Question

我有一个如下定义的cassandra表：

create table if not exists test(
    id int,
    readDate timestamp,
    totalreadings text,
    readings text,
    PRIMARY KEY(meter_id, date)
) WITH CLUSTERING ORDER BY(date desc);

该读数包含定期（30分钟）收集的所有数据快照的地图以及全天的汇总数据。

数据如下：

id=8, readDate=Tue Dec 20 2016, totalreadings=220.0, readings={0=9.0, 1=0.0, 2=9.0, 3=5.0, 4=2.0, 5=7.0, 6=1.0, 7=3.0, 8=9.0, 9=2.0, 10=5.0, 11=1.0, 12=1.0, 13=2.0, 14=4.0, 15=4.0, 16=7.0, 17=7.0, 18=5.0, 19=4.0, 20=9.0, 21=6.0, 22=8.0, 23=4.0, 24=6.0, 25=3.0, 26=5.0, 27=7.0, 28=2.0, 29=0.0, 30=8.0, 31=9.0, 32=1.0, 33=8.0, 34=9.0, 35=2.0, 36=4.0, 37=5.0, 38=4.0, 39=7.0, 40=3.0, 41=2.0, 42=1.0, 43=2.0, 44=4.0, 45=5.0, 46=3.0, 47=1.0}]]
id=8, readDate=Tue Dec 21 2016, totalreadings=221.0, readings={0=9.0, 1=0.0, 2=9.0, 3=5.0, 4=2.0, 5=7.0, 6=1.0, 7=3.0, 8=9.0, 9=2.0, 10=5.0, 11=1.0, 12=1.0, 13=2.0, 14=4.0, 15=4.0, 16=7.0, 17=7.0, 18=5.0, 19=4.0, 20=9.0, 21=6.0, 22=8.0, 23=4.0, 24=6.0, 25=3.0, 26=5.0, 27=7.0, 28=2.0, 29=0.0, 30=8.0, 31=9.0, 32=1.0, 33=8.0, 34=9.0, 35=2.0, 36=4.0, 37=5.0, 38=4.0, 39=7.0, 40=3.0, 41=2.0, 42=1.0, 43=2.0, 44=4.0, 45=5.0, 46=3.0, 47=1.0}]]
id=8, readDate=Tue Dec 22 2016, totalreadings=219.0, readings={0=9.0, 1=0.0, 2=9.0, 3=5.0, 4=2.0, 5=7.0, 6=1.0, 7=3.0, 8=9.0, 9=2.0, 10=5.0, 11=1.0, 12=1.0, 13=2.0, 14=4.0, 15=4.0, 16=7.0, 17=7.0, 18=5.0, 19=4.0, 20=9.0, 21=6.0, 22=8.0, 23=4.0, 24=6.0, 25=3.0, 26=5.0, 27=7.0, 28=2.0, 29=0.0, 30=8.0, 31=9.0, 32=1.0, 33=8.0, 34=9.0, 35=2.0, 36=4.0, 37=5.0, 38=4.0, 39=7.0, 40=3.0, 41=2.0, 42=1.0, 43=2.0, 44=4.0, 45=5.0, 46=3.0, 47=1.0}]]
id=8, readDate=Tue Dec 23 2016, totalreadings=224.0, readings={0=9.0, 1=0.0, 2=9.0, 3=5.0, 4=2.0, 5=7.0, 6=1.0, 7=3.0, 8=9.0, 9=2.0, 10=5.0, 11=1.0, 12=1.0, 13=2.0, 14=4.0, 15=4.0, 16=7.0, 17=7.0, 18=5.0, 19=4.0, 20=9.0, 21=6.0, 22=8.0, 23=4.0, 24=6.0, 25=3.0, 26=5.0, 27=7.0, 28=2.0, 29=0.0, 30=8.0, 31=9.0, 32=1.0, 33=8.0, 34=9.0, 35=2.0, 36=4.0, 37=5.0, 38=4.0, 39=7.0, 40=3.0, 41=2.0, 42=1.0, 43=2.0, 44=4.0, 45=5.0, 46=3.0, 47=1.0}]]

java pojo类如下所示：

public class Test{

    private int id;
    private Date readDate;
    private String totalreadings;   
    private Map<Integer, Double> readings;
//setters
//getters
}

我正在尝试查找每个快照的所有阅读的最后4天聚合平均值。所以从逻辑上讲，我有4个最近4天的测试对象列表，每个测试对象都有一个包含间隔读数的地图。

是否有一种简单的方法可以在4天内找到类似快照条目的聚合。例如，我想聚合特定数据快照（1,2,3,4,5,6等），而不是总聚合。

Answer 1

在改变你的表结构后，问题可以在Cassandra中完全解决。 - 主要是我把你的读数放到地图上。

create table  test(
  id int,
  readDate timestamp,
  totalreadings float,
  readings map<int,float>,
  PRIMARY KEY(id, readDate)
) WITH CLUSTERING ORDER BY(readDate desc);

现在我使用CQL输入了一些数据：

insert into test (id,readDate,totalReadings, readings ) values (8 '2016-12-20', 220.0, {0:9.0, 1:0.0, 2:9.0, 3:5.0, 4:2.0, 5:7.0, 6:1.0, 7:3.0, 8:9.0, 9:2.0, 10:5.0, 11:1.0, 12:1.0, 13:2.0, 14:4.0, 15:4.0, 16:7.0, 17:7.0, 18:5.0, 19:4.0, 20:9.0, 21:6.0, 22:8.0, 23:4.0, 24:6.0, 25:3.0, 26:5.0, 27:7.0, 28:2.0, 29:0.0, 30:8.0, 31:9.0, 32:1.0, 33:8.0, 34:9.0, 35:2.0, 36:4.0, 37:5.0, 38:4.0, 39:7.0, 40:3.0, 41:2.0, 42:1.0, 43:2.0, 44:4.0, 45:5.0, 46:3.0, 47:1.0});
insert into test (id,readDate,totalReadings, readings ) values (8, '2016-12-21', 221.0,{0:9.0, 1:0.0, 2:9.0, 3:5.0, 4:2.0, 5:7.0, 6:1.0, 7:3.0, 8:9.0, 9:2.0, 10:5.0, 11:1.0, 12:1.0, 13:2.0, 14:4.0, 15:4.0, 16:7.0, 17:7.0, 18:5.0, 19:4.0, 20:9.0, 21:6.0, 22:8.0, 23:4.0, 24:6.0, 25:3.0, 26:5.0, 27:7.0, 28:2.0, 29:0.0, 30:8.0, 31:9.0, 32:1.0, 33:8.0, 34:9.0, 35:2.0, 36:4.0, 37:5.0, 38:4.0, 39:7.0, 40:3.0, 41:2.0, 42:1.0, 43:2.0, 44:4.0, 45:5.0, 46:3.0, 47:1.0});

要从地图中提取单个值，我创建了一个用户定义函数（UDF）。此UDF选择包含读数的地图的正确值aut。有关UDF的更多信息，请参阅Cassandra docs on UDF。请注意，默认情况下在cassandra中禁用UDF，因此您需要修改cassandra.yaml以包含enable_user_defined_functions: true

create function map_item(readings map<int,float>, idx int) called on null input returns float language java as ' return readings.get(idx);';

创建功能后，您可以将平均值计算为

select avg(map_item(readings, 7)) from test where readDate > '2016-12-20' allow filtering;

给了我： system.avg（betterconnect.map_item（读物，7）） ------------------------------------------------- 3

您可能希望提供where子句的日期和索引（在我的示例中为7）作为应用程序中的参数。

每个类似快照的时间序列数据聚合

1 个答案: