r - 如何按两个字段分组并按日期计算每种类型的数据？

Question

我有一个数据集（前 100 行）：

structure(list(department = structure(c(21L, 14L, 4L, 11L, 21L, 
12L, 15L, 11L, 3L, 18L, 4L, 20L, 25L, 3L, 3L, 13L, 19L, 22L, 
18L, 16L, 16L, 16L, 16L, 4L, 20L, 12L, 4L, 27L, 1L, 6L, 16L, 
1L, 13L, 13L, 25L, 18L, 8L, 23L, 10L, 16L, 4L, 21L, 2L, 5L, 18L, 
10L, 23L, 4L, 7L, 5L, 14L, 15L, 19L, 23L, 11L, 4L, 15L, 6L, 12L, 
11L, 23L, 14L, 15L, 11L, 18L, 24L, 27L, 27L, 20L, 5L, 1L, 19L, 
4L, 10L, 4L, 26L, 3L, 14L, 15L, 12L, 22L, 14L, 20L, 25L, 2L, 
23L, 15L, 13L, 4L, 18L, 26L, 13L, 5L, 10L, 1L, 6L, 10L, 22L, 
5L, 14L), .Label = c("Beauty", "Boutique advisor", "Boutique advisors", 
"Boutique Stylist", "Clean Beauty Expert", "Conseiller en boutique", 
"Design Consultant", "Designer Trade Specialist", "Food", "Furniture", 
"In-store Design Expert", "In-store experts", "In-Store Sales Professional", 
"In-Store Style Experts", "John Hardy", "Jos. A. Bank LIVE!", 
"Levi's Stylists", "Lighting & Home Accessories", "Men's Wearhouse LIVE!", 
"Menswear", "Personal advisors", "Styliste en boutique", "Vendeurs", 
"Wine", "Women's Accessories", "Women's shoes", "Womenswear"), class = "factor"), 
    type = c("Completed", "Missed", "Missed", "Missed", "Missed", 
    "Missed", "Missed", "Completed", "Completed", "Missed", "Missed", 
    "Completed", "Completed", "Completed", "Completed", "Completed", 
    "Completed", "Completed", "Completed", "Missed", "Completed", 
    "Missed", "Completed", "Missed", "Missed", "Completed", "Missed", 
    "Missed", "Missed", "Completed", "Missed", "Completed", "Missed", 
    "Completed", "Missed", "Missed", "Completed", "Missed", "Missed", 
    "Completed", "Completed", "Missed", "Completed", "Missed", 
    "Completed", "Missed", "Missed", "Completed", "Missed", "Completed", 
    "Completed", "Missed", "Completed", "Missed", "Completed", 
    "Completed", "Missed", "Missed", "Missed", "Missed", "Completed", 
    "Missed", "Completed", "Completed", "Completed", "Missed", 
    "Missed", "Completed", "Missed", "Completed", "Completed", 
    "Missed", "Completed", "Completed", "Missed", "Missed", "Completed", 
    "Completed", "Completed", "Completed", "Missed", "Completed", 
    "Completed", "Completed", "Completed", "Completed", "Completed", 
    "Completed", "Completed", "Completed", "Completed", "Missed", 
    "Missed", "Completed", "Completed", "Completed", "Missed", 
    "Completed", "Missed", "Completed"), date = structure(c(17889, 
    17890, 17893, 17893, 17892, 17892, 17893, 17893, 17892, 17888, 
    17892, 17889, 17888, 17893, 17888, 17889, 17891, 17892, 17893, 
    17891, 17889, 17888, 17892, 17889, 17889, 17892, 17888, 17889, 
    17893, 17892, 17893, 17892, 17891, 17893, 17888, 17891, 17892, 
    17891, 17892, 17888, 17891, 17893, 17893, 17892, 17890, 17888, 
    17888, 17889, 17891, 17893, 17893, 17890, 17890, 17892, 17889, 
    17892, 17889, 17889, 17888, 17888, 17893, 17893, 17893, 17891, 
    17888, 17892, 17892, 17893, 17891, 17888, 17889, 17891, 17889, 
    17890, 17891, 17888, 17889, 17888, 17890, 17893, 17889, 17889, 
    17893, 17889, 17892, 17891, 17889, 17892, 17888, 17891, 17893, 
    17890, 17890, 17889, 17893, 17889, 17889, 17888, 17889, 17892
    ), class = "Date"), count = c(7L, 9L, 8L, 3L, 5L, 4L, 5L, 
    10L, 1L, 3L, 5L, 18L, 3L, 7L, 1L, 17L, 277L, 10L, 14L, 50L, 
    520L, 92L, 791L, 6L, 7L, 4L, 2L, 1L, 3L, 3L, 145L, 17L, 10L, 
    42L, 1L, 1L, 1L, 2L, 7L, 627L, 3L, 6L, 4L, 3L, 3L, 2L, 1L, 
    2L, 1L, 20L, 41L, 4L, 283L, 1L, 14L, 5L, 2L, 1L, 3L, 3L, 
    7L, 12L, 36L, 9L, 14L, 1L, 6L, 13L, 1L, 14L, 12L, 16L, 3L, 
    2L, 6L, 7L, 4L, 21L, 3L, 5L, 5L, 22L, 12L, 5L, 1L, 5L, 23L, 
    36L, 13L, 12L, 12L, 9L, 4L, 6L, 6L, 4L, 1L, 4L, 1L, 32L)), row.names = c(NA, 
100L), class = "data.frame")

