Question

如何使用dplyr根据多列的递减值为每一列创建等级？以下代码似乎没有考虑desc

中的arrange_at参数

rank_f <- function(ds, cols, fs){
          ds %>%arrange_at(desc(vars(!!!cols)))%>%
            mutate_at(vars(!!!cols), funs(!!!fs))
        }


USArrests %>%tibble::rownames_to_column()%>%
     rank_f(quos((Murder:Rape)),quos(min_rank))->ranked


head(USArrests) 
#              Murder Assault UrbanPop Rape
# Alabama      13.2     236       58 21.2
# Alaska       10.0     263       48 44.5
# Arizona       8.1     294       80 31.0
# Arkansas      8.8     190       50 19.5
# California    9.0     276       91 40.6
# Colorado      7.9     204       78 38.7

head(ranked)
#        rowname Murder Assault UrbanPop Rape
# 1    Alabama     44      35       16   29
# 2     Alaska     35      43        7   49
# 3    Arizona     29      47       39   43
# 4   Arkansas     31      31        9   24
# 5 California     32      44       50   48
# 6   Colorado     28      33       38   47

费率越高，排名就应该越低。

Answer 1

我们需要将其放置在funs

内

out1 <- USArrests %>%
           tibble::rownames_to_column() %>% 
           arrange_at(vars(Murder:Rape), funs(desc))

在每列上应用desc进行检查

out2 <-  USArrests %>% 
             tibble::rownames_to_column() %>% 
             arrange(desc(Murder), desc(Assault), desc(UrbanPop), desc(Rape))
identical(out1, out2)
#[1] TRUE

基于上述内容，我们可以在rank_f

中进行更改

out3 <-  out2 %>%
             mutate_at(vars(Murder:Rape), min_rank) 
rank_f <- function(ds, cols, fs){
          ds %>%
               arrange_at(vars(!!!cols), funs(desc))%>%
               mutate_at(vars(!!!cols), funs(!!!fs))
        }
out4 <- USArrests %>%
             tibble::rownames_to_column()%>%
             rank_f(quos((Murder:Rape)),quos(min_rank))

identical(out3, out4)
#[1] TRUE

更新

根据OP的评论，我们不需要做任何arrange，我们可以通过将列值转换为负数来直接应用min_rank

USArrests %>% 
   tibble::rownames_to_column() %>% 
   mutate_at(vars(Murder:Rape), funs(min_rank(-.)))

使用dplyr为一组列创建降序排列

1 个答案:

更新