计算一组列的按行加权和

huangapple
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英文:

Calculate the row-wise weighted sum for a set of columns

问题

以下是代码的翻译部分：

  1. 我有如下的数据框：
  3.     > library(tidyverse)
  4.     > dd <- tibble(a = rep(1,10), b = rep(1,10), c = rep(1,10))
  5.     > dd
  6.     # A tibble: 10 × 3
  7.            a     b     c
  8.        <dbl> <dbl> <dbl>
  9.      1     1     1     1
  10.      2     1     1     1
  11.      3     1     1     1
  12.      4     1     1     1
  13.      5     1     1     1
  14.      6     1     1     1
  15.      7     1     1     1
  16.      8     1     1     1
  17.      9     1     1     1
  18.     10     1     1     1
  20. 以及一个权重向量：
  22.     > weight <- c(1, 5, 10)
  23.     > weight
  24.     [1]  1  5 10
  26. 当我想要计算数据框的所有列的加权行和时，我执行以下操作：
  28.     > dd %>% mutate(m = rowSums(map2_dfc(dd, weight,`*`)))
  29.     # A tibble: 10 × 4
  30.            a     b     c     m
  31.        <dbl> <dbl> <dbl> <dbl>
  32.      1     1     1     1    16
  33.      2     1     1     1    16
  34.      3     1     1     1    16
  35.      4     1     1     1    16
  36.      5     1     1     1    16
  37.      6     1     1     1    16
  38.      7     1     1     1    16
  39.      8     1     1     1    16
  40.      9     1     1     1    16
  41.     10     1     1     1    16
  43. 但是我不知道如何计算数据框的**子集**的加权行和。我尝试了下面的代码，但结果混乱不堪：
  45.     > dd %>% rowwise() %>% mutate(m = rowwise(map2_dfc(c_across(b:c), weight[2:3],`*`)))
  46.     New names:
  47.     • `` -> `...1`
  48.     • `` -> `...2`
  49.     New names:
  50.     • `` -> `...1`
  51.     • `` -> `...2`
  52.     New names:
  53.     • `` -> `...1`
  54.     • `` -> `...2`
  55.     New names:
  56.     • `` -> `...1`
  57.     • `` -> `...2`
  58.     New names:
  59.     • `` -> `...1`
  60.     • `` -> `...2`
  61.     New names:
  62.     • `` -> `...1`
  63.     • `` -> `...2`
  64.     New names:
  65.     • `` -> `...1`
  66.     • `` -> `...2`
  67.     New names:
  68.     • `` -> `...1`
  69.     • `` -> `...2`
  70.     # A tibble: 10 × 4
  71.     # Rowwise: 
  72.            a     b     c m$...1 $...2
  73.        <dbl> <dbl> <dbl>  <dbl>  <dbl>
  74.      1     1     1     1      5    10
  75.      2     1     1     1      5    10
  76.      3     1     1     1      5    10
  77.      4     1     1     1      5    10
  78.      5     1     1     1      5    10
  79.      6     1     1     1      5    10
  80.      7     1     1     1      5    10
  81.      8     1     1     1      5    10
  82.      9     1     1     1      5    10
  83.     10     1     1     1      5    10
  85. 请问有人可以给我一些关于如何解决这个问题的提示吗？
英文:

I have, say, the following data frame:

  1. &gt; library(tidyverse)
  2. &gt; dd &lt;- tibble(a = rep(1,10), b = rep(1,10), c = rep(1,10))
  3. &gt; dd
  4. # A tibble: 10 &#215; 3
  5. a     b     c
  6. &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
  7. 1     1     1     1
  8. 2     1     1     1
  9. 3     1     1     1
  10. 4     1     1     1
  11. 5     1     1     1
  12. 6     1     1     1
  13. 7     1     1     1
  14. 8     1     1     1
  15. 9     1     1     1
  16. 10     1     1     1

and a vector of weights:

  1. &gt; weight &lt;- c(1, 5, 10)
  2. &gt; weight
  3. [1]  1  5 10

when I want to calculate the row-wise weighted sum for all the columns of the dataframe together, I do this:

  1. &gt; dd %&gt;% mutate(m = rowSums(map2_dfc(dd, weight,`*`)))
  2. # A tibble: 10 &#215; 4
  3. a     b     c     m
  4. &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
  5. 1     1     1     1    16
  6. 2     1     1     1    16
  7. 3     1     1     1    16
  8. 4     1     1     1    16
  9. 5     1     1     1    16
  10. 6     1     1     1    16
  11. 7     1     1     1    16
  12. 8     1     1     1    16
  13. 9     1     1     1    16
  14. 10     1     1     1    16

but I don't know how to calculate the row-wise weighted sum for a subset of the data frame. I tried the code below, but it gives messy results:

  1. &gt; dd %&gt;% rowwise() %&gt;% mutate(m = rowwise(map2_dfc(c_across(b:c), weight[2:3],`*`)))
  2. New names:
  3. • `` -&gt; `...1`
  4. • `` -&gt; `...2`
  5. New names:
  6. • `` -&gt; `...1`
  7. • `` -&gt; `...2`
  8. New names:
  9. • `` -&gt; `...1`
  10. • `` -&gt; `...2`
  11. New names:
  12. • `` -&gt; `...1`
  13. • `` -&gt; `...2`
  14. New names:
  15. • `` -&gt; `...1`
  16. • `` -&gt; `...2`
  17. New names:
  18. • `` -&gt; `...1`
  19. • `` -&gt; `...2`
  20. New names:
  21. • `` -&gt; `...1`
  22. • `` -&gt; `...2`
  23. New names:
  24. • `` -&gt; `...1`
  25. • `` -&gt; `...2`
  26. New names:
  27. • `` -&gt; `...1`
  28. • `` -&gt; `...2`
  29. New names:
  30. • `` -&gt; `...1`
  31. • `` -&gt; `...2`
  32. # A tibble: 10 &#215; 4
  33. # Rowwise: 
  34. a     b     c m$...1 $...2
  35. &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;  &lt;dbl&gt; &lt;dbl&gt;
  36. 1     1     1     1      5    10
  37. 2     1     1     1      5    10
  38. 3     1     1     1      5    10
  39. 4     1     1     1      5    10
  40. 5     1     1     1      5    10
  41. 6     1     1     1      5    10
  42. 7     1     1     1      5    10
  43. 8     1     1     1      5    10
  44. 9     1     1     1      5    10
  45. 10     1     1     1      5    10

Can someone please give me a hint as to how to approach this problem?

答案1

得分: 4

这是矩阵相乘。您的原始代码等同于 as.matrix(dd) %*% weight。在 mutate 内部的子集中，您可以这样做：

  1. dd %>% mutate(m = (across(b:c) %>% as.matrix()) %*% weight[1:2])
英文:

This is matrix multiplication. Your original is equivalent to as.matrix(dd) %*% weight. For a subset inside mutate you can do this:

  1. dd %&gt;% mutate(m = (across(b:c) %&gt;% as.matrix()) %*% weight[1:2])

答案2

得分: 2

使用 tidyverse 方法，我们可以创建一个命名向量用于 'weight'，通过列 'b' 到 'c' 进行循环 across，根据列名（cur_column()）来选择 'weight' 值，进行乘法并获得 rowSums。

  1. library(dplyr)
  2. names(weight) <- names(dd)
  3. dd %>% 
  4.    mutate(m = rowSums(across(b:c,  ~ .x * weight[cur_column()])))

-output

  1. # A tibble: 10 × 4
  2.        a     b     c     m
  3.    <dbl> <dbl> <dbl> <dbl>
  4.  1     1     1     1    15
  5.  2     1     1     1    15
  6.  3     1     1     1    15
  7.  4     1     1     1    15
  8.  5     1     1     1    15
  9.  6     1     1     1    15
  10.  7     1     1     1    15
  11.  8     1     1     1    15
  12.  9     1     1     1    15
  13. 10     1     1     1    15

或者如果我们想要使用 rowwise（不推荐，因为它速度较慢）

  1. dd %>% 
  2.   rowwise %>% 
  3.   mutate(m = sum(c_across(b:c) * weight[2:3])) %>% 
  4.   ungroup

或者使用 crossprod

  1. dd %>%
  2.    mutate(m = crossprod(t(pick(b:c)), weight[2:3])[,1])

或者使用 base R

  1. dd$m <-  rowSums(dd[2:3] * weight[2:3][col(dd[2:3])])
英文:

Using tidyverse methods, we can create a named vector for 'weight', loop across the columns 'b' to 'c', subset the 'weight' value based on the column name (cur_column()), multiply and get the rowSums

  1. library(dplyr)
  2. names(weight) &lt;- names(dd)
  3. dd %&gt;% 
  4. mutate(m = rowSums(across(b:c,  ~ .x * weight[cur_column()])))

-output

  1. # A tibble: 10 &#215; 4
  2. a     b     c     m
  3. &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
  4. 1     1     1     1    15
  5. 2     1     1     1    15
  6. 3     1     1     1    15
  7. 4     1     1     1    15
  8. 5     1     1     1    15
  9. 6     1     1     1    15
  10. 7     1     1     1    15
  11. 8     1     1     1    15
  12. 9     1     1     1    15
  13. 10     1     1     1    15

Or if we want to use rowwise (not recommended as it is slower)

  1. dd %&gt;% 
  2. rowwise %&gt;%
  3. mutate(m = sum(c_across(b:c) * weight[2:3])) %&gt;%
  4. ungroup

Or use crossprod

  1. dd %&gt;%
  2. mutate(m = crossprod(t(pick(b:c)), weight[2:3])[,1])

Or with base R

  1. dd$m &lt;-  rowSums(dd[2:3] * weight[2:3][col(dd[2:3])])

huangapple
  • 本文由 发表于 2023年4月4日 10:13:53
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  • dplyr
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  • tidyverse
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