我需要它看起来像这样（按部门（行）和每天每种类型的相应计数（列）分组）：

目前，我有两种方法来解决这个问题，这两种方法都不能产生预期的结果，但我怀疑我很接近，因为解决方案似乎介于两者之间。

第一种方法：

library(dplyr) # For the purpose of this reproducible example should you need it

dept %>%
group_by(
    department
  ) %>% 
  summarise(
    missed = sum(type == "Missed"),
    completed = sum(type == "Completed"),
    missed_pct = missed / (missed + completed)
  )

这给了我这个：

# A tibble: 7 x 4
  department          missed completed missed_pct
  <fct>                <int>     <int>      <dbl>
1 Beauty                   2         5      0.286
2 Food                     0         1      0    
3 Menswear                 4         6      0.4  
4 Wine                     1         1      0.5  
5 Women's Accessories      2         5      0.286
6 Women's shoes            3         5      0.375
7 Womenswear               4         5      0.444

第二种方法：

library(dplyr) # For the purpose of this reproducible example should you need it

dept %>%
  group_by(
    department,
    date
  ) %>% 
  summarise(
    missed = sum(type == "Missed"),
    completed = sum(type == "Completed"),
    missed_pct = missed / (missed + completed)
  )

这给了我这个：

# A tibble: 28 x 5
# Groups:   department [?]
   department date       missed completed missed_pct
   <fct>      <date>      <int>     <int>      <dbl>
 1 Beauty     2018-12-23      0         1        0  
 2 Beauty     2018-12-24      0         1        0  
 3 Beauty     2018-12-26      0         1        0  
 4 Beauty     2018-12-27      1         1        0.5
 5 Beauty     2018-12-28      1         1        0.5
 6 Food       2018-12-27      0         1        0  
 7 Menswear   2018-12-23      1         1        0.5
 8 Menswear   2018-12-24      1         1        0.5
 9 Menswear   2018-12-25      0         1        0  
10 Menswear   2018-12-26      1         1        0.5

我怎样才能做到这一点？

标签： r

Answer 1

使用dplyr：

library(dplyr)
dept %>%
  group_by(department, date, type) %>% 
  summarise(count = sum(count, na.rm = T)) %>%   # data had a few duplicate rows
  spread(type, count, fill = 0) %>% 
  as.data.frame() %>% 
  group_by(department, date) %>% 
  mutate(missed_pct = Missed / (Missed + Completed)) %>%
  melt(id.vars = c("department", "date")) %>% 
  spread(date, value) %>% 
  rename(type = variable)

让我知道您是否希望对此进行解释。基本上只是多次使用 spread 和 melt 来确保数据在每个阶段都按照我们想要的方式构建（我建议运行到每个 melt/spread 并注意 data.frame 的形状。）