计算不包括 0 值的列

Question

我有以下数据（实际上跨越多年，但对于这个例子，我只包括 4 个月）

id  created_at  staff
--------------------------------
1   2010-01-01  Coder
2   2010-01-15  Developer
3   2010-03-01  Data Analyst
4   2010-01-20  Developer
5   2010-03-13  Data Analyst
6   2010-04-05  Tester
7   2010-04-01  Tester
8   2010-04-04  Business Analyst
9   2010-01-22  Business Analyst
10  2010-01-25  Coder

然后我使用以下查询按月计算员工数量。

select staff,
    sum(case when date_trunc('month', created_at) = date '2010-01-01' then 1 else 0 end) as cnt_2010_01,
    sum(case when date_trunc('month', created_at) = date '2010-02-01' then 1 else 0 end) as cnt_2010_02,
    sum(case when date_trunc('month', created_at) = date '2010-03-01' then 1 else 0 end) as cnt_2010_03,
    sum(case when date_trunc('month', created_at) = date '2010-04-01' then 1 else 0 end) as cnt_2010_04
from mytable
group by staff

下面是上述 SQL 生成的输出示例。

staff               2010-01   2010-02   2010-03   2010-04   
----------------------------------------------------------
Coder               2         0         0         0
Developer           2         0         0         0
Data Analyst        0         0         2         0
Tester              0         0         0         2
Business Analyst    1         0         0         1

我现在正在尝试计算具有值的列数。我正在寻找的预期结果是

staff               months   2010-01   2010-02   2010-03   2010-04  
--------------------------------------------------------------------------
Coder               1        2         0         0         0
Developer           1        2         0         0         0
Data Analyst        1        0         0         2         0
Tester              1        0         0         0         2
Business Analyst    2        1         0         0         1

Answer 1

只需使用count(distinct):

select staff,
       count(distinct date_trunc('month', created_at)) as num_months,
       sum(case when date_trunc('month', created_at) = date '2010-01-01' then 1 else 0 end) as cnt_2010_01,
       sum(case when date_trunc('month', created_at) = date '2010-02-01' then 1 else 0 end) as cnt_2010_02,
       sum(case when date_trunc('month', created_at) = date '2010-03-01' then 1 else 0 end) as cnt_2010_03,
       sum(case when date_trunc('month', created_at) = date '2010-04-01' then 1 else 0 end) as cnt_2010_04
from mytable
group by staff

Answer 2

您可以使用 count(distinct):

来计算不同的月份

select staff,
    count(distinct date_trunc('month', created_at)) as cnt_month,
    sum(case when date_trunc('month', created_at) = date '2010-01-01' then 1 else 0 end) as cnt_2010_01,
    sum(case when date_trunc('month', created_at) = date '2010-02-01' then 1 else 0 end) as cnt_2010_02,
    sum(case when date_trunc('month', created_at) = date '2010-03-01' then 1 else 0 end) as cnt_2010_03,
    sum(case when date_trunc('month', created_at) = date '2010-04-01' then 1 else 0 end) as cnt_2010_04
from mytable
group by staff

Answer 3

你还需要使用 count(distinct):

select staff,
    count(distinct date_trunc('month', created_at)) as months
    sum(case when date_trunc('month', created_at) = date '2010-01-01' then 1 else 0 end) as cnt_2010_01,
    sum(case when date_trunc('month', created_at) = date '2010-02-01' then 1 else 0 end) as cnt_2010_02,
    sum(case when date_trunc('month', created_at) = date '2010-03-01' then 1 else 0 end) as cnt_2010_03,
    sum(case when date_trunc('month', created_at) = date '2010-04-01' then 1 else 0 end) as cnt_2010_04
from mytable
group by staff

计算不包括 0 值的列

Counting columns excluding 0 value

sql

datetime

count

distinct

amazon-redshift