使用tidyverse在R中合并两个tibble。

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英文:

Combine two tibbles in R using tidyverse

问题

  1. 我有两个tibbles
  3. x <- tibble(A = letters[1:4],
  4.             B = letters[5:8])
  5. y <- tibble(A = letters[9:12],
  6.             B = letters[13:16])
  8. 我想要得到
  10. # 一个tibble: 4 × 2
  11.   A     B    
  12.   <chr> <chr>
  13. 1 a,i   i,m  
  14. 2 b,j   j,n  
  15. 3 c,k   k,o  
  16. 4 d,l   l,p 
  18. 我需要一段代码来实现，而不是硬编码列名，因为我必须处理可能有从一个到多个列的tibbles。这两个tibbles将始终具有相同数量的列和行。如果可能的话，我更愿意避免使用for循环。
  20. 谢谢,
  21. Marco
英文:

I have 2 tibbles

  1. x &lt;- tibble(A = letters[1:4],
  2.             B = letters[5:8])
  3. y &lt;- tibble(A = letters[9:12],
  4.             B = letters[13:16])

I'd like to obtain

  1. # A tibble: 4 &#215; 2
  2.   A     B    
  3.   &lt;chr&gt; &lt;chr&gt;
  4. 1 a,i   i,m  
  5. 2 b,j   j,n  
  6. 3 c,k   k,o  
  7. 4 d,l   l,p

I need a code that does it without hardcoding the column names because I have to work on tibbles that can have from one to many columns. The 2 tibbles will always have the same number of columns and rows. I prefer to avoid for loop if possible.

Thank you,
Marco

答案1

得分: 3

一个选项可能是：

  1. x %>%
  2.   mutate(across(everything(), ~ paste(., y[[cur_column()]], sep = ",")))
  3.   
  4.   A     B    
  5.   <chr> <chr>
  6. 1 a,i   e,m  
  7. 2 b,j   f,n  
  8. 3 c,k   g,o  
  9. 4 d,l   h,p
英文:

One option could be:

  1. x %&gt;%
  2.  mutate(across(everything(), ~ paste(., y[[cur_column()]], sep = &quot;,&quot;)))
  4.   A     B    
  5.   &lt;chr&gt; &lt;chr&gt;
  6. 1 a,i   e,m  
  7. 2 b,j   f,n  
  8. 3 c,k   g,o  
  9. 4 d,l   h,p

答案2

得分: 2

使用paste的基本R选项。

  1. z <- as.data.frame(x)
  2. z[] <- paste(unlist(x), unlist(y), sep = ",")
  3. # 如果您需要tibble作为输出。
  4. tibble::tibble(z)
  6. # 一个tibble: 4 × 2
  7. #  A     B    
  8. #  <chr> <chr>
  9. #1 a,i   e,m  
  10. #2 b,j   f,n  
  11. #3 c,k   g,o  
  12. #4 d,l   h,p

(Note: I've left the code part unchanged and only translated the surrounding text.)

英文:

A base R option using paste.

  1. z &lt;- as.data.frame(x)
  2. z[] &lt;- paste(unlist(x), unlist(y), sep = &quot;,&quot;)
  3. # If you need tibble as output. 
  4. tibble::tibble(z)
  6. # A tibble: 4 &#215; 2
  7. #  A     B    
  8. #  &lt;chr&gt; &lt;chr&gt;
  9. #1 a,i   e,m  
  10. #2 b,j   f,n  
  11. #3 c,k   g,o  
  12. #4 d,l   h,p

答案3

得分: 2

你可以使用 purr 来将这两个数据框作为列的集合进行映射：

  1. purrr::map2_df(x, y, paste, sep=",")
英文:

You can use purr to map over the two data.frames as collections of columns

  1. purrr::map2_df(x, y, paste, sep=&quot;,&quot;)

答案4

得分: 0

这是另一种选项，不依赖于x和y的顺序相同：

  1. library(purrr)
  3. data.frame(map(set_names(names(x)), ~ paste(x[[.x]], y[[.x]], sep = ",")))
英文:

Here is another option that does not rely on x and y being in the same order:

  1. library(purrr)
  3. data.frame(map(set_names(names(x)), ~ paste(x[[.x]], y[[.x]], sep = &quot;,&quot;)))

huangapple
  • 本文由 发表于 2023年7月12日 21:28:58
  • 转载请务必保留本文链接：https://go.coder-hub.com/76671127.html
  • r
  • tidyverse
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