Как я могу сгруппировать по двум полям и посчитать каждый тип данных по дате появления?

Question

У меня есть набор данных (первые 100 строк):

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")

Мне нужно, чтобы это выглядело так (сгруппировано по отделу (строки) и соответствующему количеству каждого типа в день (столбцы)):

В настоящее время у меня есть два подхода к этому, ни один из которых не дает желаемого результата, но я подозреваю, что я близок, потому что решение, кажется, лежит где-то между ними.

Первый подход:

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)
  )

Что дает мне это:

# 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

Второй подход:

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)
  )

Что дает мне это:

# 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

Как я могу это сделать?

Answer 1

Использование 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)

Дайте мне знать, если вы хотите, чтобы все это объяснили. По сути, достаточно многократного использования распределения и плавления, чтобы гарантировать, что данные структурированы так, как мы хотим на каждом этапе (я бы посоветовал подойти к каждому плавлению / распространению и отметить, как формируется data.frame.)

Answer 2

Вместо группировки вам нужно изменить форму данных из длинного в широкий формат. Это называется «кастинг».

library(reshape2)
dcast(dept, department + type ~ date, fun.aggregate = sum)

Что дает:

         department      type 2018-12-23 2018-12-24 2018-12-25 2018-12-26 2018-12-27 2018-12-28
1            Beauty Completed          0         12          0          0         17          6
2            Beauty    Missed          0          0          0          0          0          3
3  Boutique advisor Completed          0          0          0          0          1          4
4 Boutique advisors Completed          1          4          0          0          1          7
5  Boutique Stylist Completed         13          5          0          3          5          0
6  Boutique Stylist    Missed          2          6          0          6          5          8

Ваше изображение также показывает% строку. Вам это нужно?

Редактировать: чтобы добавить строки процентов, рассчитайте их перед изменением формы:

dept %>%
# create the percentage rows by grouping by department/date/type. Later we will combine these rows back with the original data
group_by(department, date, type) %>%
# add a column n with the sum of count in each group
summarise(n=sum(count)) %>%
# do 2 separate things:
#   - add a percent column
#   - change all the values in the type column to have a % at the end so they don't get mixed up with the original values later
mutate(percent = n * 100 / sum(n), type = paste(type, "%")) %>%
# remove all rows except the percent ones
filter(type == "Missed %") %>%
# remove the temporary 'n' column we created earlier, and rename the percent column to 'count' so it can go through the 'dcast' function later without any problems
select(department, type, count = percent, "date") %>%
# append with the original data
bind_rows(dept) %>%
# cast the data with the date column used as columns
# and fill it with the sum of the 'count' column
# the percentage rows we created earlier will pass through the function unharmed as there is only one of them in each department/type/date
dcast(department + type ~ date, fun.aggregate = sum, value.var = "count")

Что дает:

         department      type 2018-12-23 2018-12-24 2018-12-25 2018-12-26 2018-12-27 2018-12-28
1            Beauty Completed          0         12          0          0         17    6.00000
2            Beauty    Missed          0          0          0          0          0    3.00000
3            Beauty  Missed %          0          0          0          0          0   33.33333
4  Boutique advisor Completed          0          0          0          0          1    4.00000
5 Boutique advisors Completed          1          4          0          0          1    7.00000
6  Boutique Stylist Completed         13          5          0          3          5    0.00000