基于R实现电影节评分多列排序及跨年份数据适配问询

背景与需求 #

我负责管理每年举办“最佳影片”评选的电影节，每年参赛影片最多50部，评委数量不固定。计划使用R语言分析各位评委的评分，生成Markdown报告分发评委。

示例原始数据 #

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)

核心转换要求 #

需将Mean（平均分）和TrimmedMean（截尾平均分）视为额外“评委”，针对每位评委（含这两列）按评分从高到低排序，仅保留影片名称，生成如下格式的排序后数据框：

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)

数据集与扩展性需求 #

主数据集包含过去5年的数据，可通过Year字段子集化。希望无需修改代码，仅指定年份即可分析对应年份的数据，代码需适配不同年份的影片数量、评委数量差异。

现有资源 #

我已准备好分析用的R代码和CSV文件，为保护隐私，评委姓名已改为Judge 1、Judge 2等。

内容的提问来源于stack exchange，提问作者Bill_Lussier