如何使用现有的数值向量来派生新变量

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英文:

how to derive new variables from existing using a vector of values

问题

这是我所拥有的数据：

  1. temp <- tibble::tribble(
  2. ~x1, ~x2, ~x3, ~x4, ~x5,
  3. 1, 2, 3, 'AA', 'BB',
  4. 2, 3, 4, 'AB', 'CC',
  5. 3, 4, 5, 'AC', 'DD',
  6. 4, 5, 6, 'AD', 'EE',
  7. 5, 6, 7, 'AE', 'FF',
  8. 6, 7, 8, 'AF', 'GG'
  9. ) %>% dplyr::mutate(dplyr::across(x4,as.character), dplyr::across(x5,as.character))

我想通过与值为c(1,2,3)的temp_val向量相乘来导出新变量'px1'、'px2'、'px3'。

  1. temp_val <- c(1,2,3)

我尝试了以下代码：

  1. temp_val <- c(1,2,3)
  2. nam1 <- c('px1','px2','px3')
  3. temp5 <- temp %>% mutate(across(where(is.numeric), ~ . * temp_val, .names = '{nam1}'))

我得到了以下数据：

如何使用现有的数值向量来派生新变量

我的期望是将向量值视为列，并使用这些列创建多个变量。

如何使用现有的数值向量来派生新变量

英文:

here is the data what i have

  1. temp &lt;- tibble::tribble(
  2. ~x1, ~x2, ~x3, ~x4, ~x5,
  3. 1, 2, 3, &#39;AA&#39;, &#39;BB&#39;,
  4. 2, 3, 4, &#39;AB&#39;, &#39;CC&#39;,
  5. 3, 4, 5, &#39;AC&#39;, &#39;DD&#39;,
  6. 4, 5, 6, &#39;AD&#39;, &#39;EE&#39;,
  7. 5, 6, 7, &#39;AE&#39;, &#39;FF&#39;,
  8. 6, 7, 8, &#39;AF&#39;, &#39;GG&#39;
  9. ) %&gt;% dplyr::mutate(dplyr::across(x4,as.character), dplyr::across(x5,as.character))

I want to derive new variables 'px1', 'px2', 'px3' by multiplying with a vector temp_val with values c(1,2,3)

  1. temp_val &lt;- c(1,2,3)

I tried the below code

  1. temp_val &lt;- c(1,2,3)
  2. nam1 &lt;- c(&#39;px1&#39;,&#39;px2&#39;,&#39;px3&#39;)
  3. temp5 &lt;- temp %&gt;% mutate(across(where(is.numeric), ~ . * temp_val, .names = &#39;{nam1}&#39;))

I get below data

如何使用现有的数值向量来派生新变量

my expectation, is to have vector values considered as columns and multiple variables with those columns

如何使用现有的数值向量来派生新变量

答案1

得分: 3

我们可以创建一个命名向量，然后根据从列名循环追加p得到的名称，对temp_val的元素进行子集化。

  1. library(dplyr)
  2. library(stringr)
  3. names(temp_val) <- nam1
  4. temp %>% 
  5.   mutate(across(where(is.numeric), ~ 
  6.   .x * temp_val[str_c("p", cur_column())], .names = "p{.col}"))

-output

  1. # A tibble: 6 × 8
  2.      x1    x2    x3 x4    x5      px1   px2   px3
  3.   <dbl> <dbl> <dbl> <chr> <chr> <dbl> <dbl> <dbl>
  4. 1     1     2     3 AA    BB        1     4     9
  5. 2     2     3     4 AB    CC        2     6    12
  6. 3     3     4     5 AC    DD        3     8    15
  7. 4     4     5     6 AD    EE        4    10    18
  8. 5     5     6     7 AE    FF        5    12    21
  9. 6     6     7     8 AF    GG        6    14    24

或者如果名称顺序相同，可以使用map2的另一种选项。

  1. library(purrr)
  2. map2_dfc(temp %>% 
  3.   select(where(is.numeric)), temp_val, `*`) %>% 
  4.  setNames(nam1)%>% 
  5.  bind_cols(temp, .)
  6. # A tibble: 6 × 8
  7.      x1    x2    x3 x4    x5      px1   px2   px3
  8.   <dbl> <dbl> <dbl> <chr> <chr> <dbl> <dbl> <dbl>
  9. 1     1     2     3 AA    BB        1     4     9
  10. 2     2     3     4 AB    CC        2     6    12
  11. 3     3     4     5 AC    DD        3     8    15
  12. 4     4     5     6 AD    EE        4    10    18
  13. 5     5     6     7 AE    FF        5    12    21
  14. 6     6     7     8 AF    GG        6    14    24
英文:

We may create a named vector and then subset the elements of the temp_val based on the name derived from appending p to the column name looped (cur_column())

  1. library(dplyr)
  2. library(stringr)
  3. names(temp_val) &lt;- nam1
  4. temp %&gt;% 
  5.   mutate(across(where(is.numeric), ~ 
  6.   .x * temp_val[str_c(&quot;p&quot;, cur_column())], .names = &quot;p{.col}&quot;))

-output

  1. # A tibble: 6 &#215; 8
  2.      x1    x2    x3 x4    x5      px1   px2   px3
  3.   &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt; &lt;chr&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
  4. 1     1     2     3 AA    BB        1     4     9
  5. 2     2     3     4 AB    CC        2     6    12
  6. 3     3     4     5 AC    DD        3     8    15
  7. 4     4     5     6 AD    EE        4    10    18
  8. 5     5     6     7 AE    FF        5    12    21
  9. 6     6     7     8 AF    GG        6    14    24

Or another option if the names are in the same order, then use map2

  1. library(purrr)
  2. map2_dfc(temp %&gt;% 
  3.   select(where(is.numeric)), temp_val, `*`) %&gt;% 
  4.  setNames(nam1)%&gt;% 
  5.  bind_cols(temp, .)
  6. # A tibble: 6 &#215; 8
  7.      x1    x2    x3 x4    x5      px1   px2   px3
  8.   &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt; &lt;chr&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
  9. 1     1     2     3 AA    BB        1     4     9
  10. 2     2     3     4 AB    CC        2     6    12
  11. 3     3     4     5 AC    DD        3     8    15
  12. 4     4     5     6 AD    EE        4    10    18
  13. 5     5     6     7 AE    FF        5    12    21
  14. 6     6     7     8 AF    GG        6    14    24

答案2

得分: 2

除了 @akrun 的回答之外，您还可以使用逐行计算：

  1. library(dplyr)
  2. library(tidyr)
  3. temp <- tibble::tribble(
  4.   ~x1, ~x2, ~x3, ~x4, ~x5,
  5.   1, 2, 3, 'AA', 'BB',
  6.   2, 3, 4, 'AB', 'CC',
  7.   3, 4, 5, 'AC', 'DD',
  8.   4, 5, 6, 'AD', 'EE',
  9.   5, 6, 7, 'AE', 'FF',
  10.   6, 7, 8, 'AF', 'GG'
  11. ) %>% dplyr::mutate(dplyr::across(x4, as.character), dplyr::across(x5, as.character))
  13. temp_val <- c(1, 2, 3)
  14. nam1 <- c('px1', 'px2', 'px3')
  16. temp %>% 
  17.   rowwise() %>% 
  18.   transmute(newvars = list(c_across(where(is.numeric)) * temp_val)) %>% 
  19.   unnest_wider(newvars) %>% 
  20.   setNames(nam1) %>% 
  21.   bind_cols(temp, .)

在 2023-01-08 由 reprex 包 (v2.0.1) 创建

英文:

In addition to @akrun's answer, you could also use a row-wise calculation:

  1. library(dplyr)
  2. #&gt; 
  3. #&gt; Attaching package: &#39;dplyr&#39;
  4. #&gt; The following objects are masked from &#39;package:stats&#39;:
  5. #&gt; 
  6. #&gt;     filter, lag
  7. #&gt; The following objects are masked from &#39;package:base&#39;:
  8. #&gt; 
  9. #&gt;     intersect, setdiff, setequal, union
  10. library(tidyr)
  11. temp &lt;- tibble::tribble(
  12.   ~x1, ~x2, ~x3, ~x4, ~x5,
  13.   1, 2, 3, &#39;AA&#39;, &#39;BB&#39;,
  14.   2, 3, 4, &#39;AB&#39;, &#39;CC&#39;,
  15.   3, 4, 5, &#39;AC&#39;, &#39;DD&#39;,
  16.   4, 5, 6, &#39;AD&#39;, &#39;EE&#39;,
  17.   5, 6, 7, &#39;AE&#39;, &#39;FF&#39;,
  18.   6, 7, 8, &#39;AF&#39;, &#39;GG&#39;
  19. ) %&gt;% dplyr::mutate(dplyr::across(x4,as.character), dplyr::across(x5,as.character))
  22. temp_val &lt;- c(1,2,3)
  23. nam1 &lt;- c(&#39;px1&#39;,&#39;px2&#39;,&#39;px3&#39;)
  26. temp %&gt;% 
  27.   rowwise() %&gt;% 
  28.   transmute(newvars = list(c_across(where(is.numeric))*temp_val)) %&gt;% 
  29.   unnest_wider(newvars) %&gt;% 
  30.   setNames(nam1) %&gt;% 
  31.   bind_cols(temp, .)
  32. #&gt; # A tibble: 6 &#215; 8
  33. #&gt;      x1    x2    x3 x4    x5      px1   px2   px3
  34. #&gt;   &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt; &lt;chr&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
  35. #&gt; 1     1     2     3 AA    BB        1     4     9
  36. #&gt; 2     2     3     4 AB    CC        2     6    12
  37. #&gt; 3     3     4     5 AC    DD        3     8    15
  38. #&gt; 4     4     5     6 AD    EE        4    10    18
  39. #&gt; 5     5     6     7 AE    FF        5    12    21
  40. #&gt; 6     6     7     8 AF    GG        6    14    24

<sup>Created on 2023-01-08 by the reprex package (v2.0.1)</sup>

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  • 本文由 发表于 2023年1月9日 01:27:36
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