r - 按组合划分的出现频率(2 × 2)

Question

我拥有有关客户及其访问过的商店(至少一次)的数据。

Customer | Store

1   A
1   B
2   A
2   C
3   A
4   A
4   B
4   C

我想知道有多少用户访问了每2 家商店的组合。

如何转换先前的数据结构(使用 R)以获得以下结构？

Store 1 | Store 2 | Nb_Customer
A           B         2     (Customer 1 & 4 visited store A & B )
A           C         2     (Customer 2 & 4 visited store A & C)

编辑 关于 Henrik 的解决方案:如您所见，我对配对有疑问。

 # number of visits for each customer in each store 
> df <- data.frame(Customer=c(1,1,2,2,3,4,4,4), Store=c('A', 'B', 'A', 'C', 'A', 'A', 'B', 'C'))
> # number of visits for each customer in each store 
> tt <- with(df, table(df$Customer, df$Store))
> tt

    A B C
  1 1 1 0
  2 1 0 1
  3 1 0 0
  4 1 1 1
> 
> # number of stores
> n <- with(df, length(unique(df$Store)))
> n
[1] 3
> 
> # all pairs of column numbers, to be selected from the table tt
> cols <- with(df, combn(n, 2))
> cols
     [,1] [,2] [,3]
[1,]    1    1    2
[2,]    2    3    3
> 
> # pairs of stores
> pair <- t(with(df, combn(unique(df$Store), 2)))
> pair
     [,1] [,2]
[1,] "A"  "B" 
[2,] "1"  "3" 
[3,] "2"  "3" 

Answer 1

另一种可能性:

# number of visits for each customer in each store 
tt <- with(df, table(Customer, Store))
tt

# number of stores
n <- with(df, length(unique(Store)))
n

# all pairs of column numbers, to be selected from the table tt
cols <- with(df, combn(n, 2))
cols

# pairs of stores
pair <- t(with(df, combn(unique(Store), 2)))
pair

# select pairs of columns from tt
# count number of rows for which each customer has visited more than one store
# combine the counts with names of stores from 'pairs' to a data frame
ll <- lapply(seq(ncol(cols)), function(x){
  tt2 <- tt[ , cols[ , x]]
  n_cust <- sum(rowSums(tt2) > 1)
  data.frame(store1 = pair[x, 1], store2 = pair[x, 2], n_cust = n_cust)
})
ll

# convert list to data frame
df2 <- do.call(rbind, ll)
df2

#   store1 store2 n_cust
# 1      A      B      2
# 2      A      C      2
# 3      B      C      1