Question

我一直在计算带有子选择和一些聚合函数的NULL和非NULL列

CREATE TEMPORARY TABLE citizens(name text, country text,profession text,postalcode text);

INSERT INTO citizens VALUES
 ('Fred', 'USA', 'Professor', NULL),
 ('Amy', 'USA', 'Professor', NULL),
 ('Ted', 'USA', 'Professor', 90210),
 ('Barb', 'USA', 'Lawyer', 10248),
 ('Wally', 'USA', 'Lawyer', NULL),
 ('Fred', 'Canada', 'Professor', 'S0H'),
 ('Charles', 'Canada', 'Professor', 'S4L'),
 ('Nancy', 'Canada', 'Lawyer', NULL),
 ('Linda', 'Canada', 'Professor', NULL),
 ('Steph', 'France', 'Lawyer', 75008 ),
 ('Arnold', 'France', 'Lawyer', 75008 ),
 ('Penny', 'France', 'Lawyer', 75008 ),
 ('Harry', 'France', 'Lawyer', NULL);

SELECT country,
    profession,
    MAX(have_postalcode::int*num) AS num_have,
    MAX((1-have_postalcode::int)*num) AS num_not_have
FROM
    (
    SELECT country, profession,
        COUNT(*) AS num,
        (postalcode IS NOT NULL) AS have_postalcode
    FROM citizens
    GROUP BY country, profession, have_postalcode
    ) AS d
GROUP BY country, profession

结果

USA     Professor   1   2
Canada  Lawyer      0   1
USA     Lawyer      1   1
France  Lawyer      3   1
Canada  Professor   2   1

但似乎应该有一种更流畅的方式（例如，让我感到痛苦的是MAX只是用来抓住一个非常重要的值）。有人有个好主意吗？

Answer 1

SELECT country, profession,
        COUNT(postalcode) AS num_have
      , (COUNT(*) - COUNT(postalcode)) AS num_not_have
FROM citizens
GROUP BY country, profession;

http://sqlfiddle.com/#!1/17a9d/15

Answer 2

SELECT  country
,       profession
,       sum(case when postalcode is not null then 1 end) as num_have
,       sum(case when postalcode is null then 1 end) as num_not_have
FROM    citizens
GROUP BY 
        country
,       profession

Answer 3

SELECT Country, Profession, 
    count(Country) as num_have, count(*) - count(PostalCode) as num_not_have
FROM citizens
GROUP BY Country, Profession

光滑的方式来计算NULL和非NULL行？

3 个答案: