Question

考虑下表和伪查询：

Distinct Customers
WHERE
Most common PaymentMethod = 'CreditCard'
AND
Most common DeliveryService = '24hr'

Customer    TransID   PaymentMethod   DeliveryService
-----------------------------------------------------
Susan       1         CreditCard        24hr
Susan       2         CreditCard        24hr
Susan       3         Cash              24hr
John        4         CreditCard        48hr
John        5         CreditCard        48hr
Diane       6         CreditCard        24hr
Steve       7         Paypal            24hr
Steve       8         CreditCard        48hr
Steve       9         Paypal            24hr


Should return (2) records:

Customer
---------
Susan
Diane

另一种看待它的方法是我想排除少数民族案件，即： 我不想归还'史蒂夫'，因为虽然他曾经使用过一次信用卡，但他一般不会这样做，我只关心大多数行为，跨多个列。

实际上，有更多的列（10s）需要应用相同的原则，因此我采用的技术至少可以扩展到搜索100k的记录。

Answer 1

CREATE TABLE #D
(
Customer   VARCHAR(50), TransID INT,  PaymentMethod  VARCHAR(50),  DeliveryService VARCHAR(50)
)
INSERT INTO #D VALUES
('Susan',1,'CreditCard','24hr'),
('Susan',2,'CreditCard','24hr'),
('Susan',3,'Cash','24hr'),
('John ',4,'CreditCard','48hr'),
('John ',5,'CreditCard','48hr'),
('Diane',6,'CreditCard','24hr'),
('Steve',7,'Paypal','24hr'),
('Steve',8,'CreditCard','48hr'),
('Steve',9,'Paypal','24hr')


;with cte as
(
SELECT *,row_number() OVER (PARTITION BY PaymentMethod,Customer ORDER BY TransID) AS RN FROM #D
)
select DISTINCT Customer FROM cte  where    PaymentMethod = 'CreditCard'
AND DeliveryService = '24hr' and rn>1

Answer 2

试试这个..

CREATE TABLE #TEMP (Customer varchar(20),TransID INT, PaymentMethod varchar(20),DeliveryService VARCHAR(10))
    INSERT INTO #TEMP VALUES
    ('Susan',1,'CreditCard','24hr'),
    ('Susan',2,'CreditCard','24hr'),
    ('Susan',3,'Cash','24hr'),
    ('John',4,'CreditCard','48hr'),
    ('John',5,'CreditCard','48hr'),
    ('Diane',6,'CreditCard','24hr'),
    ('Steve',7,'Paypal','24hr'),
    ('Steve',8,'CreditCard','48hr'),
    ('Steve',9,'Paypal','24hr');



SELECT DISTINCT Customer FROM (
SELECT ROW_NUMBER () OVER (PARTITION BY PaymentMethod,Customer ORDER BY Customer) AS RNPaymentMethod,
ROW_NUMBER () OVER (PARTITION BY DeliveryService,Customer ORDER BY Customer) AS RNDeliveryService,Customer,TransID,PaymentMethod,DeliveryService FROM #TEMP) X
WHERE  X.PaymentMethod = 'CreditCard' AND X.DeliveryService = '24hr' AND X.RNPaymentMethod=1 AND X.RNDeliveryService=1

PS：我还为送货服务保留了额外的行号，因为您提到我们需要查看多列中的多数行为。

希望这有帮助！

Answer 3

好的，让我试试。

根据这个问题，您需要知道最常见的事件，我认为您必须声明一个完全返回此函数的函数：

对于这个例子，我使用了临时表的相同值，但是我创建了一个永久表，如果没有，我就无法创建和测试这些函数。我真的相信这些功能可以优化，但我没有时间做更多。

使用功能，您可以修改公式，并使其符合您的标准。

create function most_common_payment(@customer varchar(100))
returns varchar(100)
as
begin
    declare @total int, @payment varchar(100), @max_times int

    -- total records
    select @total = COUNT(*) from tempD where Customer=@customer;
    if @total = 0 return ''

    -- max ocurrences payment method
    select top 1 @payment = PaymentMethod, @max_times = count(*)
    from tempd 
    where Customer = @customer
    group by Customer, PaymentMethod
    order by COUNT(*) desc;
    if  @max_times <= 1 return '';

    -- percentatge
    if ((@max_times * 100) / @total) < 50 set @payment = '';

    return @payment;
end
go

和DeliveryService相同

crate function most_common_delivery(@customer varchar(100))
returns varchar(100)
as
begin

    declare @total int, @delivery varchar(100), @max_times int

    -- total records
    select @total = COUNT(*) from tempD where Customer=@customer;
    if @total = 0 return ''

    -- max ocurrences payment method
    select top 1 @delivery = DeliveryService, @max_times = count(*)
    from tempd 
    where Customer = @customer
    group by Customer, DeliveryService
    order by COUNT(*) desc;
    if  @max_times <= 1 return '';

    -- percentatge
    if ((@max_times * 100) / @total) < 50 set @delivery = '';

    return @delivery;
end

好的，现在我可以查询所需的结果：

select distinct
    Customer
    ,dbo.most_common_payment(tempd.Customer) as MostCommonPayment
    ,dbo.most_common_delivery(tempd.Customer) as MostCommonDelivery
from
    tempd
where
    dbo.most_common_payment(tempd.Customer) = 'CreditCard'
    and dbo.most_common_delivery(tempd.Customer) = '24hr'

这就是结果：

Customer   MostCommonPayment    MostCommonDelivery
--------   -----------------    ------------------
Susan      CreditCard           24hr

没有过滤器

Customer   MostCommonPayment    MostCommonDelivery
--------   -----------------    ------------------
Diane
John       CreditCard           48hr
Steve      Paypal               24hr
Susan      CreditCard           24hr

Answer 4

一种方法使用窗口函数和聚合：

with cp as (
     select customerid, paymentmethod, count(*) as cnt,
            rank() over (partition by customerid order by count(*) desc) as seqnum
     from t
     group by customerid, paymentmethod
    ),
    cd as (
     select customerid, deliveryservice, count(*) as cnt
            rank() over (partition by customerid over by count(*) desc) as seqnum
     from t
     group by customerid, deliveryservice
    )
select cp.customerid
from cp join
     cd
     on cp.customerid = cd.customerid
where (cp.seqnum = 1 and cp.PaymentMethod = 'CreditCard') and
      (cd.seqnum = 1 and cd.DeliveryService = '24hr');

因为您需要两个不同维度的排名，我认为您需要两个子查询（或等效的）。

根据其他列中最常见的值（不包括外围值）过滤结果

4 个答案: