SAS/SQL-在铁路行程数据中识别/分组回程

我在Oracle中有一个表，其中包含铁路行程。有些行程的行程类型为“S”，这意味着它们是单次行程。然而，一些客户基本上使用这些单程旅行进行回程（例如，购买一张从伦敦到曼彻斯特的单程票和一张从曼彻斯特到伦敦的单程票）。我需要能够识别这种情况，并以某种方式将两次旅行分为一行。更为复杂的是，“旅行原因”记录在事务级别，两次旅行可能有不同的事务ID——因此我需要能够在新行中保留此变量的两个值。我今天一直在努力解决这个问题，还没有找到一个可以接受的解决方案，所以我想在这里征求意见

以下是示例数据：

Cust_ID Journey_ID  Origin     Destination  Type    Date        Reason
100     100001      London       Manchester S       15/01/2014  Family
100     100100      Manchester   London     S       16/01/2014  Family
100     110023      London       Manchester S       25/01/2014  Family
100     114000      Manchester   London     S       29/01/2014  Holiday
100     129345      London       Norwich    S       02/02/2014  Business
100     134578      Norwich      London     S       15/02/2014  Business
100     145843      London       Manchester S       01/03/2014  Family
100     147893      Manchester   London     S       04/03/2014  Family
200     157878      Birmingham   London     S       04/04/2014  Friends
200     159899      London       Birmingham S       06/04/2014  Friends

我想创建如下内容：

Cust_ID Journey_ID  Origin    Destination   Date            Reason1      Reason2
100     100001      London    Manchester    15/01/2014      Family       Family
100     110023      London    Manchester    25/01/2014      Family       Holiday
100     129345      London    Norwich       02/02/2014      Business     Business
100     145843      London    Manchester    01/03/2014      Family       Family
200     157878      Birmingham    London    04/04/2014      Friends      Friends

我可用的工具是Oracle SQL Developer和SAS。如果您有任何关于这方面的想法，我们将不胜感激！该表包含数百万条记录，因此效率是一个问题

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编辑：忘记在表中包含事务\u id。出境和回程旅行可能相同，也可能不同

这里有一个解决方案，尽管使用哈希表可能有更好的方法。本质上，我的方法是首先将返回旅程的事务id更改为与出站旅程相同，然后输出所需结果就相当简单了。如果需要保留回程的原始事务id，则必须将数据输出到新的数据集，而不是使用MODIFY，然后添加到“R”类型的其他行程中。我添加了一些额外的数据行，包括两次回程（一次发生在两次单程之间）和一次没有回程的单程。希望这有帮助

/* create dummy dataset */
data have;
input Cust_ID Journey_ID Origin $ :12. Destination $ :12. Type $ Date :ddmmyy10. Reason $ Trans_id;
format date date9.;
datalines;
100     100001      London       Manchester S       15/01/2014  Family      1
100     100100      Manchester   London     S       16/01/2014  Family      2
100     110023      London       Manchester S       25/01/2014  Family      3
100     114000      Manchester   London     S       29/01/2014  Holiday     4
100     100300      London       Exeter     R       31/01/2014  Business    5
100     100300      Exeter       London     R       01/02/2014  Business    5
100     129345      London       Norwich    S       02/02/2014  Business    6
100     130300      Norwich      Ipswich    R       11/02/2014  Business    7
100     130300      Ipswich      Norwich    R       12/02/2014  Business    7
100     134578      Norwich      London     S       15/02/2014  Business    8
100     145843      London       Manchester S       01/03/2014  Family      9
100     147893      Manchester   London     S       04/03/2014  Family      10
100     148123      London       Brighton   S       06/03/2014  Family      11
200     157878      Birmingham   London     S       04/04/2014  Friends     12
200     159899      London       Birmingham S       06/04/2014  Friends     13
;
run;

/* sort data */
proc sort data=have;
by cust_id date trans_id;
run;

/* update transaction id for return journey to be same as outbound journey */
data have;
/* dataset entered twice to enable BY group processing */
modify have (where=(type='S')) have (where=(type='S')); 
by cust_id date trans_id;
retain _tid _orig _dest _pair;
if first.cust_id or _pair=1 then do;
        _tid = trans_id;
        _orig = origin;
        _dest = destination;
        _pair=0;
end;
/* does destination and origin match origin and destination of previous journey */
else if origin=_dest and destination=_orig then do;
    trans_id = _tid;
    _pair=1;
end;
run;

/* need to sort again as some trans_id's have changed */
proc sort data=have;
by cust_id trans_id descending date;
run;

/* output required data with both reason */
data want (keep=cust_id journey_id origin destination type date reason1 reason2);
set have;
by cust_id trans_id descending date;
length reason1 reason2 $12;
retain reason2;
if first.trans_id and not last.trans_id then reason2 = reason;
else if last.trans_id then do;
    reason1 = reason;
    output;
    call missing(reason1,reason2);
end;
run;