通过R中另一列的成对组合来计算一列的唯一值 [英] Count unique values of a column by pairwise combinations of another column in R

问题描述

假设我有以下数据框：

   ID Code
1   1    A
2   1    B
3   1    C
4   2    B
5   2    C
6   2    D
7   3    C
8   3    A
9   3    D
10  3    B
11  4    D
12  4    B

我想通过代码列的成对组合来获得 ID列的唯一值的计数：

I would like to get the count of unique values of the column "ID" by pairwise combinations of the column "Code":

  Code.Combinations Count.of.ID
1              A, B           2
2              A, C           2
3              A, D           1
4              B, C           3
5              B, D           3
6              C, D           2

我已经搜索了解决方案在线，到目前为止还无法达到预期的效果。
任何帮助将不胜感激。谢谢！

I have searched for solution(s) online, so far haven't been able to achieve the desired result. Any help would be appreciated. Thanks!

推荐答案

这是 data.table 的解决方法问题。使用 combn 函数选择所有可能的代码组合，然后为每个唯一的 CodeComb 计数ID：

Here is a data.table way to solve the problem. Use combn function to pick up all possible combinations of Code and then count ID for each unique CodeComb:

library(data.table)
setDT(df)[, .(CodeComb = sapply(combn(Code, 2, simplify = F), 
                                function(cmb) paste(sort(cmb), collapse = ", "))), .(ID)]
# list all combinations of Code for each ID
         [, .(IdCount = .N), .(CodeComb)]    
# count number of unique id for each code combination

#    CodeComb IdCount
# 1:     A, B       2
# 2:     A, C       2
# 3:     B, C       3
# 4:     B, D       3
# 5:     C, D       2
# 6:     A, D       1

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