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在使用PG數(shù)據(jù)庫(kù)的這段時(shí)間,總結(jié)了三種刪除重復(fù)數(shù)據(jù)的方法,其中最容易想到的就是最常規(guī)的刪除方法,但此方法性能較差,刪數(shù)據(jù)耗時(shí)較久,雖容易實(shí)現(xiàn),但性能太差,影響寫數(shù)據(jù)的速率。
另外就是被使用的group by刪除方法,效率較高。
還有一種是剛發(fā)現(xiàn)的,還沒(méi)有驗(yàn)證,現(xiàn)在就總結(jié)下這三種刪除方法,并驗(yàn)證各自的執(zhí)行效率。
首先創(chuàng)建一張基礎(chǔ)表,并插入一定量的重復(fù)數(shù)據(jù)。
test=# create table deltest(id int, name varchar(255));
CREATE TABLE
test=# create table deltest_bk (like deltest);
CREATE TABLE
test=# insert into deltest select generate_series(1, 10000), 'ZhangSan';
INSERT 0 10000
test=# insert into deltest select generate_series(1, 10000), 'ZhangSan';
INSERT 0 10000
test=# insert into deltest_bk select * from deltest;
常規(guī)刪除方法
最容易想到的方法就是判斷數(shù)據(jù)是否重復(fù),對(duì)于重復(fù)的數(shù)據(jù)只保留ctid最?。ɑ蜃畲螅┑哪菞l數(shù)據(jù),刪除其他的數(shù)據(jù)。
test=# explain analyse delete from deltest a where a.ctid <> (select min(t.ctid) from deltest t where a.id=t.id);
QUERY PLAN
-----------------------------------------------------------------------------------------------------------------------------
Delete on deltest a (cost=0.00..195616.30 rows=1518 width=6) (actual time=67758.866..67758.866 rows=0 loops=1)
-> Seq Scan on deltest a (cost=0.00..195616.30 rows=1518 width=6) (actual time=32896.517..67663.228 rows=10000 loops=1)
Filter: (ctid <> (SubPlan 1))
Rows Removed by Filter: 10000
SubPlan 1
-> Aggregate (cost=128.10..128.10 rows=1 width=6) (actual time=3.374..3.374 rows=1 loops=20000)
-> Seq Scan on deltest t (cost=0.00..128.07 rows=8 width=6) (actual time=0.831..3.344 rows=2 loops=20000)
Filter: (a.id = id)
Rows Removed by Filter: 19998
Total runtime: 67758.931 ms
test=# select count(*) from deltest;
count
-------
10000
(1 行記錄)
可以看到,id相同的數(shù)據(jù),保留ctid最小的那條,其他的刪除。相當(dāng)于把deltest表中的數(shù)據(jù)刪掉一半,耗時(shí)達(dá)到67s多。相當(dāng)慢。
group by刪除方法
第二種方法為group by方法,通過(guò)分組找到ctid最小的數(shù)據(jù),然后刪除其他數(shù)據(jù)。
test=# truncate table deltest;
TRUNCATE TABLE
test=# insert into deltest select * from deltest_bk;
INSERT 0 20000
test=# explain analyse delete from deltest a where a.ctid not in (select min(ctid) from deltest group by id);
QUERY PLAN
----------------------------------------------------------------------------------------------------------------------------------
Delete on deltest a (cost=131.89..2930.46 rows=763 width=6) (actual time=30942.496..30942.496 rows=0 loops=1)
-> Seq Scan on deltest a (cost=131.89..2930.46 rows=763 width=6) (actual time=10186.296..30814.366 rows=10000 loops=1)
Filter: (NOT (SubPlan 1))
Rows Removed by Filter: 10000
SubPlan 1
-> Materialize (cost=131.89..134.89 rows=200 width=10) (actual time=0.001..0.471 rows=7500 loops=20000)
-> HashAggregate (cost=131.89..133.89 rows=200 width=10) (actual time=10.568..13.584 rows=10000 loops=1)
-> Seq Scan on deltest (cost=0.00..124.26 rows=1526 width=10) (actual time=0.006..3.829 rows=20000 loops=1)
Total runtime: 30942.819 ms
(9 行記錄)
test=# select count(*) from deltest;
count
-------
10000
(1 行記錄)
可以看到同樣是刪除一半的數(shù)據(jù),使用group by的方式,時(shí)間節(jié)省了一半。但仍含需要30s,下面試一下第三種刪除操作。
新的刪除方法
在postgres修煉之道這本書中,作者提到一種效率較高的刪除方法, 在這里驗(yàn)證一下,具體如下:
test=# truncate table deltest;
TRUNCATE TABLE
test=# insert into deltest select * from deltest_bk;
INSERT 0 20000
test=# explain analyze delete from deltest a where a.ctid = any(array (select ctid from (select row_number() over (partition by id), ctid from deltest) t where t.row_number > 1));
QUERY PLAN
----------------------------------------------------------------------------------------------------------------------------------
Delete on deltest a (cost=250.74..270.84 rows=10 width=6) (actual time=98.363..98.363 rows=0 loops=1)
InitPlan 1 (returns $0)
-> Subquery Scan on t (cost=204.95..250.73 rows=509 width=6) (actual time=29.446..47.867 rows=10000 loops=1)
Filter: (t.row_number > 1)
Rows Removed by Filter: 10000
-> WindowAgg (cost=204.95..231.66 rows=1526 width=10) (actual time=29.436..44.790 rows=20000 loops=1)
-> Sort (cost=204.95..208.77 rows=1526 width=10) (actual time=12.466..13.754 rows=20000 loops=1)
Sort Key: deltest.id
Sort Method: quicksort Memory: 1294kB
-> Seq Scan on deltest (cost=0.00..124.26 rows=1526 width=10) (actual time=0.021..5.110 rows=20000 loops=1)
-> Tid Scan on deltest a (cost=0.01..20.11 rows=10 width=6) (actual time=82.983..88.751 rows=10000 loops=1)
TID Cond: (ctid = ANY ($0))
Total runtime: 98.912 ms
(13 行記錄)
test=# select count(*) from deltest;
count
-------
10000
(1 行記錄)
看到上述結(jié)果,真讓我吃驚了一把,這么快的刪除方法還是首次看到,自己真實(shí)孤陋寡聞,在這里要膜拜一下修煉之道這本書的大神作者了。
補(bǔ)充:pgsql 刪除表中重復(fù)數(shù)據(jù)保留其中的一條
1.在表中(表名:table 主鍵:id)增加一個(gè)字段rownum,類型為serial
2.執(zhí)行語(yǔ)句:
delete from table where rownum not in(
select max(rownum) from table group by id
)
3.最后刪除rownum
以上為個(gè)人經(jīng)驗(yàn),希望能給大家一個(gè)參考,也希望大家多多支持。如有錯(cuò)誤或未考慮完全的地方,望不吝賜教。
文章來(lái)源:腳本之家
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