我正在使用兩個資料集 - 一組有成對的專案：

original <- data.frame(label1 = c("cat", "cat", "dog", "dog", "cat", "tiger", "tiger", "cow"),
                      label2 = c("dog", "dog", "cat", "cat", "dog", "cow", "cow", "tiger"))
original
  label1 label2
1    cat    dog
2    cat    dog
3    dog    cat
4    dog    cat
5    cat    dog
6  tiger    cow
7  tiger    cow
8    cow  tiger

第二個資料集包含第一組專案的索引代碼：

index <- data.frame(item = c("cat", "dog", "tiger", "cow"),
                    code = c(1, 0, 1, 0))
index
   item code
1   cat    1
2   dog    0
3 tiger    1
4   cow    0

我正在尋找一種方式來創建兩個新列：tag0和tag1，使它看起來像這樣：

new <- data.frame(label1 = c("cat", "cat", "dog", "dog", "cat", "tiger", "tiger", "cow"),
                  label2 = c("dog", "dog", "cat", "cat", "dog", "cow", "cow", "tiger"),
                  tag1 = c("cat", "cat", "cat", "cat", "cat", "tiger", "tiger", "tiger"),
                  tag0 = c("dog", "dog", "dog", "dog", "dog", "cow", "cow", "cow"))
new
  label1 label2  tag1 tag0
1    cat    dog   cat  dog
2    cat    dog   cat  dog
3    dog    cat   cat  dog
4    dog    cat   cat  dog
5    cat    dog   cat  dog
6  tiger    cow tiger  cow
7  tiger    cow tiger  cow
8    cow  tiger tiger  cow

tag0指的是對應的標簽code=0，tag1指的code=1是indexdataframe中對應的標簽。

任何人都可以幫助我提供tidyverse基于解決方案的解決方案嗎？

uj5u.com熱心網友回復：

這里有兩個解決方案tidyverse。雖然第一個適用于這種特殊情況，但我更喜歡第二個，它更優雅且可擴展。

解決方案 1：JOIN每個Alabel*

首先匯入tidyverse并生成您的資料集original和index.

library(tidyverse)


# ...
# Code to generate 'original' and 'index' datasets.
# ...

然后應用此作業流程。

original %>%
  # Uniquely identify each row (for pivoting later).
  mutate(row_id = row_number()) %>%
  # Match 'label1' to the tags.
  left_join(
    index,
    by = c("label1" = "item"),
    keep = TRUE
  ) %>%
  # Match 'label2' to the tags.
  left_join(
    index,
    by = c("label2" = "item"),
    keep = TRUE,
    suffix = c(".1", ".2")
  ) %>%
  # Pivot 'item.1 | ... | item.n | code.1 | ... | code.n' into a consolidated
  # 'item | code' form.
  pivot_longer(
    cols = matches("^(item|code)\\.(\\d )?$"),
    names_pattern = "^(item|code)\\.(\\d )?$",
    names_to = c(".value", NA)
  ) %>%
  # Pivot back into a 'tag1 | tag0' form.
  pivot_wider(
    values_from = item,
    names_from = code,
    names_glue = "tag{code}"
  ) %>%
  # Omit unique identifier.
  select(!row_id)

結果

鑒于這里復制的original和index資料集

original <- data.frame(
  label1 = c("cat", "cat", "dog", "dog", "cat", "tiger", "tiger", "cow"),
  label2 = c("dog", "dog", "cat", "cat", "dog", "cow", "cow", "tiger")
)

index <- data.frame(
  item = c("cat", "dog", "tiger", "cow"),
  code = c(1, 0, 1, 0)
)

該解決方案應產生以下結果：

# A tibble: 8 x 4
  label1 label2 tag1  tag0 
  <chr>  <chr>  <chr> <chr>
1 cat    dog    cat   dog  
2 cat    dog    cat   dog  
3 dog    cat    cat   dog  
4 dog    cat    cat   dog  
5 cat    dog    cat   dog  
6 tiger  cow    tiger cow  
7 tiger  cow    tiger cow  
8 cow    tiger  tiger cow  

筆記

如果您的original資料集有任何其他label*列，您將需要為這些列中的每一列執行額外JOIN的操作。

解決方案 2：單 CROSS JOIN

這是一個更優雅的作業流程，它也更靈活：它適用于任意數量的label*列 inoriginal和任意一組codes in index。

original %>%
  # Uniquely identify each row (for pivoting later).
  mutate(row_id = row_number()) %>%
  # Perform a cross-join compare every 'item' to every 'label*'.
  full_join(
    index,
    by = character()
  ) %>%
  # Keep only those rows where 'item' matches a 'label*'.
  rowwise() %>%
  filter(item %in% c_across(matches("^label\\d "))) %>%
  # Pivot into a 'tag1 | tag0' form.
  pivot_wider(
    values_from = item,
    names_from = code,
    names_glue = "tag{code}"
  ) %>%
  # Omit unique identifier.
  select(!row_id)

結果

結果保持不變。

# A tibble: 8 x 4
  label1 label2 tag1  tag0 
  <chr>  <chr>  <chr> <chr>
1 cat    dog    cat   dog  
2 cat    dog    cat   dog  
3 dog    cat    cat   dog  
4 dog    cat    cat   dog  
5 cat    dog    cat   dog  
6 tiger  cow    tiger cow  
7 tiger  cow    tiger cow  
8 cow    tiger  tiger cow  

筆記

唯一的缺點是它必須執行 a CROSS JOIN，這可能會阻礙更大資料集的性能。

uj5u.com熱心網友回復：

另一種可能的解決方案：

library(tidyverse)

original <- data.frame(label1 = c("cat", "cat", "dog", "dog", "cat", "tiger", "tiger", "cow"),
                       label2 = c("dog", "dog", "cat", "cat", "dog", "cow", "cow", "tiger"))

index <- data.frame(item = c("cat", "dog", "tiger", "cow"),
                    code = c(1, 0, 1, 0))

original %>% 
  full_join(index, by=c("label1" = "item")) %>% 
  full_join(index, by=c("label2" = "item")) %>% 
  mutate(tag1 = if_else(code.x == 1, label1, label2)) %>% 
  mutate(tag2 = if_else(code.y == 1, label1, label2)) %>% 
  select(!starts_with("code"))

#>   label1 label2  tag1 tag2
#> 1    cat    dog   cat  dog
#> 2    cat    dog   cat  dog
#> 3    dog    cat   cat  dog
#> 4    dog    cat   cat  dog
#> 5    cat    dog   cat  dog
#> 6  tiger    cow tiger  cow
#> 7  tiger    cow tiger  cow
#> 8    cow  tiger tiger  cow

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