Question

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

我可以使用以下方法获得“专业知识”的总数：

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

但是，我想要的是专业知识价值组合的计数。换句话说，有多少“ id”具有两种专业知识的相同组合，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

Answer 1

linked answer中的latemail's comment创建矩阵

crossprod(table(df) > 0)

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

OP希望使用长格式的数据帧。

1）交叉加入

下面是一个data.table解决方案，它使用了CJ()（交叉联接）功能：

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]与data.table或t(combn(df$expertise, 2))的{{1}}等效。

2）自加入

这是另一个使用 self-join 的变体：

combinat::combn2(df$expertise)

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

Answer 2

一种解决方案不如交叉产品表方法有效，但易于理解：

library(tidyr)

df %>% group_by(id) %>%
    summarize(expertise = list(combn(sort(expertise), 2, FUN = paste, collapse = '_'))) %>%
    unnest(expertise) %>%
    group_by(expertise) %>%
    summarize(count = n()) %>%
    separate(expertise, c('expertise1', 'expertise2'), sep = '_')

# # A tibble: 3 x 3
#   expertise1 expertise2 count
#   <chr>      <chr>      <int>
# 1 julia      python         3
# 2 julia      r              2
# 3 python     r              2

数据帧R中值组合的计数

2 个答案:

1）交叉加入

2）自加入