Sorting SAS中的复杂连接

我正在尝试加入以下两个数据集：

data testA;
input categorical $3. value;
*order = _n_;
datalines;
Dog. 
M  7
F  5
Cat.
M  4
F  2
;
run;

data testA;
set testA;
order=_n_;
run;

data testB;
input categorical $2. value;
datalines;
Dog. 
F  3
Cat.
M  1
F  2
;
run;

proc sql;
create table final as
select a.*,b.* from testA a left join testB b on 
a.categorical=b.categorical
order by order;
quit;

我期望的输出如下：

data testA;
input categorical $ value value2;
datalines;
Dog . .
M 7 .
F 5 3
Cat . .
M 4 1
F 2 2
;
run;

---

我遇到的问题是1）“分类”id没有按字母顺序排序，我不想改变它的顺序2）因为有两个Ms和Fs，我不知道如何在不重命名mf的情况下进行连接，以使其唯一3）它可能是一个内部连接，因为值中可能包含的内容可能不包含在值中2

如果您的数据具有该类别值作为散布的行，您需要创建第三列来保存在通过数据集时发现的值。对于讨论，将此新列命名为

组

——它也将是分类的，并且在其他类别列的“上方”具有层次结构。它是执行复杂联接所需的“合成”类别，将从最终结果中丢弃

want

join将是一个简单的“黑匣子”，包括分组、合并、鬼鬼祟祟的数学运算和行和的组和

示例代码创建了一个表

fulljoin_peek

，该表不是结果所需的，但将提供对通过黑盒的数据流的深入了解。该代码还处理类别在组中重复的“真实数据”情况

样本数据：

data testA;
input categorical $3. value;
datalines;
Dog .   * missing means categorical is really group
M  7
F  5
Cat .
M  4
F  2
Rat .   * B does not have rat
T  5
Bat .   * Bat has two M (repeated category) need to be summed
M  7
M  3
Fly .
M  5
F  6
;
run;

data testB;
input categorical $3. value;
datalines;
Dog .   * only one category
F  3
Cat .
M  1
F  2
Cow .   * A does not have cow
X  7
Bat .   * Bat has two F (repeated category) need to be summed
F  7
F  13
Fly .   * F M order different than A
F  16
M  20
;
run;

增强数据有一个组列和有关原始顺序的信息：

data A2;
  set testA;
  if value = . then do;
    * presume missing is the 'discovery' of when the 
    * group value has to be assigned;
    group = categorical; retain group;
    group_order + 1;  
    value_order = 0;
  end;
  value_order + 1;
  format group_order value_order 4.;
run;

data B2; 
  set testB;
  if value = . then do;
    * presume missing is the 'discovery' of when the 
    * group value has to be assigned;
    group = categorical; retain group; 
    group_order + 1;
    value_order = 0;
  end;
  value_order + 1;
  format group_order value_order 4.;
run;

联接操作（数据窥视）

想要加入（回答）

---

从测试中移除狗…F。你怎么知道你应该以中频还是调频的顺序输出？您需要另一个信息源来设置排序。您是否仍在尝试加入人口统计配置文件中每个变量的摘要？如果是这样，还有另一个选项，修改它，使其具有所有值，即使它不存在。@Reeza，是的，我仍在尝试加入人口统计配置文件中的摘要。当这些值不存在时，我会修改宏来修改它们吗？我会在这个周末看一看，然后再给你回复。我有另一个地方可以立即使用，但它被埋入临床试验计划中，所以我需要找到它。@Reeza，非常感谢！我将不胜感激。我能够进行合并，但这是一个比我想象的更为手动的过程。

* this full join shows how data matches up for the answer
* the answer will use grouping, coalescing, summing and adding;
proc sql;
  create table fulljoin_peek as
  select
    coalesce (A.categorical, B.categorical) as want_categorical
  , sum(A.value,B.value) as want_value format=4.
  , A.group as A_group
  , B.group as B_group
  , A.group_order as A_group_order
  , B.group_order as B_group_order
  , A.categorical as A_cat
  , B.categorical as B_cat
  , A.value as A_value
  , B.value as B_value
  , A.value_order as A_value_order
  , B.value_order as B_value_order
  from
    A2 as A
  full join 
    B2 as B
  on 
    A.group = B.group
    and A.categorical = B.categorical
;

proc sql;
  create table

    want (drop=group_order value_order) as

  select 
    coalesce (A.categorical, B.categorical) as want_categorical
  , min (coalesce (A.group_order-1e6,B.group_order)) as group_order
  , min (coalesce (A.value_order-1e6,B.value_order)) as value_order   %* -1e6 forces A order to have precedence ;
  , sum ( sum (A.value,B.value) ) as value
  from
    A2 as A
  full join 
    B2 as B
  on 
    A.group = B.group
    and A.categorical = B.categorical
  group by 
    A.group, want_categorical
  order by 
    group_order,  value_order
  ;