Question

我有数据框“类别”，“ID”，“得分（t）”，我想获得“Rank（t）”：

Category    ID          Score.08.2007   Score.09.2007    Rank.08.2007    Rank.09.2007   ...
Orange      FSGBR070N3  0.16            ...              5               ...
Orange      FSGBR070N3  0.05            ...              7               ...
Orange      FSGBR070N3  0.11                             6
Orange      FS00008L4G  0.28                             1
Orange      FS00008VLD  0.27                             2
Orange      FS00008VLD  0.27                             2
Orange      FS00008VLD  0.27                             2
Orange      FS00009SQX  -2.03                            8
Orange      FS00009SQX  NA                          
Orange      FSUSA0A1KW  NA          
Orange      FSUSA0A1KW  NA  
Orange      FSUSA0A1KX  NA  
Orange      FSUSA0A1KY  NA  
Orange      FS0000B389  NA  
Banana      FS000092GP  96.25                            1
Banana      FS000092GP  96.25                            1
Banana      FS000092GP  96.25                            1
Banana      FS000092GP  52.33                            4
Banana      FS0000ATLN  31.73                            5
Banana      FSUSA0AVMF  1.38                             7
Banana      FSGBR058O8  1.37                             8
Banana      FSGBR05845  2.24                             6

排名基于每个“类别”中“得分”的降序排序。我努力捕捉的附加规范是，当存在相同的得分和相同的ID时，对于具有不同值的以下得分，指定等级来自先前ID的等级加上共享相同ID的ID的等级。得分（示例中的排名输出列应该明确这一点。）

NA不应该获得排名：

na.last = NA

我已经开始为排名创建一个矩阵，然后我可能需要sort（），但是我很难在时间序列和附加规范中捕获这个...无法找到这样具体的现有问题。帮助赞赏！

time_series <- c("08.2007","09.2007","10.2007",...)
abs_ranks_mat <- as.data.frame(mat.or.vec(nrow(ID),length(time_series)))

Answer 1

解决方案使用dplyr。 df是来自@ trosendal的例子。 df3是最终输出。

关键是使用min_rank函数来创建排名。 mutate_at允许我们指定我们做或不想进行排名的列。之后，我们可以更改列名并与原始数据框合并。

library(dplyr)

df <- df %>% mutate(RowID = 1:n())

df2 <- df %>%
  group_by(Category) %>%
  mutate_at(vars(-ID, -RowID), funs(min_rank(desc(.)))) %>%
  ungroup() %>%
  select(-Category, -ID) %>%
  setNames(., gsub("Score", "Rank", colnames(.)))

df3 <- df %>% 
  left_join(df2, by = "RowID") %>%
  select(-RowID)

Answer 2

您的数据：

df <- structure(list(Category = c("Orange", "Orange", "Orange", "Orange", 
"Orange", "Orange", "Orange", "Orange", "Orange", "Orange", "Orange", 
"Orange", "Orange", "Orange", "Banana", "Banana", "Banana", "Banana", 
"Banana", "Banana", "Banana", "Banana"), ID = c("FSGBR070N3", 
"FSGBR070N3", "FSGBR070N3", "FS00008L4G", "FS00008VLD", "FS00008VLD", 
"FS00008VLD", "FS00009SQX", "FS00009SQX", "FSUSA0A1KW", "FSUSA0A1KW", 
"FSUSA0A1KX", "FSUSA0A1KY", "FS0000B389", "FS000092GP", "FS000092GP", 
"FS000092GP", "FS000092GP", "FS0000ATLN", "FSUSA0AVMF", "FSGBR058O8", 
"FSGBR05845"), Score.08.2007 = c(0.16, 0.05, 0.11, 0.28, 0.27, 
0.27, 0.27, -2.03, NA, NA, NA, NA, NA, NA, 96.25, 96.25, 96.25, 
52.33, 31.73, 1.38, 1.37, 2.24), Score.09.2007 = c(0.16, 0.05, 
0.14, 0.22, 0.23, 0.27, 0.27, -2.03, NA, NA, 0.14, NA, 0.56, 
NA, 96.25, 93.25, 96.25, 51.33, 31.73, 1.38, 1.37, 2.24)), .Names = c("Category", 
"ID", "Score.08.2007", "Score.09.2007"), row.names = c(NA, -22L
), class = "data.frame")

循环分数并在每个类别中生成排名：

for(i in names(df)[grep("Score", names(df))]) {
    df[,paste0("rank", i)] <- do.call("c", lapply(unique(df$Category), function(x){
        a <- floor(rank(df[df$Category == x, i], na.last = TRUE))
        a[is.na(df[df$Category == x, i])] <- NA
        a <- max(a, na.rm = TRUE) - a + 1
        return(a)
    }))
}

df

R

2 个答案: