此纯sql代码在5分钟内（约8核/ 32 gb ram）将大约35000条记录分组。

This pure sql code grouped about 35000 records in 5 minutes (8 cores/32 gb ram). Enjoy.

--table with RELATIONS, idea is to place every related item in a bucket
create table RELATIONS
(
    bucket int,        -- initially 0
    bucketsub int,    -- initially 0
    relnr1 float,    
    relnr2 float    -- relnr1 = a, relnr2 = b means a and b are related
)

create table ##BucketRelnrs ( relnr float ); --table functions as temp list
declare @bucket int; 
declare @bucketsub int;
declare @nrOfUpdates int;
declare @relnr float;

set @bucket=0;
set @relnr=0;
WHILE @relnr>=0 and @bucket<50000 --to prevent the while loop from hanging.
BEGIN
    set @bucket = @bucket+1
    set @bucketsub=1;

    set @relnr = (select isnull(min(relnr1),-1) from RELATIONS where bucket=0); --pick the smallest relnr that has not been assigned a bucket yet
    set @nrOfUpdates = (select count(*) from RELATIONS where bucket=0 and (relnr1=@relnr or relnr2=@relnr));
    update RELATIONS set bucket=@bucket, bucketsub=@bucketsub where bucket=0 and (relnr1=@relnr or relnr2=@relnr);
    set @bucketsub = @bucketsub+1;

    WHILE @nrOfUpdates>0 and @bucketsub<=10    --to prevent the inner while loop from hanging, actually determines the number of iterations
    BEGIN
        --refill temp list with newly found related relnrs
        truncate table ##BucketRelnrs;
        insert into ##BucketRelnrs
        select distinct relnr1 from RELATIONS where bucket=@bucket
        union select distinct relnr2 from RELATIONS where bucket=@bucket;

        --calculate the number of relations that will be updated next, if zero then stop iteration
        set @nrOfUpdates =
        (
            select count(*) from RELATIONS where bucket=0
            and (relnr1 in (select relnr from ##BucketRelnrs) or relnr2 in (select relnr from ##BucketRelnrs))
        );

        --update the RELATIONS table
        update RELATIONS set bucket=@bucket, bucketsub=@bucketsub where bucket=0
        and (relnr1 in (select relnr from ##BucketRelnrs) or relnr2 in (select relnr from ##BucketRelnrs));

        set @bucketsub = @bucketsub+1;
    END
END

drop table ##BucketRelnrs; --clean temp table