Question

我非常接近在一组面板数据框中进行排名练习的解决方案（了解围绕此问题的类似问题，例如：Specific group rankings in R）。

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                             3
Orange      FS00008VLD  0.27                             4
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                            2
Banana      FS000092GP  96.25                            3
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

代码：

for (i in 4:ncol(MRAR)){
  eq_ranks[i] <- lapply(unique(MRAR$Morningstar.Category),function(x)
    {
     a <- rank(MRAR[MRAR$Morningstar.Category == x, i], na.last="keep")
     return(a)
  })
}

.error：

Error in `[<-.data.frame`(`*tmp*`, i, value = list(c(NA, 1047, NA, NA,  : 
replacement element 1 has 3159 rows, need 3530

我也看过ave方法，但ave语法似乎不允许na.last =＆＃34; keep＆＃34;需求。但我也开发了一种dplyr方法：

aux <- as.vector(cbind(names(ER)))

eq_ranks <- function(MRAR,group_by){
  group_by %>%
    group_by(!!Morningstar.Category) %>% 
mutate_at(MRAR,quo(eq_rank=rank(MRAR)), vars(aux))
}

再次，这将NA值视为＆＃34; last＆＃34; （而不是＆＃34;保持＆＃34;），语义逻辑似乎是正确的。

真的很感谢你的帮助， Wilhelm Fantastisch。

Answer 1

不确定这是否正是您想要的，但下面有data.frame这对我有用。希望它有所帮助

df <- data.frame(Category=c(rep("Orange",10), rep("Banana",10)),
             Score.08.2007=c(runif(6),rep(NA,4),runif(4),rep(NA,2),runif(4)),
             Score.09.2017=c(runif(5),rep(NA,3),runif(2),runif(4),rep(NA,4),runif(2)),
             stringsAsFactors=F)

dplyr解决方案

library(dplyr)

eq_ranks <- function(theseCols, newCols, df){
               theseCols <- enquo(theseCols)
               df1 <- df %>%
                       group_by(Category) %>%
                       mutate_at(vars(!!theseCols), funs(rank(., na.last="keep"))) %>%
                       ungroup() %>%
                       select(-Category) %>%
                       setNames(newCols)
               df2 <- cbind(df, df1)
               return(df2)
            }

aux <- colnames(df)[-1]
newCols <- sub("Score", "Rank", aux)
eq_ranks(aux,newCols,df)

输出

structure(list(Category = c("Orange", "Orange", "Orange", "Orange", 
"Orange", "Orange", "Orange", "Orange", "Orange", "Orange", "Banana", 
"Banana", "Banana", "Banana", "Banana", "Banana", "Banana", "Banana", 
"Banana", "Banana"), Score.08.2007 = c(0.757087148027495, 0.202692255144939, 

0.711121222469956, 0.121691921027377, 0.245488513959572, 0.14330437942408, 
NA, NA, NA, NA, 0.239629415096715, 0.0589343772735447, 0.642288258532062, 
0.876269212691113, NA, NA, 0.778914677444845, 0.79730882588774, 
0.455274453619495, 0.410084082046524), Score.09.2017 = c(0.810870242770761, 
0.604933290276676, 0.654723928077146, 0.353197271935642, 0.270260145887733, 
NA, NA, NA, 0.99268406117335, 0.633493264438584, 0.213208135217428, 
0.129372348077595, 0.478118034312502, 0.924074469832703, NA, 
NA, NA, NA, 0.59876096714288, 0.976170694921166), Rank.08.2007 = c(6, 
3, 5, 1, 4, 2, NA, NA, NA, NA, 2, 1, 5, 8, NA, NA, 6, 7, 4, 3
), Rank.09.2017 = c(6, 3, 5, 2, 1, NA, NA, NA, 7, 4, 2, 1, 3, 
5, NA, NA, NA, NA, 4, 6)), .Names = c("Category", "Score.08.2007", 
"Score.09.2017", "Rank.08.2007", "Rank.09.2017"), row.names = c(NA, 
-20L), class = "data.frame")

在R中for循环中的组内排名

1 个答案:

dplyr解决方案

输出