如何防止dplyr::select合并名称而不是分配新名称？

huangapple
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英文:

How to prevent dplyr::select from combining names rather than assigning a new name?

问题

  1. # 我试图基于另一个命名的向量选择列，并同时为该列分配一个新名称。但是，dplyr似乎会合并名称，我在文档中找不到停止此操作的选项。
  3. data <- data.frame(day = 1,
  4.                    week = 2,
  5.                    n_in_year = 365)
  7. new_values <- c(period = "day",
  8.  n = 365)
  10. # 这会合并名称而不是分配新名称
  11. data %>%
  12.   dplyr::select(time = new_values[1])
  14. # 例如
  15. #   time...period
  16. # 1             1
  18. # 我希望它的行为像这样
  19. data %>%
  20.   dplyr::select(new_values[1]) %>%
  21.   dplyr::rename(time = period)
英文:

I'm trying to select columns based on another named vector and assign a new name for that column at the same time. However dplyr appears to combine the names and I can't see an option to stop this in the documentation.

  1. # dplyr ‘1.1.2’
  2. # R Version 4.3.0
  4. data &lt;- data.frame(day = 1,
  5.                    week = 2,
  6.                    n_in_year = 365)
  8. new_values &lt;- c(period = &quot;day&quot;,
  9.  n = 365)
  11. # This combines the names rather than assigning a new name
  12. data %&gt;%
  13.   dplyr::select(time = new_values[1])
  15. # e.g
  16. #   time...period
  17. # 1             1
  19. # I want it to behave like this
  20. data %&gt;%
  21.   dplyr::select(new_values[1]) %&gt;%
  22.   dplyr::rename(time = period)

答案1

得分: 2

问题

您的问题是new_values仍然保留了其names()：

  1. data %>% dplyr::select(time = new_values[1])
  2. #>   time...period
  3. #> 1             1
  5. data %>% dplyr::select(time = unname(new_values)[1])
  6. #>   time
  7. #> 1    1

这种行为是有意为之的，用于dplyr::select()所使用的"整洁选择"。传递一个命名的character向量（如new_values）的列名将允许程序化用户在各种层次中“组合”和“传播”列名。下面的文档以symbol而不是字符串来说明这一点：

  1. mtcars %>% select_loc(foo = c(mpg, cyl))
  2. #> foo1 foo2 
  3. #>    1    2
  1. mtcars %>% select_loc(foo = c(bar = mpg, baz = cyl))
  2. #> foo...bar foo...baz 
  3. #>         1         2
  1. mtcars %>% select_loc(foo = c(bar = c(mpg, cyl)))
  2. #> foo...bar1 foo...bar2 
  3. #>          1          2

解决方案

虽然unname()可以完成任务，但最好直接使用[[来提取没有名称（period）的值...

  1. #                                       |---|
  2. data %>% dplyr::select(time = new_values[[1]])
  3. #>   time
  4. #> 1    1
  6. #                                       |----------|
  7. data %>% dplyr::select(time = new_values[["period"]])
  8. #>   time
  9. #> 1    1

...或者更好的办法是将new_values制作成一个list，这样其值（如365）不会全部被强制转换为字符串（如"365"）在一个character向量中：

  1. # 原始的'new_values'作为向量...
  2. new_values <- c(period = "day", n = 365)
  3. new_values
  4. #> period      n 
  5. #>  "day"  "365" 
  7. # ...以及新的'new_values'作为列表：
  8. new_values <- list(period = "day", n = 365)
  9. new_values
  10. #> $period
  11. #> [1] "day"
  12. #> 
  13. #> $n
  14. #> [1] 365
  17. # 轻松选择()您想要的：        |-----|
  18. data %>% dplyr::select(time = new_values$period)
  19. #>   time
  20. #> 1    1
英文:

Issue

Your issue is that new_values still has its names():

  1. data %&gt;% dplyr::select(time = new_values[1])
  2. #&gt;   time...period
  3. #&gt; 1             1
  5. data %&gt;% dplyr::select(time = unname(new_values)[1])
  6. #&gt;   time
  7. #&gt; 1    1

This behavior is intentional, for the "tidy selection" used by dplyr::select(). Passing a named character vector (like new_values) of column names will allow a programmatic user to "combine" and "propagate" column names in various hierarchies. This is illustrated by the documentation below, with symbols rather than strings:

> mtcars %>% select_loc(foo = c(mpg, cyl))
> #> foo1 foo2
> #> 1 2

> mtcars %>% select_loc(foo = c(bar = mpg, baz = cyl))
> #> foo...bar foo...baz
> #> 1 2

> mtcars %>% select_loc(foo = c(bar = c(mpg, cyl)))
> #> foo...bar1 foo...bar2
> #> 1 2

Solution

While unname() does the job, you're better off just using [[ to extract the value without the name (period)...

  1. #                                       |---|
  2. data %&gt;% dplyr::select(time = new_values[[1]])
  3. #&gt;   time
  4. #&gt; 1    1
  6. #                                       |----------|
  7. data %&gt;% dplyr::select(time = new_values[[&quot;period&quot;]])
  8. #&gt;   time
  9. #&gt; 1    1

...or better yet, making new_values a list, so its values (like 365) are not all coerced to strings (like &quot;365&quot;) in a character vector:

  1. # Original &#39;new_values&#39; as a vector...
  2. new_values &lt;- c(period = &quot;day&quot;, n = 365)
  3. new_values
  4. #&gt; period      n 
  5. #&gt;  &quot;day&quot;  &quot;365&quot; 
  7. # ...and new &#39;new_values&#39; as a list:
  8. new_values &lt;- list(period = &quot;day&quot;, n = 365)
  9. new_values
  10. #&gt; $period
  11. #&gt; [1] &quot;day&quot;
  12. #&gt; 
  13. #&gt; $n
  14. #&gt; [1] 365
  17. # Easily select() what you want:        |-----|
  18. data %&gt;% dplyr::select(time = new_values$period)
  19. #&gt;   time
  20. #&gt; 1    1

答案2

得分: 1

请检查下面更新的代码：

  1. # dplyr ‘1.1.2’
  2. # R Version 4.3.0
  4. data <- data.frame(day = 1,
  5.                    week = 2,
  6.                    n_in_year = 365)
  8. new_values <- c(period = "day",
  9.                 n = 365)
  11. # 这个组合了名称，而不是分配新名称
  12. data %>%
  13.   dplyr::select(time = new_values[[1]])
  15.   time
  16. 1    1
  18. # 我希望它的行为像这样
  19. data %>%
  20.   dplyr::select(new_values[[1]]) %>%
  21.   dplyr::rename(time = day)

请注意，我保留了代码中的英文部分，只翻译了注释和一些注释中的内容。

英文:

Please check the updated code below

  1. # dplyr ‘1.1.2’
  2. # R Version 4.3.0
  4. data &lt;- data.frame(day = 1,
  5.                    week = 2,
  6.                    n_in_year = 365)
  8. new_values &lt;- c(period = &quot;day&quot;,
  9.                 n = 365)
  11. # This combines the names rather than assigning a new name
  12. data %&gt;%
  13.   dplyr::select(time = new_values[[1]])
  15.   time
  16. 1    1
  18. # I want it to behave like this
  19. data %&gt;%
  20.   dplyr::select(new_values[[1]]) %&gt;%
  21.   dplyr::rename(time = day)

huangapple
  • 本文由 发表于 2023年6月27日 20:14:37
  • 转载请务必保留本文链接：https://go.coder-hub.com/76564753.html
  • dplyr
  • r
  • select
  • tidyverse
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