ????? 的行，但在函数外部，此代码有效）。

relative_hist_count <- function(df, myfactors) {
  require(ggplot2)
  require(plyr)
  require(reshape2)

  # ddply it to get one column for each bin of the histogram
  myhistRTcounts <- ddply(df, c(myfactors,"correct"), histRTcounts)

  # transform it in long format
  myhistRTcounts.long = melt(myhistRTcounts, id.vars =c(myfactors,"correct"), variable.name="bin", value.name = 'mycount')

  # rename the bin names with the ms value they correspond to
  levels(myhistRTcounts.long$bin) <- seq(5, 800, by=10)[-1]-5

  # make them numeric and not a factor anymore
  myhistRTcounts.long$bin = as.numeric(levels(myhistRTcounts.long$bin))[myhistRTcounts.long$bin]

  # cast to have count_correct and count_incorrect as columns
  # ??????????????????????? problem when putting that into a function
  # Here I was not able to figure out how to combine myfactors to the other variables in the call
  myhistRTcount.short = dcast(myhistRTcounts.long, subjectname + Visibility + bin ~ correct)
  names(myhistRTcount.short)[4:5] <- c("countinc","countcor")

  # compute relative counts
  myhistRTcounts.rel <- ddply(myhistRTcount.short, myfactors, transform, 
                          incorrect = countinc / sum(countinc+countcor),
                          correct = countcor / sum(countinc+countcor)
  )
  myhistRTcounts.rel = subset(myhistRTcounts.rel,select=c(-countinc,-countcor))

  myhistRTcounts.rel.long = melt(myhistRTcounts.rel, id.vars = c(myfactors,"bin"), variable.name = 'correct', value.name = 'mycount')

  # ??????????????????????? idem here, problem when putting that into a function to call myfactors
  ggplot(data=myhistRTcounts.rel.long, aes(x=bin, y=mycount, color=factor(correct))) + geom_line() + facet_grid(Visibility ~ subjectname) + xlim(0, 600) + theme_bw()

  return(myhistRTcounts.rel.long)

将其应用于数据的调用

new.df = relative_hist_count(my.data, myfactors = c("subjectname","Visibility"))

所以首先，我需要你的帮助才能使它作为一个函数工作，并有可能在 dcast() 和 ggplot() 中使用 myfactors 变量。

但更重要的是，我几乎可以肯定这个函数可以用更少的步骤以更优雅、最直接的方式编写。

预先感谢您的帮助！

4

2 回答 2

0

也许这有助于设置数据？

countfun <- function(x,...) {
  res <- hist(x,plot=FALSE,...)
  data.frame(counts=res$counts,
             break1=res$breaks[-length(res$breaks)],
             break2=res$breaks[-1])
}

library(plyr)
plot.dat <- ddply(my.data,.(Visibility),function(df){
  res <- ddply(df,.(correct),function(df2) {countfun(df2$RT,breaks=seq(100, 600, by=10))})
  res$freq2 <- res$counts/nrow(df)
  res
})

您可能需要将整个parse, eval,as.formula东西概括为任意因素。我现在没有时间。

但是，如果您打算进行概括，最好修改hist函数以接受一个参数以用作计数的因素。

于 2013-01-31T10:54:42.113 回答
0

感谢 Roland，我没想过要写一个自制的 hist 函数。请在下面找到它：

RelativeHistRT <- function (df, breaks = seq(5,800,10)) 
{
  distrib.correct   = hist(df$RT[df$correct==1], breaks, right=FALSE, plot=FALSE)
  distrib.incorrect = hist(df$RT[df$correct==0], breaks, right=FALSE, plot=FALSE)

  n.total = sum(distrib.correct$counts) + sum(distrib.incorrect$counts)

  data.frame(bin_mids  = distrib.correct$mids,
         correct   = distrib.correct$counts / n.total,
         incorrect = distrib.incorrect$counts / n.total)
}

并将其应用于我的原始 data.frame 并获得我正在寻找的内容：

myhistRTcounts <- ddply(my.data, .(subjectname,Visibility), RelativeHistRT)

这确实要短得多，并且完全符合我的要求。

于 2013-02-01T08:25:11.787 回答