如何为 R 中的每一行分配数据？

Question

我有一个如下所示的数据集：

df <- tribble(
  ~id,  ~price, ~type, ~number_of_book,        
  "1",    10,     "X",        3,    
  "1",     2,     "X",        1, 
  "1",     5,     "Y",        1,         
  "2",     7,     "X",        4,
  "2",     6,     "X",        1,
  "2",     6,     "Y",        2, 
  "3",     2,     "X",        4,
  "3",     8,     "X",        2,
  "3",     1,     "Y",        4,
  "3",     9,     "Y",        5,
)

现在，我要回答这个问题：对于每个 id 和每个选定的价格组，X 占书籍的百分比，Y 占书籍的百分比？换句话说，每个id和价格组的书籍类型分布是怎样的？

要做到这一点，首先我需要在脑海中将这个数据集可视化：

agg_df <- tribble(
  ~type,     ~id,       ~less_than_two,    ~two-five,  ~five-six, ~more_than_six,     
    "X",      "1",              1,               0,           0,            3,
    "Y",      "1",              0,               1,           0,            0,
    "X",      "2",              0,               0,           1,            4,
    "Y",      "2",              0,               0,           2,            2,
    "X",      "3",              4,               0,           0,            2,
    "Y",      "3",              4,               0,           0,            5,
)

然后，这将是我想要的数据集：

desired_df <- tribble(
  ~type,     ~id,       ~less_than_two,  ~three-five,  ~five-six, ~more_than_six,     
  "X",      "1",            "100%",           "0%",          "0%",       "100%",
  "Y",      "1",              "0%",         "100%",          "0%",         "0%",
  "X",      "2",              "0%",           "0%",       "33.3%",      "66.6%",
  "Y",      "2",              "0%",           "0%",       "66.6%",       "33.3%",
  "X",      "3",             "50%",           "0%",          "0%",      "28.5%",
  "Y",      "3",             "50%",           "0%",          "0%",       "71.4%",
)

这个期望的数据集告诉我，当 id 为“3”并且价格箱超过 6 美元时，有两本 X 类型的书，但有五本 Y 类型的书。所以，这是分布：X(28.5%) 和 Y(71.4%)。

注意：我在这里有一个类似的问题，但现在我无法解决更复杂的操作：How to manipulate (aggregate) the data in R?

如果您能帮助我，我将不胜感激。提前致谢。

Answer 1

我们可以使用findInterval 将price 划分为不同的组，计算number_of_book 的sum 对于每个id、type 和price_group，然后计算每个id 的比率和price_group。最后，我们使用pivot_wider获取更宽格式的数据。

library(dplyr)

df %>% 
  mutate(price_group = c("less_than_two", "three_five", "five_six", "more_than_six")
                      [findInterval(price, c(2, 5, 6), left.open = TRUE) + 1]) %>% 
  group_by(id, type, price_group) %>%
  summarise(number_of_book = sum(number_of_book)) %>% 
  group_by(id, price_group) %>% 
  mutate(n = number_of_book/sum(number_of_book) * 100) %>% 
  select(-number_of_book) %>% 
  tidyr::pivot_wider(names_from = price_group, values_from = n, 
                     values_fill = list(n = 0))

#  id    type  less_than_two more_than_six three_five five_six
#  <chr> <chr>         <dbl>         <dbl>      <dbl>    <dbl>
#1 1     X               100         100            0      0  
#2 1     Y                 0           0          100      0  
#3 2     X                 0         100            0     33.3
#4 2     Y                 0           0            0     66.7
#5 3     X                50          28.6          0      0  
#6 3     Y                50          71.4          0      0  

Answer 2

我们可以在'price'列上创建一个带有cut的bin组，按'id'、'grp'分组，通过将'number_of_book'除以'number_of_book'的sum来创建百分比，然后重新整形为“宽”格式

library(dplyr)
library(tidyr)
df %>% 
  group_by(id,grp = cut(price, breaks = c(-Inf, 2, 5, 6, Inf), 
    c('less_than_two', 'three-five', 'five-six', 'more_than_six')), add = TRUE) %>%
  mutate(Perc = 100 *number_of_book/sum(number_of_book)) %>%
  select(-price, -number_of_book) %>%
  mutate(rn = row_number()) %>%
  pivot_wider(names_from = grp, values_from = Perc, values_fill = list(Perc = 0)) %>%
  select(-rn)
# A tibble: 6 x 6
# Groups:   id [3]
#  id    type  more_than_six less_than_two `three-five` `five-six`
#  <chr> <chr>         <dbl>         <dbl>        <dbl>      <dbl>
#1 1     X             100             100            0        0  
#2 1     Y               0               0          100        0  
#3 2     X             100               0            0       33.3
#4 2     Y               0               0            0       66.7
#5 3     X              28.6            50            0        0  
#6 3     Y              71.4            50            0        0  

Answer 3

也许不是完美的解决方案，但另一种方法是使用case_when 来定义不同的类别：

library(tidyverse)
df %>% group_by(id, type, price) %>% 
  mutate(Less2 = case_when(price <= 2 ~ cumsum(number_of_book)),
                                            Three_Five = case_when(price %in% 3:5 ~ cumsum(number_of_book)),
                                            Five_six = case_when(price %in% 5:6 ~ cumsum(number_of_book)),
                                            More_six = case_when(price >6 ~ cumsum(number_of_book))) %>% 
  replace(is.na(.),0) %>% 
  ungroup(.) %>% 
  group_by(id, type) %>% 
  summarise_at(vars(Less2:More_six), ~sum(.)) %>%
  ungroup(.) %>%
  group_by(id) %>%
  mutate_at(vars(Less2:More_six), ~ replace_na(./sum(.), 0)) 

# A tibble: 6 x 6
# Groups:   id [3]
  id    type  Less2 Three_Five Five_six More_six
  <chr> <chr> <dbl>      <dbl>    <dbl>    <dbl>
1 1     X       100          0      0      100  
2 1     Y         0        100    100        0  
3 2     X         0          0     33.3    100  
4 2     Y         0          0     66.7      0  
5 3     X        50          0      0       28.6
6 3     Y        50          0      0       71.4
```