根据 R 中的列表列表改变列

Question

我有一个数据框，我想 gather 使其为高格式，然后 mutate 在另一列上，其值基于列表中另一列的字符串的成员资格列表。例如，我有以下数据框和列表列表：

dummy_data <- data.frame("id" = 1:20,"test1_10" = sample(1:100, 20),"test2_11" = sample(1:100, 20),
"test3_12" = sample(1:100, 20),"check1_20" = sample(1:100, 20),
"check2_21" = sample(1:100, 20),"sound1_30" = sample(1:100, 20),
"sound2_31" = sample(1:100, 20),"sound3_32" = sample(1:100, 20))

dummylist <- list(c('test1_','test2_','test3_'),c('check1_','check2_'),c('sound1_','sound2_','sound3_'))
names(dummylist) <- c('shipments','arrivals','departures')

然后我像这样收集数据框：

dummy_data <- dummy_data %>%
  gather("part", "number", 2:ncol(.))

我想要做的是添加一个列，该列具有在 dummylist 中找到的列表的名称，其中部分列中下划线之前的字符串是一个成员。我可以这样做：

dummydata <- dummydata %>%
mutate(Group = case_when(
    str_extract(part,'.*_') %in% dummylist[[1]] ~ names(dummylist[1]),
    str_extract(part,'.*_') %in% dummylist[[2]] ~ names(dummylist[2]),
    str_extract(part,'.*_') %in% dummylist[[3]] ~ names(dummylist[3])
  ))

但是，dummylist 中的每个 list/group 需要一个单独的 str_extract 行。而我的真实数据远不止 3 lists/groups。所以我想知道是否有更有效的方法来执行此变异步骤以获取列表的名称？

非常感谢任何帮助，谢谢！

Answer 1

将 'dummylist' 转换为两列数据集后，使用 regex_left_join 可能更容易

library(fuzzyjoin)
library(dplyr)
library(tidyr)
library(tibble)
dummy_data %>% 
   # // reshape to long format - pivot_longer instead of gather
   pivot_longer(cols = -id, names_to = 'part', values_to = 'number') %>% 
   # // join with the tibble/data.frame converted dummylist
   regex_left_join(dummylist %>%
        enframe(name = 'Group', value = 'part') %>% 
        unnest(part)) %>% 
   rename(part = part.x) %>%
   select(-part.y)

-输出

# A tibble: 160 × 4
      id part      number Group     
   <int> <chr>      <int> <chr>     
 1     1 test1_10      72 shipments 
 2     1 test2_11      62 shipments 
 3     1 test3_12      17 shipments 
 4     1 check1_20     89 arrivals  
 5     1 check2_21     54 arrivals  
 6     1 sound1_30     39 departures
 7     1 sound2_31     94 departures
 8     1 sound3_32     95 departures
 9     2 test1_10      77 shipments 
10     2 test2_11       4 shipments 
# … with 150 more rows

Answer 2

如果您事先准备好 查找 table，则不需要任何额外的库，但是 dplyr 和 tidyr：

lookup <- sapply(
    names(dummylist),
    \(nm) { setNames(rep(nm, length(dummylist[[nm]])), dummylist[[nm]]) }
    ) |>
    setNames(nm = NULL) |>
    unlist()    

lookup

#      test1_       test2_       test3_      check1_      check2_      sound1_      sound2_      sound3_ 
# "shipments"  "shipments"  "shipments"   "arrivals"   "arrivals" "departures" "departures" "departures"

现在您只需 gsub 即时翻译您的 part，在通常的 mutate() 动词中：

dummy_data |>
    pivot_longer(-id, names_to = 'part', values_to = 'number') |>
    mutate(group = lookup[gsub('^(\w+_).*$', '\1', part)])
    
# # A tibble: 160 × 4
#      id part      number group     
#   <int> <chr>      <int> <chr>     
# 1     1 test1_10      91 shipments 
# 2     1 test2_11      74 shipments 
# 3     1 test3_12      46 shipments 
# 4     1 check1_20     62 arrivals  
# 5     1 check2_21      7 arrivals  
# 6     1 sound1_30     35 departures
# 7     1 sound2_31     23 departures
# 8     1 sound3_32     84 departures
# 9     2 test1_10      59 shipments 
# 10    2 test2_11      73 shipments 
# # … with 150 more rows

根据 R 中的列表列表改变列

Mutate column based on list of lists in R

r

dataframe

dplyr

tidyr