你可以使用从数据框中提取的值来指定在R中要进行子集操作的列。

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英文:

How can I use a value extracted from a dataframe to specify columns to subset in R?

问题

Sure, here's the translated code snippet:

  1. 我有一个数据框，我想在函数内部对其进行子集化，以便只保留两列都为1或NA的行。对于df：
  3. df <- data.frame(a = c(1,1,0,NA,0,1), 
  4.                  b = c(0,1,0,1,0, NA),
  5.                  c = c(0,0,0,0,0,0))
  7. 我想要的结果是：
  9.    a  b  c
  10. 2  1  1  0
  11. 4 NA  1  0
  12. 6  1 NA  0
  14. 我遇到的问题是，我有许多列的名称会变化。因此，这个方法效果很好：
  16. subset(df, (is.na(a) | a == 1) & (is.na(b) | b == 1))
  18. 但是，当列名'a'和'b'在函数操作过程中变为'd'和'f'时，这个方法就失效了。通过列索引指定的方法更加健壮：
  20. subset(df, (is.na(df[,1]) | df[,1] == 1) & (is.na(df[,2]) | df[,2] == 1))
  22. 但这样做很麻烦，而且如果先前的处理步骤出错，导致列'c'在'a'或'b'之前，那么我可能会选择错误的列进行子集化。
  24. 我还有另一个指定要进行子集化的列名的数据框：
  26. cro_df <- data.frame(pop = c('c1', 'c2'),
  27.                      p1 = c('a', 'd'),
  28.                      p2 = c('b', 'f'))
  30. 我想能够从该数据框中提取列名，以在我的子集化函数中使用，例如：
  32. col1 <- cro_df[cro_df[, 'pop'] == 'c1', 'p1']
  33. subset(df, is.na(col1) | col1 == 1)
  35. 这将返回一个空数据框。我已经尝试将col1转换为符号和因子，但没有成功：
  37. subset(df, as.symbol(col1) == 1)
  38. subset(df, sym(col1) == 1)
  39. subset(df, as.factor(col1) == 1)
  41. 它们都返回：
  43. [1] a b c
  44. <0 rows> (or 0-length row.names)
  46. 是否有一种方法可以使用第二个数据框cro_df来指定要进行子集化的列？
英文:

I have a dataframe that I want to subset inside a function so that only rows where both columns are either 1 or NA remain. For df:

  1. df &lt;- data.frame(a = c(1,1,0,NA,0,1), 
  2.                  b = c(0,1,0,1,0, NA),
  3.                  c = c(0,0,0,0,0,0))

I want:

  1.    a  b  c
  2. 2  1  1  0
  3. 4 NA  1  0
  4. 6  1 NA  0

The problem I'm having is I have many columns with names that change. So this works well:

  1. subset(df, (is.na(a) | a == 1) &amp; (is.na(b) | b == 1))

but when column names 'a' and 'b' become 'd' and 'f' during the operation of the function it breaks. Specifying by column index works more robustly:

  1. subset(df, (is.na(df[,1]) | df[,1] == 1) &amp; (is.na(df[,2]) | df[,2] == 1))

But is cumbersome, and if a previous processing step messes up and column 'c' ends up before 'a' or 'b' I end up subsetting by the wrong columns.

I also have another dataframe that specifies what the column names to subset by will be:

  1. cro_df &lt;- data.frame(pop = c(&#39;c1&#39;, &#39;c2&#39;),
  2.                      p1 = c(&#39;a&#39;, &#39;d&#39;),
  3.                      p2 = c(&#39;b&#39;, &#39;f&#39;))
  4.   pop p1 p2
  5. 1  c1  a  d
  6. 2  c2  b  f

I would like to be able to extract the column names from that dataframe to use in my subset function, e.g.:

  1. col1 &lt;- cro_df[cro_df[,&#39;pop&#39;]==&#39;c1&#39;, &#39;p1&#39;]
  2. subset(df, is.na(col1) | col1 == 1)

This returns an empty dataframe. I have tried turning col1 into a symbol and a factor with no success:

  1. subset(df, as.symbol(col1) == 1)
  2. subset(df, sym(col1) == 1)
  3. subset(df, as.factor(col1) == 1)

And they all return:

  1. [1] a b c
  2. &lt;0 rows&gt; (or 0-length row.names)

Is there a way I can specify my columns to subset using the second dataframe cro_df?

答案1

得分: 1

你可以使用dplyr包中的filter和if_all。

按照你认为最适合的方式选择要筛选的列的名称。在我的案例中，我创建了一个变量cols，其中包含'a'和'b'。

然后，我检查cols中所有列名是否都满足条件，并使用filter筛选满足if_all语句为TRUE的行：

  1. library(dplyr) # packageVersion("dplyr") >= 1.1.0
  3. cols <- c('a', 'b')
  4. filter(df, if_all(all_of(cols), \(x) is.na(x) | x == 1))
  5. #>    a  b c
  6. #> 1  1  1 0
  7. #> 2 NA  1 0
  8. #> 3  1 NA 0

如果你将不同的列名分配给cols，你可以重用相同的代码。

英文:

You can use filter and if_all from the dplyr package.

Select in the manner you find best suited for your case the names of the columns you want to filter. In my case I just created a variable cols that contains &#39;a&#39; and &#39;b&#39;.

Then I check all_of the column names in cols and filter the rows if_all statements are TRUE:

  1. library(dplyr) # packageVersion(&quot;dplyr&quot;) &gt;= 1.1.0
  3. cols &lt;- c(&#39;a&#39;, &#39;b&#39;)
  4. filter(df, if_all(all_of(cols), \(x) is.na(x) | x == 1))
  5. #&gt;    a  b c
  6. #&gt; 1  1  1 0
  7. #&gt; 2 NA  1 0
  8. #&gt; 3  1 NA 0

If you assign different column names to cols you can reuse the same code.

答案2

得分: 1

以下是您请求的代码的中文翻译：

  1. # 加载必要的包
  2. library(dplyr)
  3. library(purrr)
  5. # 创建第一个数据框
  6. df <- data.frame(a = c(1,1,0,NA,0,1), 
  7.                  b = c(0,1,0,1,0, NA),
  8.                  c = c(0,0,0,0,0,0))
  10. # 添加第二个具有不同列名的数据框
  11. df2 <- data.frame(d = c(1,1,0,NA,0,1), 
  12.                   f = c(0,1,0,1,0, NA),
  13.                   c = c(0,0,0,0,0,0))
  15. # 使用dplyr::if_all()在dplyr::filter()中应用筛选条件
  16. df |>
  17.   filter(if_all(c(a, b), \(x) is.na(x) | x == 1))
  18. # 输出：
  19. #   a  b c
  20. # 1  1  1 0
  21. # 2 NA  1 0
  22. # 3  1 NA 0
  24. # 创建自定义函数以适应不同的列名
  25. custom_filter <- 
  26.   function(data, v1, v2) {
  27.     filter(data,
  28.            if_all(c({{v1}}, {{v2}}), \(x) is.na(x) | x == 1))
  29.   }
  31. # 示例如何使用自定义函数
  32. custom_filter(df, a, b)
  33. # 输出：
  34. #   a  b c
  35. # 1  1  1 0
  36. # 2 NA  1 0
  37. # 3  1 NA 0
  38. custom_filter(df2, d, f)
  39. # 输出：
  40. #   d  f c
  41. # 1  1  1 0
  42. # 2 NA  1 0
  43. # 3  1 NA 0
  45. # 使用cro_df数据框和将所有数据框放入list()中，以便通过所有数据框并应用筛选条件的编程方式（purrr::map2()）。
  46. cro_df <- data.frame(pop = c('c1', 'c2'),
  47.                      p1 = c('a', 'd'),
  48.                      p2 = c('b', 'f'))
  50. cro_l <- 
  51.   cro_df |>
  52.   split(1:nrow(cro_df))
  54. data_l <- list(df, df2)
  56. map2(data_l,
  57.      cro_l,
  58.      \(x, y) custom_filter(
  59.        x, y$p1, y$p2
  60.      ))
  61. # 输出：
  62. # [[1]]
  63. #   a  b c
  64. # 1  1  1 0
  65. # 2 NA  1 0
  66. # 3  1 NA 0
  67. # 
  68. # [[2]]
  69. #   d  f c
  70. # 1  1  1 0
  71. # 2 NA  1 0
  72. # 3  1 NA 0
英文: 
  1. library(dplyr)
  2. library(purrr)
  3.   
  4. df &lt;- data.frame(a = c(1,1,0,NA,0,1), 
  5.                  b = c(0,1,0,1,0, NA),
  6.                  c = c(0,0,0,0,0,0))

Let’s add a second data frame with different column names.

  1. df2 &lt;- data.frame(d = c(1,1,0,NA,0,1), 
  2.                  f = c(0,1,0,1,0, NA),
  3.                  c = c(0,0,0,0,0,0))

We can use dplyr::if_all() in dplyr::filter() to apply the filter.

  1. df |&gt; 
  2.   filter(if_all(c(a, b), \(x) is.na(x) | x == 1))
  3. #&gt;    a  b c
  4. #&gt; 1  1  1 0
  5. #&gt; 2 NA  1 0
  6. #&gt; 3  1 NA 0

Using that idea we now write a custom function to accomodate for changing
column names.

  1. custom_filter &lt;- 
  2.   function(data, v1, v2) {
  3.     filter(data,
  4.            if_all(c({{v1}}, {{v2}}), \(x) is.na(x) | x == 1))
  5.   }

Here is how that can work.

  1. custom_filter(df, a, b)
  2. #&gt;    a  b c
  3. #&gt; 1  1  1 0
  4. #&gt; 2 NA  1 0
  5. #&gt; 3  1 NA 0
  6. custom_filter(df2, d, f)
  7. #&gt;    d  f c
  8. #&gt; 1  1  1 0
  9. #&gt; 2 NA  1 0
  10. #&gt; 3  1 NA 0

Using your cro_df dataframe and by placing all dataframes in a list()
we can now programmatically (purrr::map2()) go through all of the
dataframes and apply the filter.

  1. cro_df &lt;- data.frame(pop = c(&#39;c1&#39;, &#39;c2&#39;),
  2.                      p1 = c(&#39;a&#39;, &#39;d&#39;),
  3.                      p2 = c(&#39;b&#39;, &#39;f&#39;))
  5. cro_l &lt;- 
  6.   cro_df |&gt; 
  7.   split(1:nrow(cro_df))
  9. data_l &lt;- list(df, df2)
  11. map2(data_l,
  12.      cro_l,
  13.      \(x, y) custom_filter(
  14.        x, y$p1, y$p2
  15.      ))
  16. #&gt; [[1]]
  17. #&gt;    a  b c
  18. #&gt; 1  1  1 0
  19. #&gt; 2 NA  1 0
  20. #&gt; 3  1 NA 0
  21. #&gt; 
  22. #&gt; [[2]]
  23. #&gt;    d  f c
  24. #&gt; 1  1  1 0
  25. #&gt; 2 NA  1 0
  26. #&gt; 3  1 NA 0

答案3

得分: 0

以下是翻译好的内容：

也许这是一个不错的开始？

  1. with(cro_df[cro_df$pop == "c1",],
  2.   df[ (is.na(df[[p1]]) | df[[p1]] == 1) & (is.na(df[[p2]]) | df[[p2]] == 1), ]
  3. )
  4. #    a  b c
  5. # 2  1  1 0
  6. # 4 NA  1 0
  7. # 6  1 NA 0

FYI，subset 用于交互式使用，其帮助页面指出：

  2. 这是一个方便的函数，用于交互式使用。
  3. 对于编程，最好使用标准的子集函数，如[,]，尤其是参数 'subset' 的非标准评估可能会导致意外后果。
英文:

Perhaps this is a good start?

  1. with(cro_df[cro_df$pop == &quot;c1&quot;,],
  2.   df[ (is.na(df[[p1]]) | df[[p1]] == 1) &amp; (is.na(df[[p2]]) | df[[p2]] == 1), ]
  3. )
  4. #    a  b c
  5. # 2  1  1 0
  6. # 4 NA  1 0
  7. # 6  1 NA 0

FYI, subset is intended for interactive use, its help page says

  1. Warning:
  2. This is a convenience function intended for use interactively.
  3. For programming it is better to use the standard subsetting
  4. functions like [, and in particular the non-standard evaluation
  5. of argument ‘subset’ can have unanticipated consequences.

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  • 本文由 发表于 2023年5月17日 21:36:46
  • 转载请务必保留本文链接：https://go.coder-hub.com/76272728.html
  • r
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