v1 = [-5, 5, 10, 15, 20, 25, 100]
v2 = [-9999999, 0, 10, 30, 9999999]
res = array(ANY, v2.size()-1)
for(i in 0 : (v2.size()-1)) {
if (v2[i] == -9999999 && v2[i+1] != 9999999) cond = v1 <= v2[i+1]
else if (v2[i+1] == 9999999) cond = v1 > v2[i]
else cond = v1 > v2[i] && v1 <= v2[i+1]
res[i] = iif(cond, strReplace("C" + v2[i] + "_" + v2[i+1], "-", ""), string(NULL))
}
reduce(add, res)
结果：
假设有如下两个向量
v1 = [-5, 5, 10, 15, 20, 25, 100] v2 = [-9999999, 0, 10, 30, 9999999]
可以按照如下方案实现
select id, section from aj(table(v1 as id), table(v2 as id, ["C9999999_0", "C0_10", "C10_30", "C30_9999999", NULL] as `section), `id);
测试数据数据量：v1有100万，v2有102条
可以用 asof 实现，并利用一个字典维护数据范围：
v1 = [-5, 5, 10, 15, 20, 25, 100] v2 = [-9999999, 0, 10, 30, 9999999] d = dict(INT, STRING) d[-9999999]="C9999999_0" d[0]="C0_10" d[10]="C10_30" d[30]="C30_9999999" d[v2.at(asof(v2, v1))]
