数据帧R中值组合的计数

Question

I have a dataframe like so:

    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

I can get the overall counts of "expertise" by using:

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

However what I want is the counts for combinations of expertise values. In other words how many "id" had the same combination of two expertise eg "r" and"julia"? Here is a desired output:

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

The linked answer from latemail's comment creates a matrix

crossprod(table(df) > 0)

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

while the OP expects a dataframe in long format.

1) cross join

Below is a data.table solution which uses the CJ() ( cross join ) function:

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] is the data.table equivalent for t(combn(df$expertise, 2)) or combinat::combn2(df$expertise) .

2) self-join

Here is another variant which uses a self-join :

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 

Answer 2

A solution not as efficient as crossprod-table approach but easy to understand:

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