SQL-запрос для комбинаций без повторения

Мне нужен запрос, который можно использовать в (или в качестве) функции и получить все комбинации из n значений. И мне нужны все комбинации длины k, где k = 1..n.

Расширенный выборочный ввод и результат, так что вход имеет 3 значения вместо 2 - однако количество входных значений может варьироваться от 1 до n.

Пример: вход: таблица со значениями в одном столбце в нескольких строках

Value  (nvarchar(500))
------
Ann
John
Mark

Выход № 1: таблица со значениями, объединенными в один столбец

    Ann
    John
    Mark
    Ann,John
    John,Mark
    Ann,Mark
    Ann,John,Mark

Для относительно небольших значений n(20 в этом примере) вы можете использовать метод, который использует тот факт, что натуральные целые числа являются комбинациями битов.

Решение T-SQL

Образец данных:

DECLARE @Sample AS TABLE 
(
    item_id     tinyint IDENTITY(1,1) PRIMARY KEY NONCLUSTERED,
    item        nvarchar(500) NOT NULL,
    bit_value   AS 
                CONVERT
                (
                    integer, 
                    POWER(2, item_id - 1)
                )
                PERSISTED UNIQUE CLUSTERED
);    

INSERT @Sample
    (item)
VALUES
    (N'Ann'),
    (N'Bob'),
    (N'Charles'),
    (N'Darren'),
    (N'Eric'),
    (N'Fred'),
    (N'George'),
    (N'Harry'),
    (N'Ian'),
    (N'John'),
    (N'Keith'),
    (N'Larry'),
    (N'Mark'),
    (N'Nathan'),
    (N'Owen'),
    (N'Paul'),
    (N'Quentin'),
    (N'Ryan'),
    (N'Steve'),
    (N'Terry');

Решение:

-- Maximum integer we need
-- for all combinations of 'n' bits
DECLARE 
    @max integer = 
    POWER(2,
        (
            SELECT COUNT(*) 
            FROM @Sample AS s
        )
    ) - 1;

SELECT
    combination =
        STUFF
        (
            (
                -- Choose items where the bit is set
                -- and concatenate all matches
                SELECT ',' + s.item 
                FROM @Sample AS s
                WHERE
                    n.n & s.bit_value = s.bit_value
                ORDER BY
                    s.bit_value
                FOR XML 
                    PATH (''),
                    TYPE                    
            ).value('(./text())[1]', 'varchar(8000)'), 1, 1, ''
        )
-- A standard numbers table
-- (single column, integers from 1 to 1048576, indexed)
FROM dbo.Numbers AS N
WHERE
    N.n BETWEEN 1 AND @max;

Выходной образец:

╔════════════════════════╗
║      combination       ║
╠════════════════════════╣
║ Ann                    ║
║ Bob                    ║
║ Ann,Bob                ║
║ Charles                ║
║ Ann,Charles            ║
║ Bob,Charles            ║
║ Ann,Bob,Charles        ║
║ Darren                 ║
║ Ann,Darren             ║
║ Bob,Darren             ║
║ Ann,Bob,Darren         ║
║ Charles,Darren         ║
║ Ann,Charles,Darren     ║
║ Bob,Charles,Darren     ║
║ Ann,Bob,Charles,Darren ║
║ ...                    ║
╚════════════════════════╝

План выполнения:

Для записи 1,048,576 комбинаций в переменную на моем ноутбуке требуется 41 секунда . Благодаря принудительному параллелизму я смог сократить время выполнения до 13 секунд.DOP 8

Если вам нужна таблица Numbers, это быстрый способ ее создания:

SELECT TOP (1048576)
    n = ISNULL(CONVERT(integer, ROW_NUMBER() OVER (ORDER BY (SELECT NULL))), 0)
INTO dbo.Numbers
FROM sys.columns AS c
CROSS JOIN sys.columns AS c2
CROSS JOIN sys.columns AS c3;

CREATE UNIQUE CLUSTERED INDEX cuq
ON dbo.Numbers (n)
WITH (MAXDOP = 1, SORT_IN_TEMPDB = ON);

SQLCLR решение

Гораздо более эффективная реализация возможна в SQLCLR (SQL Server 2005 и далее):

CREATE ASSEMBLY [Demo]
    AUTHORIZATION [dbo]
    FROM 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GO
CREATE FUNCTION dbo.Combinations
(
    @ElementsCSV nvarchar (4000)
)
RETURNS TABLE
(
    Combination nvarchar (4000) NULL
)
AS EXTERNAL NAME Demo.UserDefinedFunctions.Permute;

Пример использования:

SELECT 
    f.Combination
FROM dbo.Combinations('A,B,C,D') AS f;

Выход:

╔═════════════╗
║ Combination ║
╠═════════════╣
║ A           ║
║ B           ║
║ A,B         ║
║ C           ║
║ A,C         ║
║ B,C         ║
║ A,B,C       ║
║ D           ║
║ A,D         ║
║ B,D         ║
║ A,B,D       ║
║ C,D         ║
║ A,C,D       ║
║ B,C,D       ║
║ A,B,C,D     ║
╚═════════════╝

Используя набор из 20 элементов из ранее:

DECLARE @Sample AS TABLE 
(
    item_id     tinyint IDENTITY(1,1) PRIMARY KEY CLUSTERED,
    item        nvarchar(50) NOT NULL
);

INSERT @Sample
    (item)
VALUES
    (N'Ann'),
    (N'Bob'),
    (N'Charles'),
    (N'Darren'),
    (N'Eric'),
    (N'Fred'),
    (N'George'),
    (N'Harry'),
    (N'Ian'),
    (N'John'),
    (N'Keith'),
    (N'Larry'),
    (N'Mark'),
    (N'Nathan'),
    (N'Owen'),
    (N'Paul'),
    (N'Quentin'),
    (N'Ryan'),
    (N'Steve'),
    (N'Terry');

Решение SQLCLR:

-- Create CSV input
DECLARE 
    @Elements nvarchar(4000) =
        STUFF
        (
            (
                SELECT ',' + s.item 
                FROM @Sample AS s
                ORDER BY
                    s.item_id
                FOR XML 
                    PATH (''),
                    TYPE                    
            ).value('(./text())[1]', 'varchar(8000)'), 1, 1, ''
        );

DECLARE
    @bitbucket nvarchar(4000);

SELECT
    @bitbucket = combination
FROM dbo.Combinations(@Elements);

Время выполнения составляет 2,5 секунды для 1 048 576 комбинаций на моем ноутбуке в DOP 1.

Создание входа CSV:

Нахождение комбинаций:

Исходный код C #:

using System;
using System.Collections;
using Microsoft.SqlServer.Server;

public partial class UserDefinedFunctions
{
    [SqlFunction
        (
        DataAccess=DataAccessKind.None, 
        SystemDataAccess=SystemDataAccessKind.None,
        FillRowMethodName="FillRow",
        TableDefinition="Permutation nvarchar(4000)"
        )
    ]
    public static IEnumerable Permute(string ElementsCSV)
    {
        // Split CSV
        string[] elements = ElementsCSV.Split(new char[] { ',' });

        // Highest integer needed
        int count = (int)Math.Pow(2, elements.Length) - 1;

        // Pre-computed array of 2^n values
        int[] powers = new int[elements.Length];

        for (int i = 0; i < powers.Length; i++)
        {
            powers[i] = (int)Math.Pow(2, i);
        }

        // Test integers
        for (int i = 1; i <= count; i++)
        {
            // Reset output
            string s = string.Empty;

            // Test each bit that could be set
            for (int bit = 0; bit < powers.Length && i >= powers[bit]; bit++)
            {
                if ((i & powers[bit]) == powers[bit])
                {
                    // Add the element corresponding to the set bit
                    s += elements[bit] + ',';
                }
            }

            // Return a row via enumeration
            yield return s.Substring(0, s.Length - 1);
        }
    }

    // Called by SQL Server to fetch a row
    public static void FillRow(object o, out string Permutation)
    {
        Permutation = (string)o;
    }
}