根据几列中的值对术语列表进行排名

Question

我管理着一个电影节，每年举办“最佳电影”竞赛。每年最多可以有 50 部电影和任意数量的评委。

我计划使用 R 来分析各个评委的分数，然后用 Markdown 发布一份报告，然后分发给他们。

以下是包含示例原始数据的数据框：

Film = c("Vertigo", "Fargo", "Platoon", "Beetlejuice", "Zodiac")
Mean = c( 7.45,6.98,7.5,7.2,5.82)
TrimmedMean = c(7.42, 7.04,7.36,7.05,6.26)
Judge1 = c(5.56,6.00,7.00,6.22,5.33)
Judge2 = c(6.89,8.81,8.83,7.22,6.78)
Judge3 = c(9.25,6.38,6.71,6.88,6.50)
Judge4 = c(8.00,7.67,7.44,6.89,7.22)
Judge5 = c(5.11,4.89,7.11,8.78,6.44)
Judge6 = c(9.89,8.11,7.89,7.22,2.67)
RawData = data.frame(Film,Mean,TrimmedMean,Judge1,Judge2,Judge3,Judge4,Judge5,Judge6)

出于本练习的目的，“平均”和“修剪平均”列可被视为附加“评判”。

我想根据每位评委的分数对电影进行排名，然后删除数值，留下如下所示的排序电影列表：

Mean = c('Platoon','Vertigo','Beetlejuice','Fargo','Zodiac')
TrimmedMean = c('Vertigo','Platoon','Beetlejuice','Fargo','Zodiac')
Judge1 = c('Platoon','Beetlejuice','Fargo','Vertigo','Zodiac')
Judge2 = c('Platoon','Fargo','Beetlejuice','Vertigo','Zodiac')
Judge3 = c('Vertigo','Beetlejuice','Platoon','Zodiac','Fargo')
Judge4 = c('Vertigo','Fargo','Platoon','Zodiac','Beetlejuice')
Judge5 = c('Beetlejuice','Platoon','Zodiac','Vertigo','Fargo')
Judge6 = c('Vertigo','Fargo','Platoon','Beetlejuice','Zodiac')
SortedData = data.frame(Mean,TrimmedMean,Judge1,Judge2,Judge3,Judge4,Judge5,Judge6)

我生成第一个表的主数据集包含 5 年前的数据，我可以按年份进行子集化。我希望能够分析不同年份的数据，除了指定要子集的年份之外，无需更改每年的代码。该代码应该响应不同年份的数据集，这些数据集将有不同数量的评委和电影。

有什么想法吗？

Answer 1

我认为您正在寻找的功能是

order()

。它显示了向量中值的相对顺序。

order(c(1.65, 6.92, 7.98, 3.42, 5.75), decreasing=TRUE)
# [1] 3 2 5 4 1

我们可以使用这个顺序，为每个法官查找

Film

列中的姓名。

data.frame(lapply(RawData[, -1], function(x) RawData$Film[order(x, decreasing=TRUE)]))
#          Mean TrimmedMean      Judge1      Judge2      Judge3      Judge4      Judge5      Judge6
# 1     Platoon     Vertigo     Platoon     Platoon     Vertigo     Vertigo Beetlejuice     Vertigo
# 2     Vertigo     Platoon Beetlejuice       Fargo Beetlejuice       Fargo     Platoon       Fargo
# 3 Beetlejuice Beetlejuice       Fargo Beetlejuice     Platoon     Platoon      Zodiac     Platoon
# 4       Fargo       Fargo     Vertigo     Vertigo      Zodiac      Zodiac     Vertigo Beetlejuice
# 5      Zodiac      Zodiac      Zodiac      Zodiac       Fargo Beetlejuice       Fargo      Zodiac

相同，但更优雅（我认为）和更快

SortedData <- RawData[, -1]
SortedData[] <- RawData$Film[sapply(RawData[, -1], order, decreasing=TRUE)]

SortedData
#          Mean TrimmedMean      Judge1      Judge2      Judge3      Judge4      Judge5      Judge6
# 1     Platoon     Vertigo     Platoon     Platoon     Vertigo     Vertigo Beetlejuice     Vertigo
# 2     Vertigo     Platoon Beetlejuice       Fargo Beetlejuice       Fargo     Platoon       Fargo
# 3 Beetlejuice Beetlejuice       Fargo Beetlejuice     Platoon     Platoon      Zodiac     Platoon
# 4       Fargo       Fargo     Vertigo     Vertigo      Zodiac      Zodiac     Vertigo Beetlejuice
# 5      Zodiac      Zodiac      Zodiac      Zodiac       Fargo Beetlejuice       Fargo      Zodiac