[SOLVED] is there a C++ equivalent to SQL DENSE

is there a C++ equivalent to SQL DENSE_RANK() window function?

Let's say you have a table

Customer | Product
Dave     | Sneakers
Martin   | Tooth Brush
Andrew   | Shirt
Dave     | Tooth Brush

In SQL, you can do

SELECT Customer, Product, DENSE_RANK() OVER (ORDER BY Customer) AS customer_id, DENSE_RANK() OVER (ORDER BY Product) AS product_id

Customer | Product      | customer_id | product_id
Dave     | Sneakers     |  2          | 2
Martin   | Tooth Brush  |  3          | 3
Andrew   | Shirt        |  1          | 1
Dave     | Tooth Brush  |  2          | 3

Which can be usefull if you want to split the data into three tables

Customer (customer_id, customer_name)
Product (product_id, product_name)
Customer_Product(customer_id, product_id)

Is there a function in C++ that comes close to what is achieved with DENSE_RANK() in SQL ?

If a range over a structure, with a field for every column in C++ can be seen as an equivalent of a table in SQL.

With hindsight, after the question has been closed, here is the answer:

#include <vector>
#include <set>
#include <unordered_map>
#include <ranges>
using namespace std;

int main(){
    const vector<int> v = {7,9,9,7};
    const auto dense_rank = views::zip(v | ranges::to<set>(),views::iota(0)) | ranges::to<unordered_map>() ;
    return 0;
}

Unfortunately, it requires implicit conversion from tuple to pair, so it does not compile with g++14 at the moment.

Solution

No, there isn't a function in std like that. The range algorithms that are even close have a prerequisite that the input range be sorted in the desired order, and there isn't a sorted_view to get the desired orders from your data.