How to Transform Data to Find Index with Same Value - r

Question

I intend to find customers who have bought exactly the same products,
The data I have is customers' behaviors--what they have bought.
The example that I provided is a simplified version of my data. Customers will usually buy 10 to 20 products. There are around 50 products that consumers could choose to buy.
I am really confused what is an easy way to transform my data into the output that I prefer.
Could you please give me any advice? Thanks
Input:
structure(list(Customer_ID = 1:6, Products = c("Apple, Beer, Diaper",
"Beer, Apple", "Beer, Apple, Diaper, Diaper", "Apple, Diaper",
"Diaper, Apple", "Apple, Diaper, Beer, Beer")), .Names = c("Customer_ID",
"Products"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-6L), spec = structure(list(cols = structure(list(Customer_ID = structure(list(), class = c("collector_integer",
"collector")), Products = structure(list(), class = c("collector_character",
"collector"))), .Names = c("Customer_ID", "Products")), default = structure(list(), class = c("collector_guess",
"collector"))), .Names = c("cols", "default"), class = "col_spec"))
Output:
structure(list(`Products Bought` = c("Apple, Beer, Diaper", "Apple, Diaper"
), Customer_ID = c("1, 3, 6", "4, 5")), .Names = c("Products Bought",
"Customer_ID"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-2L), spec = structure(list(cols = structure(list(`Products Bought` = structure(list(), class = c("collector_character",
"collector")), Customer_ID = structure(list(), class = c("collector_character",
"collector"))), .Names = c("Products Bought", "Customer_ID")),
default = structure(list(), class = c("collector_guess",
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

Answer 1

I am suspicious that you may want to look at structuring your data in a way that is more usable. In any case, the tidyverse can be a helpful way of thinking through your task.
As mentioned, posting code for others to start with can save them time and get you an answer faster.
library(dplyr)
library(stringr)
library(tidyr)
d <- data_frame(id=c(1,2,3,4,5,6)
, bought=c('Apple, Beer, Diaper','Apple, Beer', 'Apple, Beer, Diaper, Diaper'
, 'Apple, Diaper', 'Diaper, Apple', 'Apple, Diaper, Beer, Beer'))
d %>%
## Unnest the values & take care of white space
## - This is the better data structure to have, anyways
mutate(buy=str_split(bought,',')) %>%
unnest(buy) %>% mutate(buy=str_trim(buy)) %>% select(-bought) %>%
## Get distinct (and sort?)
distinct(id, buy) %>% arrange(id, buy) %>%
## Aggregate by id
group_by(id) %>% summarize(bought=paste(buy,collapse=', ')) %>% ungroup %>%
## Count
group_by(bought) %>% summarize(ids=paste(id,collapse=',')) %>% ungroup
EDIT: referencing this SO post for getting distinct combinations faster / cleaner in dplyr

Answer 2

Using the given input data and data.table, this can be written as (rather convoluted) "one-liner":
dcast(unique(setDT(input)[, strsplit(Products, ", "), Customer_ID])[
order(Customer_ID, V1)],
Customer_ID ~ ., paste, collapse = ", ")[
, .(Customers = paste(Customer_ID, collapse = ", ")), .(Products = .)]
# Products Customers
#1: Apple, Beer, Diaper 1, 3, 6
#2: Apple, Beer 2
#3: Apple, Diaper 4, 5
Note that the OP has dropped the second line with only one customer from
the expected output but hasn't mentioned any criteria for filtering the output in the question.
Input data
(As given by OP):
input <- structure(list(Customer_ID = 1:6, Products = c("Apple, Beer, Diaper",
"Beer, Apple", "Beer, Apple, Diaper, Diaper", "Apple, Diaper",
"Diaper, Apple", "Apple, Diaper, Beer, Beer")), .Names = c("Customer_ID",
"Products"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-6L), spec = structure(list(cols = structure(list(Customer_ID = structure(list(), class = c("collector_integer",
"collector")), Products = structure(list(), class = c("collector_character",
"collector"))), .Names = c("Customer_ID", "Products")), default = structure(list(), class = c("collector_guess",
"collector"))), .Names = c("cols", "default"), class = "col_spec"))