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r - 数据帧 R 中值组合的计数

问题描述

我有一个像这样的数据框：

    df<-structure(list(id = c("A", "A", "A", "B", "B", "C", "C", "D", 
"D", "E", "E"), expertise = c("r", "python", "julia", "python", 
"r", "python", "julia", "python", "julia", "r", "julia")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -11L), .Names = c("id", 
"expertise"), spec = structure(list(cols = structure(list(id = structure(list(), class = c("collector_character", 
"collector")), expertise = structure(list(), class = c("collector_character", 
"collector"))), .Names = c("id", "expertise")), default = structure(list(), class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

df
   id expertise
1   A         r
2   A    python
3   A     julia
4   B    python
5   B         r
6   C    python
7   C     julia
8   D    python
9   D     julia
10  E         r
11  E     julia

我可以通过使用以下方法获得“专业知识”的总体数量：

library(dplyr)    
df %>% group_by(expertise) %>% mutate (counts_overall= n())

但是，我想要的是专业值组合的计数。换句话说，有多少“id”具有相同的两种专业知识组合，例如“r”和“julia”？这是所需的输出：

df_out<-structure(list(expertise1 = c("r", "r", "python"), expertise2 = c("python", 
"julia", "julia"), count = c(2L, 2L, 3L)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -3L), .Names = c("expertise1", 
"expertise2", "count"), spec = structure(list(cols = structure(list(
    expertise1 = structure(list(), class = c("collector_character", 
    "collector")), expertise2 = structure(list(), class = c("collector_character", 
    "collector")), count = structure(list(), class = c("collector_integer", 
    "collector"))), .Names = c("expertise1", "expertise2", "count"
)), default = structure(list(), class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

df_out
  expertise1 expertise2 count
1          r     python     2
2          r      julia     2
3     python      julia     3

标签： rdataframecombinations

解决方案

来自latemail评论的链接答案创建了一个矩阵

crossprod(table(df) > 0)

         expertise
expertise julia python r
   julia      4      3 2
   python     3      4 2
   r          2      2 3

而 OP 需要一个长格式的数据帧。

1) 交叉连接

以下是data.table使用CJ()（交叉连接）功能的解决方案：

library(data.table)
setDT(df)[, CJ(expertise, expertise)[V1 < V2], by = id][
  , .N, by = .(expertise1 = V1, expertise2 = V2)]

   expertise1 expertise2 N
1:      julia     python 3
2:      julia          r 2
3:     python          r 2

CJ(expertise, expertise)[V1 < V2]是 or 的等价data.table物。t(combn(df$expertise, 2))combinat::combn2(df$expertise)

2) 自加入

这是另一个使用自连接的变体：

library(data.table)
setDT(df)[df, on = "id", allow = TRUE][
  expertise < i.expertise, .N, by = .(expertise1 = expertise, expertise2 = i.expertise)]

   expertise1 expertise2 N
1:     python          r 2
2:      julia          r 2
3:      julia     python 3

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