PostgreSql 常用聚合函数

基础准备

sql 复制代码

CREATE TEMP TABLE demo(id int, v int, t text);
INSERT INTO demo VALUES
  (1, 10, 'a'),
  (2, 20, 'b'),
  (3, NULL, 'c');

id	v	t
1	10	a
2	20	b
3	NULL	c

注意：版本基于 PG 15，全部自带，无需扩展

A. 基础统计

sql 复制代码

SELECT count(*)        AS cnt,      -- 3
       count(v)        AS cnt_v,    -- 2（NULL 不计）
       sum(v)          AS s,        -- 30
       avg(v)          AS a,        -- 15
       max(v)          AS mx,       -- 20
       min(v)          AS mn        -- 10
FROM demo;

结果：

cnt	cnt_v	s	a	mx	mn
3	2	30	15	20	10

B. 扩展统计

sql 复制代码

SELECT stddev_pop(v)   AS std_pop,   -- 5
       stddev_samp(v)  AS std_samp,  -- 7.0710678
       var_pop(v)      AS var_pop,   -- 25
       var_samp(v)     AS var_samp   -- 50
FROM demo;

结果：

std_pop	std_samp	var_pop	var_samp
5	7.07106781	25	50

C. 位运算聚合

sql 复制代码

SELECT bit_and(v)      AS bit_and,   -- 0  (10 & 20 = 0)
       bit_or(v)       AS bit_or,    -- 30 (10 | 20 = 30)
       bit_xor(v)      AS bit_xor    -- 30 (10 # 20 = 30)
FROM demo;

结果：

bit_and	bit_or	bit_xor
0	30	30

D. 字符串聚合

sql 复制代码

SELECT string_agg(t, ',')            AS str1,  -- a,b,c
       string_agg(t, ',' ORDER BY id DESC) AS str2  -- c,b,a
FROM demo;

结果：

str1	str2
a,b,c	c,b,a

E. 数组 / JSON 聚合

sql 复制代码

SELECT array_agg(v)               AS arr,        -- {10,20}
       array_agg(v ORDER BY v DESC) AS arr_desc, -- {20,10}
       json_agg(v)                AS j,          -- [10,20]
       jsonb_agg(t)               AS jb          -- ["a", "b", "c"]
FROM demo;

结果：

arr	arr_desc	j	jb
{10,20}	{20,10}	[10,20]	["a", "b", "c"]

F. 排重聚合

sql 复制代码

SELECT count(DISTINCT v)     AS cdv,  -- 2
       string_agg(DISTINCT t, '') AS dst  -- abc
FROM demo;

结果：

cdv	dst
2	abc

G. 有序集合（需要 WITHIN GROUP）

sql 复制代码

SELECT percentile_cont(0.5) WITHIN GROUP (ORDER BY v) AS median, -- 15
       percentile_disc(0.5) WITHIN GROUP (ORDER BY v) AS disc,   -- 10
       mode()              WITHIN GROUP (ORDER BY t)  AS mode_t  -- a（出现最多）
FROM demo;

结果：

median	disc	mode_t
15	10	a

H. 假设集合（同样 WITHIN GROUP）

sql 复制代码

SELECT rank(15)       WITHIN GROUP (ORDER BY v) AS rnk,  -- 2
       percent_rank(15) WITHIN GROUP (ORDER BY v) AS pct, -- 0.5
       cume_dist()      WITHIN GROUP (ORDER BY v) AS cume -- 1.0
FROM demo;

结果：

rnk	pct	cume
2	0.5	1.0