Question

我有关于客户及其访问过的商店的数据（至少一次）。

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个两个商店的组合。

如何转换以前的数据结构（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)

修改关于Henrik的解决方案：正如您所看到的，我对配对存在问题。

 # 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

Answer 2

也许这不是最有效的方法，但它有效：

df <- data.frame(Customer=c(1,1,2,2,3,4,4,4), Store=c('A', 'B', 'A', 'C', 'A', 'A', 'B', 'C'))

cmb <- t(combn(unique(as.character(df$Store)),m=2))
count <- rep(0,nrow(cmb))

for (i in unique(df$Customer)){
  for (j in 1:nrow(cmb)){
    count[j] <- count[j]+as.numeric(all(cmb[j,] %in% df$Store[df$Customer==i]))
  }
}

res <- data.frame(Store1=cmb[,1], Store2=cmb[,2], Nb_customer=count)

      Store1 Store2 Nb_customer
1      A      B           2
2      A      C           2
3      B      C           1

修改

使用关联规则，您可以这样做：

# load library arulas library(arules) #original data frame df <- data.frame(Customer=c(1,1,2,2,3,4,4,4), Store=c('A', 'B', 'A', 'C', 'A', 'A', 'B', 'C')) # create list a_list <- lapply(unique(df$Customer),function(x)df$Store[df$Customer==x]) ## set transaction names names(a_list) <- paste("Tr",unique(df$Customer), sep = "") a_list ## coerce into transactions trans <- as(a_list, "transactions") # create association rules rules <- apriori(trans, parameter=list(minlen=2, maxlen=2, ext=TRUE, originalSupport=FALSE)) # calculate frequency of pairs of stores rules@quality$abs_support <- rules@quality$support*length(trans) inspect(rules) lhs rhs support confidence lhs.support lift abs_support 1 {B} => {A} 0.5 1 0.5 1 2 2 {C} => {A} 0.5 1 0.5 1 2

abs_support是同时发生的次数

Answer 3

这样的东西？

d<-data.frame(v1=c(1,1,2,2,3,4,4,4),v2=c("A","B","A","C","A","A","B","C"))
df<-as.data.frame.matrix(table(d))
which(df$A==1 & df$B==1)
which(df$A==1 & df$C==1)
which(df$B==1 & df$C==1)

组合的发生频率（2乘2）

3 个答案: