【发布时间】：2021-05-21 11:55:07
【问题描述】：
一年中的每一天我都想要这种布局
time area disease no._of_cases
----------------------------------------
2018.01.01 Spain Influenza 46
2018.01.02 Spain Influenza 46
2018.01.03 Spain Influenza 41
2018.01.01 Spain Cholera -3
2018.01.02 Spain Cholera -1
2018.01.03 Spain Cholera -2
2018.01.01 London Influenza 55
2018.01.02 London Influenza 51
2018.01.03 London Influenza 55
2018.01.01 London Cholera 1
2018.01.02 London Cholera 2
2018.01.03 London Cholera 3
2018.01.01 France Influenza 50
2018.01.02 France Influenza 50
2018.01.03 France Influenza 55
2018.01.01 France Cholera -1
2018.01.02 France Cholera -2
2018.01.03 France Cholera 3
所以我有两种疾病，每年每个月每个不同地区的每一天都有不同的病例数
以下是一个地区有两种疾病的例子和两个月的病例数结果
Spain - Influenza
No.of cases in JAN:{floor [1339 % 30]+first 1?-3+til 7} each til 30
No.of cases in FEB:{floor [1261 % 28]+first 1?-3+til 7} each til 28
Spain - Cholera
No.of cases in JAN:{floor [0 % 30]+first 1?-3+til 7} each til 30;
No.of cases in FEB:{floor [0 % 28]+first 1?-3+til 7} each til 28;
这是一月份的示例
times:2020.01.01 + til 3;
area:`Spain
disease1:`Influenza
disease2:`Cholera
no.of_cases1:{floor [1339 % 30]+first 1?-3+til 7} each til 30;
no.of_cases2:{floor [0 % 30]+first 1?-3+til 7} each til 30;
area2:`London
no.of_cases3:{floor [1583 % 30]+first 1?-3+til 7} each til 30;
no.of_cases4:{floor [0 % 30]+first 1?-3+til 7} each til 30;
area3:`France
no.of_cases5:{floor [1583 % 30]+first 1?-3+til 7} each til 30;
no.of_cases6:{floor [0 % 30]+first 1?-3+til 7} each til 30;
a:([]time:times;area:(count [times])# area;disease:count[times]#disease1;no._of_cases:count[times]# no.of_cases1)
b:([]time:times;area:(count [times])# area;disease:count[times]#disease2;no._of_cases:count[times]# no.of_cases2)
c:([]time:times;area:(count [times])# area2;disease:count[times]#disease1;no._of_cases:count[times]# no.of_cases3)
d:([]time:times;area:(count [times])# area2;disease:count[times]#disease2;no._of_cases:count[times]# no.of_cases4)
e:([]time:times;area:(count [times])# area3;disease:count[times]#disease1;no._of_cases:count[times]# no.of_cases5)
f:([]time:times;area:(count [times])# area3;disease:count[times]#disease2;no._of_cases:count[times]# no.of_cases6)
t:a,b,c,d,e,f
有没有更简单的方法可以为一年中的每一天创建此表？
这是我要转换为 kdb 的表的示例。
area |Spain |London |France
____________________________________________________________________-
disease |Month/year |2017 2018 ... | 2017 2018 ... |2017 2018 ...
----------------------------------------------------------------------
Influenza |Jan |1 2 | |
|Feb |2 3 | |
|.... |...
|
--------------------------------------------------------------------
Cholera |Jan |2 9
|Feb |1 2
|...
【问题讨论】：
-
您好，您所构建的基础数据是否具有您想要平均值的每月总计？可能最好从那而不是所有这些单独的步骤中构建一些东西？你能显示这个的基础表吗？西班牙的1339和1261流感病例来自哪里？
-
是的，它有每月的总数，我正在以此为基础生成平均值。
-
我建议将该表读入 kdb 并尝试对其进行操作以达到您想要的结果，而不是从头开始构建表。