生成唯一的用户项跨产品组合矩阵 [英] Generate matrix of unique user-item cross-product combinations

问题描述

我试图在R中创建唯一用户的跨产品矩阵.我在SO上搜索它，但找不到我想要的东西.任何帮助表示赞赏. 我有一个大数据框(超过一百万)，并显示了一个示例:

I am trying to create a cross-product matrix of unique users in R. I searched for it on SO but could not find what I was looking for. Any help is appreciated. I have a large dataframe (over a million) and a sample is shown:

df <- data.frame(Products=c('Product a', 'Product b', 'Product a', 
                            'Product c', 'Product b', 'Product c'),
                 Users=c('user1', 'user1', 'user2', 'user1', 
                         'user2','user3'))

df的输出是:

   Products Users
1 Product a user1
2 Product b user1
3 Product a user2
4 Product c user1
5 Product b user2
6 Product c user3

我想看两个矩阵: 第一个将显示拥有这两个产品(OR)的唯一用户的数量-因此输出将类似于:

I would like to see two matrices: The first one will show the number of unique users that had either products(OR) - so the output will be something like:

            Product a   Product b   Product c
Product a                 2            3
Product b     2                        3
Product c     3           3 

第二个矩阵将是同时拥有这两种产品(AND)的唯一身份用户数:

The second matrix will be the number of unique users that had both products(AND):

            Product a   Product b   Product c
Product a                 2            1
Product b     2                        1
Product c     1           1 

感谢您的帮助.

谢谢

更新:

这里更加清楚:User1和User2使用产品a.产品b由User1和User2使用，产品c由User1和User3使用.因此，在第一个矩阵中，由于有2个唯一用户，因此产品a和产品b将为2.类似地，乘积a和乘积c将为3.在第二个矩阵中，由于我想要交集，因此乘积将为2和1. 谢谢

Here is more clarity: Product a is used by User1 and User2. Product b is used by User1 and User2 and Product c is used by User1 and User3. So in the first matrix, Product a and Product b will be 2 since there are 2 unique users. Similarly, Product a and Product c will be 3. Where as in the second matrix, they would be 2 and 1 since I want the intersection. Thanks

推荐答案

尝试

lst <- split(df$Users, df$Products)
ln <- length(lst)
m1 <-  matrix(0, ln,ln, dimnames=list(names(lst), names(lst)))
m1[lower.tri(m1, diag=FALSE)] <- combn(seq_along(lst), 2, 
               FUN= function(x) length(unique(unlist(lst[x]))))
m1[upper.tri(m1)] <- m1[lower.tri(m1)]
m1
#          Product a Product b Product c
#Product a         0         2         3
#Product b         2         0         3
#Product c         3         3         0

或使用outer

f1 <- function(u, v) length(unique(unlist(c(lst[[u]], lst[[v]]))))
res <- outer(seq_along(lst), seq_along(lst), FUN= Vectorize(f1)) *!diag(3)
dimnames(res) <- rep(list(names(lst)),2)
res
#          Product a Product b Product c
#Product a         0         2         3
#Product b         2         0         3
#Product c         3         3         0

第二种情况

tcrossprod(table(df))*!diag(3)
#            Products
#Products    Product a Product b Product c
# Product a         0         2         1
# Product b         2         0         1
# Product c         1         1         0

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