根据是否满足 tidyverse 的一系列条件对类别进行分组答案

【问题标题】：group categories according to whether they meet a sequence of conditions with tidyverse根据是否满足 tidyverse 的一系列条件对类别进行分组
【发布时间】：2020-06-11 04:30:07
【问题描述】：

我有一个关于如何根据变量是否满足特定条件重新分类的问题。也就是说，如果该类别不符合条件，则将其分配给另一个符合条件的类别。

我的数据格式如下：

data = data.frame(firm_size = c("Micro", "Small", "Medium","Big"), 
                  employees = c(5,10,100,1000))

> data 
     firm_size employees
1     Micro         5
2     Small        10
3    Medium       100
4       Big      1000

所以，如果我的条件是我必须将员工人数少于 10 人的公司分组，然后将它们与其他符合条件的类别合并

> new_data 
     firm_size    employees
1    Micro-Small     15
3    Medium          100
4    Big             1000

我要做的是编写一个函数来概括这个过程，例如，如果我的数据是这样的，它也可以工作

> data 
     firm_size employees
1     Micro       5
2     Small       8
3    Medium       9
4       Big      1000


> new_data 
     firm_size            employees
1    Micro-Small-Medium   22
4    Big                  1000

我认为这可以通过 tidyverse 的工具来完成。

提前致谢

【问题讨论】：

标签： r dplyr tidyverse purrr

【解决方案1】：

这是tally 的一种方法：

library(dplyr)
size <- 10

data %>% 
  arrange(firm_size,desc(employees)) %>%
  group_by(firm_size = c(as.character(firm_size[employees > size]),
                         rep(paste(firm_size[employees <= size], collapse = "-"),
                             sum(employees <= size)))) %>%
  tally(employees, name = "employees")
## A tibble: 3 x 2
#  firm_size   employees
#  <chr>           <dbl>
#1 Big              1000
#2 Medium            100
#3 Small-Micro        15

对于您的第二组数据：

data2 %>% 
  arrange(firm_size,desc(employees)) %>%
  group_by(firm_size = c(as.character(firm_size[employees > size]),
                         rep(paste(firm_size[employees <= size], collapse = "-"),
                             sum(employees <= size)))) %>%
  tally(employees, name = "employees")
## A tibble: 2 x 2
#  firm_size          employees
#  <chr>                  <int>
#1 Big                     1000
#2 Medium-Small-Micro        22

数据

data <- structure(list(firm_size = structure(c(3L, 4L, 2L, 1L), .Label = c("Big", 
"Medium", "Micro", "Small"), class = "factor"), employees = c(5, 
10, 100, 1000)), class = "data.frame", row.names = c(NA, -4L))


data2 <- structure(list(firm_size = structure(c(3L, 4L, 2L, 1L), .Label = c("Big", 
"Medium", "Micro", "Small"), class = "factor"), employees = c(5L, 
8L, 9L, 1000L)), class = "data.frame", row.names = c("1", "2", 
"3", "4"))

【讨论】：

【解决方案2】：

你可以使用很棒的 forcats 包

library(tidyverse)

data <- data.frame(
  firm_size = c("Micro", "Small", "Medium", "Big", "Small"),
  employees = c(5, 10, 100, 1000, 10)
)


# If you need n groups


data %>%
  mutate(firm_size2 = firm_size %>% as_factor() %>% fct_lump(n = 2, w = employees)) %>% 
  group_by(firm_size2) %>% 
  summarise(sum_emp = sum(employees),.groups = "drop")
#> # A tibble: 3 x 2
#>   firm_size2 sum_emp
#>   <fct>        <dbl>
#> 1 Medium         100
#> 2 Big           1000
#> 3 Other           25


# If you need at least x on the sum of a vector

data %>%
  mutate(firm_size2 = firm_size %>% as_factor() %>% fct_lump_min(min = 10, w = employees)) %>% 
  group_by(firm_size2) %>% 
  summarise(sum_emp = sum(employees),.groups = "drop")
#> # A tibble: 4 x 2
#>   firm_size2 sum_emp
#>   <fct>        <dbl>
#> 1 Small           20
#> 2 Medium         100
#> 3 Big           1000
#> 4 Other            5

^{由reprex package (v0.3.0) 于 2020-06-11 创建}

【讨论】：

【解决方案3】：

另一种解决方案，设置为自定义函数：

library(tidyverse)

mymerge <- function(dat, min) {
  merged_dat <- dat %>% 
    filter(if_else(employees <= min, TRUE, FALSE)) %>% 
    summarize(firm_size = str_flatten(firm_size, collapse = " - "),
              employees = sum(employees))

  dat %>% 
    filter(if_else(employees <= min, FALSE, TRUE)) %>% 
    bind_rows(merged_dat)
}

mymerge(data, 30)
      firm_size employees
1        Medium       100
2           Big      1000
3 Micro - Small        15

mymerge(data, 300)
               firm_size employees
1                    Big      1000
2 Micro - Small - Medium       115

【讨论】：