如果您能给我一些提示，我将非常感激如何做或做什么以及在哪里可以解决以下任务：
例如，我有一个人员样本 (PID)，他们面临的选择集包括一个选定的汽车品牌和其他考虑的汽车替代品（两个或一个，因为一些受访者已经命名了他们认为除了买了车，其中一些 - 只有一个）。
PID decision alternative brand
1 1 1 BMW
1 0 2 AUDI
1 0 3 Mercedes
2 1 1 AUDI
2 0 2 Mercedes
2 0 3 Land_Rover
3 1 1 Mercedes
3 0 2 BMW
3 0 3 VW
4 1 1 VW
4 0 2 AUDI
5 1 1 BMW
6 1 1 AUDI
6 0 2 VW
6 0 3 VW
7 1 1 Mercedes
7 0 2 AUDI
我想计算有多少受访者在他们的选择集中拥有品牌 i，在选择集中也拥有品牌 j。作为上述数据的说明，我想得到下表：
AUDI BMW Land_Rover Mercedes VW
AUDI 0 1 1 3* 2
BMW 0 0 2 1
Land Rover 0 1 0
Mercedes 0 1
VW 1**
其内容如下： * 选择集中有 3 位选择梅赛德斯的受访者，他们的选择集中也有奥迪。** 选择集中有 1 名大众汽车的受访者，他们再次将大众汽车命名为考虑的替代品（对品牌的某种忠诚度）。
如果你能告诉我我可以用什么程序来做，我将不胜感激。我总共有 46 个品牌。
提前谢谢了。
弗拉达
PS如果曾经回答过类似的问题，我将不胜感激答案的链接，并为我无法找到它而提前道歉。
我的解决方案：
注意：它的计数有点不同。它计算有多少人购买了特定汽车，并将其他品牌命名为替代品。通过这种方式，我将看看哪些品牌在考虑范围内相互竞争。
*计算选择集中的汽车组合；
*Clean-Up: Delete unneccessary datasets in the work library;
proc datasets lib=work kill nolist memtype=data;
quit;
*Clear the output window;
ods html close; /* close previous */
ods html; /* open new */
*Counting cars´ combinations in the choice set;
data have;
input PID decision alternative brand $;
datalines;
1 1 1 BMW
1 0 2 AUDI
1 0 3 Mercedes
2 1 1 AUDI
2 0 2 Mercedes
2 0 3 Land_Rover
3 1 1 Mercedes
3 0 2 BMW
3 0 3 VW
4 1 1 VW
4 0 2 AUDI
5 1 1 BMW
6 1 1 AUDI
6 0 2 VW
6 0 3 VW
7 1 1 Mercedes
7 0 2 AUDI
;;;;
run;
data code_brand;
input brand $ code_brand;
datalines;
AUDI 1
BMW 2
Land_Rover 3
Mercedes 4
VW 5
;;
run;
data have_wide; set have;
by pid;
keep pid brand1 brand2 brand3;
retain brand1 brand2 brand3;
array abrand( 3) $ 20 brand1 brand2 brand3;
if first.pid then do;
do i=1 to 3;
abrand(i)=" ";
end;
end;
abrand(alternative)=brand;
if last.pid then output;
run;
proc freq data=have_wide noprint;
table brand1*brand2 /out=brand1_2;
run;
proc freq data=have_wide noprint;
table brand1*brand3 /out=brand1_3;
run;
proc sql;
create table temp1 as
select t1.brand1, t1.brand2, t1.count as count_1_2, t2.brand3, t2.count as count_1_3,
(t1.count+t2.count) as total
from brand1_2 t1 left join brand1_3 t2
on t1.brand1=t2.brand1 and t1.brand2=t2.brand3;
create table cs_count as
select t1.brand1 as first_car, t1.brand2 as alternative_car,
(case when t1.total is missing then t1.count_1_2
else t1.total end) as cs_count,
t2.code_brand as code_brand2
from temp1 t1 left join code_brand t2
on t1.brand2=t2.brand
order by brand1, brand2;
/* Reshaping a Dataset from long to wide format with multiple variables*/
proc transpose data=cs_count out=cs_count_wide prefix=b;
by first_car;
id code_brand2;
var cs_count;
run;
proc sql;
create table final as
select t2.code_brand as counter,
(case when t1.first_car is missing then t2.brand
else t1.first_car end) as first_car,
cats('b',t2.code_brand) as code_brand1,
t1.b1, t1.b2, . as b3, t1.b4, t1.b5
from cs_count_wide (keep= first_car b:) t1 full join code_brand t2
on t1.first_car=t2.brand
order by t2.code_brand;
data final;
set final;
drop counter first_car;
run;
proc iml;
use final;
read all var{code_brand1} into name; *create separate vector of brand names;
read all var _num_ into data; *create separate matrix of data observations;
n = nrow(data);
p = ncol(data);
lower = j(n, p, 0);
do i = 2 to n;
cols = 1:i-1;
lower[i, cols] = data[i, cols];
end; *extracts lower diagonal matrix with 0 values at the diagonal;
print lower;
upper = j(n, p, 0);
do i = 1 to n;
cols=i:p;
upper[i, cols] = data[i, cols];
end; *extracts upper diagonal matrix keeping the values of the diagonal;
lower=lower`; *transpose the lower diagonal matrix into upper diagonal matrix;
A=lower+upper; *calculates the sum of the upper diagonal matrix and the transpose of lower diagonal matrix;
CAR=name`;
c=name;
create test_end from A[colname=c];
append from A;
close test_end; *creates dataset from the matrix A;
create test_name var {"CAR"};
append;
close test_name; *creates dataset from the column vector of brand names;
quit;
*The merged dataset represents an upper diagonal symmetric matix;
data final;
merge test_name test_end;
run;