dplyr中变量之间的条件匹配答案

【问题标题】：conditional matching between variables in dplyrdplyr中变量之间的条件匹配
【发布时间】：2020-12-09 11:49:58
【问题描述】：

我试图在另一列中找到具有某些或所有可能值的列中的观察值。在这个小标题里

parties <- tibble(class = c("R","R","R","R","R","K","K","K","K","K","K",
                "L","L","L","L"),
       name = c("Party1", "Party2","Party3","Party4","Party5",
               "Party2", "Party4", "Party6","Party7","Party8","Party9",
                "Party2","Party3","Party4","Party10"))

我想找到属于“R”、“K”和“L”三个类别的所有“派对”。或者通常属于“X”或“Y”类的各方。我设法找到了一个解决方案，使用group_split(class)，然后从列表中提取每个表，最后执行两个半连接。这适用于我想要所有三个classes 的聚会的情况：

parties_split <- parties %>%
  group_split(class)

parties_K <- parties_split[[1]]
parties_L <- parties_split[[2]]
parties_R <- parties_split[[3]]

semi_join(parties_K,parties_L, by = "name") %>%
  semi_join(parties_R, by = "name") %>%
  select(-class)

name
<chr>
Party2              
Party4

这在这种情况下可行，但效率不高，尤其是当需要匹配的类（或观察）的数量远大于三个时。我特别在 tidyverse 中寻找解决方案。有任何想法吗？谢谢

【问题讨论】：

标签： r dplyr tibble

【解决方案1】：

试试看：

parties %>% 
  group_by(name) %>% 
  filter("K" %in% class, 
         "R" %in% class, 
         "L" %in% class) %>% 
  summarise()

# A tibble: 2 x 1
  name  
  <chr> 
1 Party2
2 Party4

编辑：如果您想与超过 3 方合作，您也可以使用：

mask = c("K", "R", "L")
parties %>% 
  group_by(name) %>% 
  filter(all(mask %in% class)) %>% 
  summarise()

【讨论】：

【解决方案2】：

要使这项工作适用于许多组，您可以使用purrr::reduce：

library(dplyr)

parties %>%
  group_split(class) %>%
  purrr::reduce(semi_join, by = "name") %>%
  select(name)

#   name  
#  <chr> 
#1 Party2
#2 Party4

【讨论】：

【解决方案3】：

这行得通吗：

library(dplyr)
parties %>% group_by(name) %>% mutate(cnt = n()) %>% 
   group_by(class) %>% mutate(grpno = group_indices()) %>% ungroup() %>% 
    filter(cnt >= max(grpno)) %>% select(name) %>% distinct()
# A tibble: 2 x 1
  name  
  <chr> 
1 Party2
2 Party4

【讨论】：

是的！我不得不对我的真实数据集进行一些调整，因为有重复等。但它有效，谢谢！

【解决方案4】：

另一种解决方案

library(tidyverse)

parties %>% 
  group_by(class) %>% 
  distinct() %>% 
  mutate(id = 1) %>%  
  pivot_wider(name, names_from = class, values_from = id) %>% 
  rowwise() %>% 
  filter(!is.na(sum(c_across(where(is.numeric))))) %>% 
  select(name) %>% 
  ungroup()
#> # A tibble: 2 x 1
#>   name  
#>   <chr> 
#> 1 Party2
#> 2 Party4

^{由reprex package (v0.3.0) 于 2020-12-09 创建}

【讨论】：