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ddply函数忽略迭代器

plyr iterator scope r

ch-pub · 技术社区 · 8 年前

iris$Petal.Length iris$Species

分位数的数量和值是动态的,所以最终我会尝试通过概率循环或以某种方式将其矢量化。这是我的矢量化尝试,但不太奏效。

rm(list = ls())

require(plyr)

myDat <- iris

myProbs <- c(0, 0.15, 0.5, 1)

# This doesn't return the DF I'm looking for (where probabilities/names are identified)
petals_by_species <- ddply(myDat, "Species", summarize, Quantiles = quantile(Petal.Length, probs = myProbs))
petals_by_species

我尝试了一些黑客的工作,将结果合并成一些广泛的格式,如:

rm(list = ls())

require(plyr)

myDat <- iris

myProbs <- c(0, 0.15, 0.5, 1)

# So, I loop through the probabilities and combine.
for(i in 1:length(myProbs)){

  temp <- ddply(myDat, "Species", summarize, Quantiles = quantile(Petal.Length, probs = myProbs[i]))

  names(temp) <- c("Species", paste0("Prob ", myProbs[i]))

  if(i == 1){
    petals_by_species <- temp
  } else {
    petals_by_species <- merge(petals_by_species, temp)
  }
}

petals_by_species

此输出完全令人困惑……列名正确,但值不正确(每列都重复出现)。

以上列均未返回正确的值。

很明显,我并没有用正确的方式来做这件事。但现在我的好奇心被激发了,

require(plyr)

myDat <- iris

myProbs <- c(0, 0.15, 0.5, 1)

intendedOutput <- ddply(myDat, "Species", summarize, Quantiles = quantile(Petal.Length, probs = myProbs[1]))
intendedOutput

i = 1
unintendedOutput <- ddply(myDat, "Species", summarize, Quantiles = quantile(Petal.Length, probs = myProbs[i]))
unintendedOutput

我怎样才能得到 ddply 以我期望的方式识别迭代器?有不同的吗 plyr daply 没有成功。

谢谢

1 回复 | 直到 8 年前

ch-pub 8 年前

这是一张与哈德利的单独信件中的票:

rm(list = ls())

require(plyr)

myDat <- iris

myProbs <- c(0, 0.15, 0.5, 1)

# This doesn't return the DF I'm looking for (where probabilities/names are identified)
petals_by_species <- ddply(myDat, "Species", summarize, Quantiles = quantile(Petal.Length, probs = myProbs), probs = myProbs)
petals_by_species

然后我的输出是长格式的,并报告输入,如下所示: