基于名称（正则表达式）对选定列进行变异。

huangapple
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英文:

mutate on selected columns based on name (regular expression)

问题

以下是翻译好的部分：

这是一个示例数据集：

  1. ff <- data.frame(
  2.        id = c(1:4),
  3.        w1...33112..Value = c(10, 20, 30, 40),
  4.        w1...33112..Time = c(4, 3, 2, 1),
  5.        w2...33113..Value = c(1, .9, .75, .7),
  6.        w2...33113..Time = c(10, 50, 30, 20),
  7.        w3...33552..Value = c(1, 2, 3, 4),
  8.        w3...33552..Time = c(.5, .5, .9, .9),
  9.        w4...33442..Value = c(100, 50, 40, 30),
  10.        w4...33442..Time = c(2, 1, 4, 3),
  11.        w5...35692..Value = c(.5, .6, .7, .8)
  12.        )

我想对基于名称选择的列执行一些简单的操作（通常使用diff） - 列名必须包含字符串Value

下面的示例是针对两个变量的，实际数据中有几十种这样的情况。

  1. ff.2 <- ff %>% mutate (
  2.   w1.used = c(0, diff(w1...33112..Value)), 
  3.   w2.used = c(0, diff(w2...33113..Value)), 
  4. )

新列的名称应以字符字符串开头，直到第一个点和所选字符串（例如"used"）。

英文:

Here's a sample dataset:

  2. ff &lt;- data.frame(
  3.        id = c(1:4),
  4.        w1...33112..Value = c(10, 20, 30, 40),
  5.        w1...33112..Time = c(4, 3, 2, 1),
  6.        w2...33113..Value = c(1, .9, .75, .7),
  7.        w2...33113..Time = c(10, 50, 30, 20),
  8.        w3...33552..Value = c(1, 2, 3, 4),
  9.        w3...33552..Time = c(.5, .5, .9, .9),
  10.        w4...33442..Value = c(100, 50, 40, 30),
  11.        w4...33442..Time = c(2, 1, 4, 3),
  12.        w5...35692..Value = c(.5, .6, .7, .8)
  13.        )

I want to perform some simple operations (usually using diff) on columns selected based on name - the column name must contain the string Value

The example below is for two variables, and there are dozens of such cases in real data.

  1. ff.2 &lt;- ff %&gt;% mutate (
  2.   w1.used = c(0, diff(w1...33112..Value)), 
  3.   w2.used = c(0, diff(w2...33113..Value)), 
  4. )

The name of the new column should start with a string of characters up to the first dot and the selected string (for example "used").

答案1

得分: 4

你可以简单地执行以下操作：

  1. ff %>%
  2.   mutate(across(ends_with('Value'), ~c(0, diff(.)),
  3.                 .names = "{str_extract(.col, 'w[0-9]+')}.used"))
英文:

You could simply do:

  1. ff %&gt;%
  2.   mutate(across(ends_with(&#39;Value&#39;), ~c(0, diff(.)),
  3.                 .names = &quot;{str_extract(.col, &#39;w[0-9]+&#39;)}.used&quot;))

答案2

得分: 1

The mutate_at() 函数允许您基于满足特定条件的现有变量创建新变量，例如，如果它们的名称包含特定字符串。此外，str_extract() 函数将提取字符串直到第一个点以创建新的列名。

以下是完整的翻译：

mutate_at() 函数允许您基于满足特定条件的现有变量创建新变量，例如，如果它们的名称包含特定字符串。此外，str_extract() 函数将提取字符串直到第一个点以创建新的列名。以下是完整的翻译：

这里是一个示例代码，它正好可以实现您想要的功能：

  1. library(tidyverse)
  3. # 选择只包含“Value”在其名称中的列
  4. value_cols <- grep("Value", names(ff), value = TRUE)
  6. # 创建新列
  7. ff <- ff %>%
  8.   mutate_at(
  9.     .vars = value_cols,
  10.     .funs = list(
  11.       used = ~ c(0, diff(.))
  12.     )
  13.   )
  15. # 重命名新列
  16. new_col_names <- str_extract(value_cols, "^[^.]*") %>% paste0(".used")
  18. names(ff)[grep("used", names(ff))] <- new_col_names
  20. ff
英文:

The mutate_at() function allows you to create new variables based on existing variables that match certain criteria, like if their name contains a certain string. Also, str_extract() function will extract the string until the first dot to create new column names.

Here is the sample code which does exactly what you want:

  1. library(tidyverse)
  3. # Select only columns with &quot;Value&quot; in their names
  4. value_cols &lt;- grep(&quot;Value&quot;, names(ff), value = TRUE)
  6. # Create new columns
  7. ff &lt;- ff %&gt;%
  8.   mutate_at(
  9.     .vars = value_cols,
  10.     .funs = list(
  11.       used = ~ c(0, diff(.))
  12.     )
  13.   )
  15. # Rename new columns
  16. new_col_names &lt;- str_extract(value_cols, &quot;^[^.]*&quot;) %&gt;% paste0(&quot;.used&quot;)
  18. names(ff)[grep(&quot;used&quot;, names(ff))] &lt;- new_col_names
  20. ff

答案3

得分: 1

以下是代码的翻译部分：

  1. # 选取以"Value"结尾的列
  2. d <- ff[endsWith(names(ff), "Value")]
  4. # 计算差值
  5. u <- d - rbind(d[1, ], d[-nrow(d), ])
  7. # 将差值列添加到数据框中，并设置列名
  8. ff.2 <- cbind(ff, setNames(u, sub("\\..*", ".used", names(u))))

输出结果如下：

  1. > ff.2
  2.   id w1...33112..Value w1...33112..Time w2...33113..Value w2...33113..Time
  3. 1  1                10                4              1.00               10
  4. 2  2                20                3              0.90               50
  5. 3  3                30                2              0.75               30
  6. 4  4                40                1              0.70               20
  7.   w3...33552..Value w3...33552..Time w4...33442..Value w4...33442..Time
  8. 1                 1              0.5               100                2
  9. 2                 2              0.5                50                1
  10. 3                 3              0.9                40                4
  11. 4                 4              0.9                30                3
  12.   w5...35692..Value w1.used w2.used w3.used w4.used w5.used
  13. 1               0.5       0    0.00       0       0     0.0
  14. 2               0.6      10   -0.10       1     -50     0.1
  15. 3               0.7      10   -0.15       1     -10     0.1
  16. 4               0.8      10   -0.05       1     -10     0.1

希望这对您有帮助！

英文:

A base R option

  1. d &lt;- ff[endsWith(names(ff), &quot;Value&quot;)]
  2. u &lt;- d - rbind(d[1, ], d[-nrow(d), ])
  3. ff.2 &lt;- cbind(ff, setNames(u, sub(&quot;\\..*&quot;, &quot;.used&quot;, names(u))))

gives

  1. &gt; ff.2
  2.   id w1...33112..Value w1...33112..Time w2...33113..Value w2...33113..Time
  3. 1  1                10                4              1.00               10
  4. 2  2                20                3              0.90               50
  5. 3  3                30                2              0.75               30
  6. 4  4                40                1              0.70               20
  7.   w3...33552..Value w3...33552..Time w4...33442..Value w4...33442..Time
  8. 1                 1              0.5               100                2
  9. 2                 2              0.5                50                1
  10. 3                 3              0.9                40                4
  11. 4                 4              0.9                30                3
  12.   w5...35692..Value w1.used w2.used w3.used w4.used w5.used
  13. 1               0.5       0    0.00       0       0     0.0
  14. 2               0.6      10   -0.10       1     -50     0.1
  15. 3               0.7      10   -0.15       1     -10     0.1
  16. 4               0.8      10   -0.05       1     -10     0.1

huangapple
  • 本文由 发表于 2023年6月29日 19:30:04
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