传递一个函数中新列的列名，而无需花序？

发布于 2025-02-10 07:50:40 字数 1137 浏览 2 评论 0原文

我用tidyeval为这个问题做了解决方案，是否有基本R方法？

library(dplyr)

new_col <- function(df, col_name, col_vals){
  df |>
    cbind(temp_name = col_vals) |>
    rename({{col_name}} := temp_name)
}

sleep %>% 
  new_col(sample, "sample1") |>
  new_col(condition, "condition2") |>
  head()
#>   extra group ID  sample  condition
#> 1   0.7     1  1 sample1 condition2
#> 2  -1.6     1  2 sample1 condition2
#> 3  -0.2     1  3 sample1 condition2
#> 4  -1.2     1  4 sample1 condition2
#> 5  -0.1     1  5 sample1 condition2
#> 6   3.4     1  6 sample1 condition2

“ https://reprex.tidyverse.org”创建2022-06-24

^{由 tidyeval 尚不清楚的概念。此答案也使用{{}}}和此答案使用Enquo（）。我的解决方案还需要：=。}

理想情况下，我想将元数据的向量（例如，在数据框中）映射到数据范围/对象的匹配列表。

原文

I have worked a solution to this problem with tidyeval, is there a base R method?

library(dplyr)

new_col <- function(df, col_name, col_vals){
  df |>
    cbind(temp_name = col_vals) |>
    rename({{col_name}} := temp_name)
}

sleep %>% 
  new_col(sample, "sample1") |>
  new_col(condition, "condition2") |>
  head()
#>   extra group ID  sample  condition
#> 1   0.7     1  1 sample1 condition2
#> 2  -1.6     1  2 sample1 condition2
#> 3  -0.2     1  3 sample1 condition2
#> 4  -1.2     1  4 sample1 condition2
#> 5  -0.1     1  5 sample1 condition2
#> 6   3.4     1  6 sample1 condition2

^{Created on 2022-06-24 by the reprex package (v2.0.1)}

This seems indirect and depends on tidyeval concepts that are less known. This answer also uses {{}} and this answer uses enquo(). My solution also needs :=.

Ideally, I want to map vectors of metadata (say, in a dataframe) to a matching list of dataframes/objects.

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许久 2025-02-17 07:50:40

在基础r中，我们可以使用depars/替代

new_col <- function(df, col_name, col_vals){
  cn <- deparse(substitute(col_name))
  df[[cn]] <- col_vals  
   df
}

- 测试

 sleep %>% 
+   new_col(sample, "sample1") |>
+   new_col(condition, "condition2") |>
+   head()
  extra group ID  sample  condition
1   0.7     1  1 sample1 condition2
2  -1.6     1  2 sample1 condition2
3  -0.2     1  3 sample1 condition2
4  -1.2     1  4 sample1 condition2
5  -0.1     1  5 sample1 condition2
6   3.4     1  6 sample1 condition2

In base R, we can use deparse/substitute

new_col <- function(df, col_name, col_vals){
  cn <- deparse(substitute(col_name))
  df[[cn]] <- col_vals  
   df
}

-testing

 sleep %>% 
+   new_col(sample, "sample1") |>
+   new_col(condition, "condition2") |>
+   head()
  extra group ID  sample  condition
1   0.7     1  1 sample1 condition2
2  -1.6     1  2 sample1 condition2
3  -0.2     1  3 sample1 condition2
4  -1.2     1  4 sample1 condition2
5  -0.1     1  5 sample1 condition2
6   3.4     1  6 sample1 condition2

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