计算 NULL 和非 NULL 行的流畅方法

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光滑的方法（例如，MAX只是用来抓取一个不平凡的价值，这让我很痛苦）。 有人有一个很酷的想法吗？

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