SAS技术求助：计算各城市年度数值两两差值的中位数

1. 宽表转长表（若输入为宽表） #

假设你的输入表名为city_data，结构为每个年份对应一列（如y1988、y1989…y2000），首先用PROC TRANSPOSE将其转为包含city、year、value的长表：

proc transpose data=city_data out=city_long(rename=(col1=value));
    by city;
    var y1988-y2000;
run;

data city_long;
    set city_long;
    year = input(substr(_name_,2),4.); /* 提取年份数值，如从"y1988"得到1988 */
    drop _name_;
run;

2. 生成所有i<j的年度配对并计算差值 #

通过自连接筛选出每个城市中年份i<j的组合，计算year_i - year_j的差值：

proc sql;
    create table city_diff as
    select a.city, a.year as year_i, b.year as year_j, a.value - b.value as diff
    from city_long a
    inner join city_long b
    on a.city = b.city and a.year < b.year;
quit;

3. 计算每个城市差值的中位数 #

用PROC MEANS按城市分组计算差值的中位数：

proc means data=city_diff noprint;
    by city;
    var diff;
    output out=city_median median=diff_median;
run;

也可以用PROC UNIVARIATE替代，支持输出更多统计量：

proc univariate data=city_diff noprint;
    by city;
    var diff;
    output out=city_median median=diff_median;
run;

内容的提问来源于stack exchange，提问作者LR2008