如何基于属性分配分组？

Question

想象一下，我列出了51个角色，每个角色都有其6项技能固有的标准化价值。现在，我想知道是否存在一种可编程的方法来准确而平等地将这些人分配到平等的团队中，并牢记技能水平。我不确定哪种数据格式更合适，但是直觉上我认为较长的数据集将使其更容易：

df <- structure(list(unique_id = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 
5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 
7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 
10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 
12L, 12L, 12L, 12L, 13L, 13L, 13L, 13L, 13L, 13L, 14L, 14L, 14L, 
14L, 14L, 14L, 15L, 15L, 15L, 15L, 15L, 15L, 16L, 16L, 16L, 16L, 
16L, 16L, 17L, 17L, 17L, 17L, 17L, 17L, 18L, 18L, 18L, 18L, 18L, 
18L, 19L, 19L, 19L, 19L, 19L, 19L, 20L, 20L, 20L, 20L, 20L, 20L, 
21L, 21L, 21L, 21L, 21L, 21L, 22L, 22L, 22L, 22L, 22L, 22L, 23L, 
23L, 23L, 23L, 23L, 23L, 24L, 24L, 24L, 24L, 24L, 24L, 25L, 25L, 
25L, 25L, 25L, 25L, 26L, 26L, 26L, 26L, 26L, 26L, 27L, 27L, 27L, 
27L, 27L, 27L, 28L, 28L, 28L, 28L, 28L, 28L, 29L, 29L, 29L, 29L, 
29L, 29L, 30L, 30L, 30L, 30L, 30L, 30L, 31L, 31L, 31L, 31L, 31L, 
31L, 32L, 32L, 32L, 32L, 32L, 32L, 33L, 33L, 33L, 33L, 33L, 33L, 
34L, 34L, 34L, 34L, 34L, 34L, 35L, 35L, 35L, 35L, 35L, 35L, 36L, 
36L, 36L, 36L, 36L, 36L, 37L, 37L, 37L, 37L, 37L, 37L, 38L, 38L, 
38L, 38L, 38L, 38L, 39L, 39L, 39L, 39L, 39L, 39L, 40L, 40L, 40L, 
40L, 40L, 40L, 41L, 41L, 41L, 41L, 41L, 41L, 42L, 42L, 42L, 42L, 
42L, 42L, 43L, 43L, 43L, 43L, 43L, 43L, 44L, 44L, 44L, 44L, 44L, 
44L, 45L, 45L, 45L, 45L, 45L, 45L, 46L, 46L, 46L, 46L, 46L, 46L, 
47L, 47L, 47L, 47L, 47L, 47L, 48L, 48L, 48L, 48L, 48L, 48L, 49L, 
49L, 49L, 49L, 49L, 49L, 50L, 50L, 50L, 50L, 50L, 50L, 51L, 51L, 
51L, 51L, 51L, 51L), attribute = structure(c(2L, 1L, 3L, 4L, 
5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 
3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 
2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 
5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 
3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 
2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 
5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 
3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 
2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 
5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 
3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 
2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 
5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 
3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 
2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 
5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 
3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 
2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 
5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L, 2L, 1L, 3L, 4L, 5L, 6L), .Label = c("Analytics", 
"Communication", "Creativity", "Problem solving", "Programming", 
"Project management"), class = "factor"), skill_level = c(1, 
1, 2, 1, 0, 0, 1, 2, 1, 1, 1, 1, 4, 2, 2, 3, 2, 4, 2, 1, 1, 2, 
2, 2, 2, 0, 0, 3, 0, 0, 2, 3, 3, 2, 2, 1, 2, 1, 1, 2, 2, 2, 2, 
3, 3, 3, 3, 1, 3, 3, 3, 3, 1, 3, 1, 1, 1, 2, 2, 2, 4, 0, 0, 2, 
0, 0, 3, 2, 3, 3, 2, 1, 1, 3, 4, 4, 4, 3, 3, 2, 3, 3, 3, 1, 2, 
2, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 0, 2, 2, 2, 2, 3, 
1, 2, 1, 1, 1, 0, 0, 0, 3, 2, 2, 3, 4, 3, 2, 2, 2, 2, 0, 2, 2, 
2, 1, 2, 0, 0, 3, 3, 4, 3, 2, 3, 2, 1, 0, 3, 0, 2, 2, 1, 1, 2, 
1, 1, 2, 1, 1, 2, 0, 1, 2, 3, 3, 3, 2, 2, 2, 2, 1, 2, 1, 1, 2, 
1, 1, 2, 1, 1, 0, 1, 2, 2, 0, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 2, 
1, 2, 1, 1, 1, 1, 1, 0, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 1, 3, 
2, 2, 3, 0, 1, 3, 2, 3, 2, 3, 2, 1, 1, 1, 2, 0, 2, 2, 2, 2, 2, 
2, 1, 2, 2, 2, 2, 2, 0, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 0, 2, 
2, 2, 3, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 2, 0, 2, 1, 2, 2, 
2, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 1, 4, 3, 3, 3, 2, 3, 2, 
2, 2, 3, 1, 2, 2, 3, 2, 3, 1, 3)), class = c("spec_tbl_df", "tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -306L))

[我的想法是以某种方式专注于每个技能组的运行平均值，但是我不知道从哪里开始。

也许，我使问题复杂化了，可以通过一组特定的分组和排序操作来解决。坦白说，我什至不知道如何搜索诸如此类的现有分配问题，这使我放慢了速度。

谢谢。

Answer 1

您所描述的听起来像您想要进行聚类分析。这是一个使用kmeans聚类的方法，分为4组：

library(tidyr)
library(dplyr)
library(broom)

mat <- df %>% 
  pivot_wider(id_cols = unique_id, names_from = attribute, values_from = skill_level) %>%
  select(-unique_id)

clust <- kmeans(mat, 4)

df_new <- mat %>% 
  mutate(group = clust$cluster)

group_average <- clust %>% 
  tidy() %>% 
  rename(Communication = x1,
         Analytics = x2,
         Creativity = x3,
         "Problem solving" = x4,
         Programming = x5,
         "Project management" = x6)
group_average
#> # A tibble: 4 x 9
#>   Communication Analytics Creativity `Problem solvin~ Programming
#>           <dbl>     <dbl>      <dbl>            <dbl>       <dbl>
#> 1         2.11       1.94      2.22              2.33       1.94 
#> 2         2          1.22      0.944             2.22       0.667
#> 3         0.833      1.33      1.5               1          0.5  
#> 4         2.78       2.67      2.89              3          2.33 
#> # ... with 4 more variables: `Project management` <dbl>, size <int>,
#> #   withinss <dbl>, cluster <fct>