R：基于dyplr中的多个条件进行汇总

Question

我正在尝试总结一个数据框以创建两个总结：

1. 统计订单数量只出现QUOT或QUOG
2. 计算QUOT或QUOG出现的订单数量以及出现其他Holds

下面是代码的开头：

library(dplyr)


dat <- data.frame(Order = c(123,123,123,145,145,189,210,210,123,123,164), 
                  Location = c("Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Charlotte","Charlotte","Charlotte"),
                  Hold = c("QUOT","ENGR","VEND","QUOG","ENGR","QUOT","ENGR","VEND","QUOT","CUST","QUOT")
)


test <- dat %>%
  group_by(Order, Location) %>%

  .....

我一直在试图找出特定订单是否只有QUOT或QUOG ，然后它是否有QUOT或QUOG以及其他。

预期输出：

   Location Only Multiple
1   Chicago    1        2
2 Charlotte    1        1

所以对于预期的输出：

• 订单 123，芝加哥：其中包含QUOT和另一个保留（ ENGR和VEND ），因此这将被视为芝加哥的倍数
• 芝加哥 145 号订单：其中包含QUOG和另一个持有 ( ENGR )，因此这将被视为芝加哥的倍数
• 芝加哥 189 号订单：其中有QUOT且没有其他保留，因此这将被视为仅适用于芝加哥
• 订购210，芝加哥：既没有QUOT或QUOG所以这个顺序被排除在计数
• 订单 123，夏洛特：其中包含QUOT和另一个保留 ( CUST )，因此这将被视为夏洛特的倍数
• 夏洛特 164 号订单：其中有QUOT且没有其他保留，因此这将被视为仅适用于夏洛特

Answer 1

我认为这应该有效——你可能想用其他一些订单来测试这个：

library(dplyr)
library(tidyr)

dat <- data.frame(
  Order = c(123,123,123,145,145,189,210,210,123,123,164), 
  Location = c("Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Charlotte","Charlotte","Charlotte"),
  Hold = c("QUOT","ENGR","VEND","QUOG","ENGR","QUOT","ENGR","VEND","QUOT","CUST","QUOT")
)

dat %>% 
    group_by(Order, Location) %>% 
    mutate(
        quot_or_quog = Hold %in% c("QUOT", "QUOG"),
        distinct_quot_or_quog = n_distinct(quot_or_quog)
    ) %>% 
    # Remove those that do not have "QUOT" or "QUOG"
    filter(quot_or_quog) %>% 
    mutate(
        label = if_else(distinct_quot_or_quog == 1, "Only", "Multiple")
    ) %>% 
    group_by(label, add = TRUE) %>%
    summarise(num_label = n_distinct(label)) %>% 
    group_by(Location, label) %>%
    count(num_label) %>% 
    pivot_wider(
        names_from = label,
        values_from = n
    ) %>% 
    select(-num_label)
#> # A tibble: 2 x 3
#> # Groups:   Location [2]
#>   Location  Multiple  Only
#>   <fct>        <int> <int>
#> 1 Charlotte        1     1
#> 2 Chicago          2     1

^{由reprex 包(v0.3.0) 于 2020 年 2 月 24 日创建}

Answer 2

这是使用dplyr和tidyr另一个解决方案。 这次首先进行旋转，然后进行过滤和汇总以得出您的解决方案。

library(dplyr)
library(tidyr)

dat.summary <- dat %>%
  mutate(hold_count = 1) %>% 
  pivot_wider(names_from = Hold, values_from = hold_count) %>% 
  mutate(only = if_else((QUOT == 1 | QUOG == 1) & is.na(ENGR) & is.na(VEND) & is.na(CUST), 1, 0),
         multiple = if_else((QUOT == 1 | QUOG == 1) & (ENGR == 1 | VEND == 1 | CUST ==1), 1, 0)) %>% 
  group_by(Location) %>% 
  summarise(only = sum(only, na.rm = T), multiple = sum(multiple, na.rm = T))

dat.summary

给你：

# A tibble: 2 x 3
  Location   only multiple
  <fct>     <dbl>    <dbl>
1 Charlotte     1        1
2 Chicago       1        2

数据

dat <- data.frame(
  Order = c(123,123,123,145,145,189,210,210,123,123,164), 
  Location = c("Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Chicago","Charlotte","Charlotte","Charlotte"),
  Hold = c("QUOT","ENGR","VEND","QUOG","ENGR","QUOT","ENGR","VEND","QUOT","CUST","QUOT")
)