如何操作（聚合）R中的数据？

Question

我有一个如下所示的数据集：

df <- tribble(
  ~id,  ~price, ~number_of_book,        
  "1",    10,         3,        
  "1",     5,         1,         
  "2",     7,         4,
  "2",     6,         2, 
  "2",     3,         4,
  "3",     4,         1,
  "4",     5,         1,
  "4",     6,         1,
  "5",     1,         2,
  "5",     9,         3,
)

正如您在数据集中看到的，如果 id 为“1”，则有 3 本书每本书的价格为 10 美元，而 1 本书的价格为 5 美元。基本上，我想查看每个价格箱的图书数量的份额（％）。这是我想要的数据集：

df <- tribble(
  ~id,    ~less_than_three,   ~three-five,  ~five-six, ~more_than_six,     
  "1",          "0%",              "25%",     "0%",         "75%",
  "2",          "0%",              "40%",     "20%",        "40%",
  "3",          "0%",              "100%",    "0%",         "0%",  
  "4",          "0%",              "50%",     "50%",        "0%",
  "5",          "40%",             "0%",      "0%",         "60%",
)

现在，我首先对价格进行聚类。为此，我运行以下代码：

out <- cut(df$price, breaks = c(0, 3, 5, 6, 10),
           labels = c("<3","3-5","5-6", ">6")) 

out = table(out) / sum(table(out)) 

但不幸的是，由于缺乏编码知识，我无法更进一步。你能帮我得到想要的数据吗？

Answer 1

我们可以使用cut获取区间，然后使用tidyr将数据转换为宽格式，最后使用janitor添加百分比。

library(dplyr)
library(tidyr)
library(janitor)

df %>% 
  mutate(interval = cut(price, c(0,3,5,6,Inf))) %>% 
  select(-price) %>% 
  pivot_wider(names_from = interval, values_from = number_of_book) %>% 
  adorn_percentages()

#>  id (6,Inf] (3,5] (5,6] (0,3]
#>   1    0.75  0.25    NA    NA
#>   2    0.40    NA   0.2   0.4
#>   3      NA  1.00    NA    NA
#>   4      NA  0.50   0.5    NA
#>   5    0.60    NA    NA   0.4

Answer 2

使用 dplyr，您可以添加一个列 cols，该列将用于列名。然后，您可以将每个 id 中每个 col 的书籍数量相加。接下来，您可以通过将这些数字除以该 id 的总和来计算百分比，然后应用scales::percent 将格式设置为百分比而不是小数。现在您只需要 pivot_wider 提供从中获取名称和值的变量，并对列重新排序以匹配原始标签顺序。 （这比其他答案涉及更多一点，因为它说明了给定 (id, cols/interval) 对有 >1 行的情况，并且看门人简化了事情）

labels = c("less_than_three","three_to_five","five_to_six", "more_than_six")

df %>% 
  group_by(id, cols = cut(price, breaks = c(0, 3, 5, 6, 10), labels = labels)) %>% 
  summarise(n = sum(number_of_book)) %>% 
  group_by(id) %>% 
  mutate(pct = scales::percent(n/sum(n), 1)) %>% 
  pivot_wider(id_cols = id, names_from = cols, values_from = pct) %>% 
  select_at(c('id', labels)) %>% 
  ungroup

# # A tibble: 5 x 5
#   id    less_than_three three_to_five five_to_six more_than_six
#   <chr> <chr>           <chr>         <chr>       <chr>        
# 1 1     NA              25%           NA          75%          
# 2 2     40%             NA            20%         40%          
# 3 3     NA              100%          NA          NA           
# 4 4     NA              50%           50%         NA           
# 5 5     40%             NA            NA          60%       

如果您想用 0% 替换 NA（我认为这在这种情况下是有意义的，并且与问题中显示的输出相匹配），您可以使用下面评论中提到的方法。

df %>% 
  group_by(id, cols = cut(price, breaks = c(0, 3, 5, 6, 10), labels = labels)) %>% 
  summarise(n = sum(number_of_book)) %>% 
  group_by(id) %>% 
  mutate(pct = scales::percent(n/sum(n), 1)) %>% 
  pivot_wider(id_cols = id, names_from = cols, values_from = pct,
              values_fill = list(pct = '0%')) %>% 
  select_at(c('id', labels)) %>% 
  ungroup

# # A tibble: 5 x 5
#   id    less_than_three three_to_five five_to_six more_than_six
#   <chr> <chr>           <chr>         <chr>       <chr>        
# 1 1     0%              57%           0%          43%          
# 2 2     40%             0%            20%         40%          
# 3 3     0%              100%          0%          0%           
# 4 4     0%              50%           50%         0%           
# 5 5     40%             0%            0%          60%