根据wikipedia：
如果 h 他/她的 Np 论文至少 h 引用，那么科学家的索引 h 和其他（Np - h）论文各自的引用次数不超过 h 。
想象一下，我们有科学家，论文，CITATIONS表格，科学家和论文之间有1-n关系，PAPERS和CITATION TABLES之间有1-n关系。如何编写一个SQL语句来计算科学家表中每个科学家的h分数？
为了展示我在这里做的一些研究工作，每篇论文的SQL计算引用次数：
SELECT COUNT(CITATIONS.id) AS citations_count
FROM PAPERS
LEFT OUTER JOIN CITATIONS ON (PAPERS.id = CITATIONS.paper_id)
GROUP BY PAPERS.id
ORDER BY citations_count DESC;
答案 0 :(得分：3)
h值正在做的是以两种方式计算引用次数。让我们说科学家有以下引用次数：
10
8
5
5
2
1
让拥有多次或多次引用的数字，以及两者之间的差异：
10 1 9
8 2 6
5 3 2
5 3 2
2 5 -3
1 6 -5
您想要的数字是0。在这种情况下，数字是4。
数字为4的事实使得这很难，因为它不在原始数据中。这使得计算更难，因为您需要生成数字表。
以下使用SQL Server语法生成包含100个数字的表：
with numbers as (
select 1 as n
union all
select n+1
from numbers
where n < 100
),
numcitations as (
SELECT p.scientistid, p.id, COUNT(c.id) AS citations_count
FROM PAPERS p LEFT OUTER JOIN
CITATIONS c
ON p.id = c.paper_id
GROUP BY p.scientist, p.id
),
hcalc as (
select scientistid, numbers.n,
(select count(*)
from numcitations nc
where nc.scientistid = s.scientistid and
nc.citations_count >= numbers.n
) as hval
from numbers cross join
(select scientistid from scientist) s
)
select *
from hcalc
where hval = n;
编辑：
有一种方法可以在不使用数字表的情况下完成此操作。 h分数是引用次数大于或等于引用次数的情况的计数。这更容易计算：
select scientistid, count(*)
from (SELECT p.scientistid, p.id, COUNT(c.id) AS citations_count,
rank() over (partition by p.scientistid, p.id order by count(c.id) desc) as ranking
FROM PAPERS p LEFT OUTER JOIN
CITATIONS c
ON p.id = c.paper_id
GROUP BY p.scientist, p.id
) t
where ranking <= citations_count
group by scientistid;
答案 1 :(得分：1)
“也许”迟到了，但这是另一种不依赖于分析功能的解决方案。 希望它对任何有类似问题的人都有用。
（我之前的回答未考虑您的数据库设计）
查询将是：
with aux as ( -- Amount of citations
select p.id, p.scientist_id, p.title, count(*) as citations
from paper p join citation c2 on (p.id = c2.paper_id)
group by 1,2,3 )
select id, name, min(h) as h_index from (
select c.id,
c.name,
aux1.title,
aux1.citations,
-- Replacing the use of analytical functions
-- Eq. to: count(*) over (parition by c.id order by p.citations desc)
( select count(*) from aux aux2
where aux2.citations >= aux1.citations
and aux2.scientist_id = aux1.scientist_id) as h
from scientist c join aux aux1 on (c.id = aux1.scientist_id)
) where h >= citations
group by id, name;
运行示例的支持表和数据将是：
create table scientist (
id integer primary key, -- Alias to ROWID SQLite
name varchar );
create table paper (
id integer primary key,
scientist_id integer not null,
title varchar );
create table citation (
id integer primary key,
paper_id integer not null,
citation_details varchar ); -- not used
-- Data insertion
insert into scientist (name) values
('Mr Doom'),
('Batman'),
('Superman');
insert into paper (scientist_id, title) values
(1,'Doomsday'), -- 12 citations
(1,'Bad Day'), -- 11 citations
(1,'Sad Day'), -- 10 citations
(1,'Programming Day'), -- 5 citations
(1,'Debugin bug #! Error'), -- 2 citations
(1,'Happy Coding'), -- 0 citations
(2,'I have money'), -- 1 citations
(3,'I have no planet'); -- 0 citations (it's not very popular)
insert into citation (paper_id) values
(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),
(2),(2),(2),(2),(2),(2),(2),(2),(2),(2),(2),
(3),(3),(3),(3),(3),(3),(3),(3),(3),(3),
(4),(4),(4),(4),(4),
(5),(5),
(7);
我使用SQLite3.8运行它。可以在我的github repository中找到源代码和SQLite示例数据库。
答案 2 :(得分：0)
这是MS SQL解决方案。
/*
The h-index is calculated by counting the number of publications for which an author has been
cited by other authors at least that same number of times. For instance, an h-index of 17 means
that the scientist has published at least 17 papers that have each been cited at least 17 times.
If the scientist's 18th most cited publication was cited only 10 times, the h-index would remain at 17.
If the scientist's 18th most cited publication was cited 18 or more times, the h-index would rise to 18.
*/
declare @num_pubs int = 4;
declare @sited_seed int = 10;
declare @publication table
(
scientist_id int,
publication_title int,
publication_cited int
);
with numbers as (
select 1 as n
union all
select n + 1
from numbers
where n < @num_pubs
)
insert into @publication select 1 as scientist_id, n as publication_title, ABS(Checksum(NewID()) % @sited_seed) as publication_cited from numbers
select * from @publication
-- data sample for scientist#1
-- scientist_id publication sited
-- 1 pub 1 2
-- 1 pub 2 0
-- 1 pub 3 1
-- 1 pub 4 3
select scientist_id, max(pub_row_number) as h_index
from (
select p.scientist_id, pub_row_number --, count(p.publication_cited)
from @publication as p
inner join
(
select scientist_id, publication_title, publication_cited,
ROW_NUMBER() OVER (PARTITION BY scientist_id ORDER BY publication_cited) as pub_row_number from @publication
-- create a unique index for publications, and using it to triangulate against the number of cited publications
-- scientist_id publication sited pub_row_number
-- 1 pub 1 2 1
-- 1 pub 2 0 2
-- 1 pub 3 1 3
-- 1 pub 4 3 4
) as c
on pub_row_number <= p.publication_cited and p.scientist_id = c.scientist_id
-- triangulation {pub_row_number <= p.publication_cited} solves two problems
-- 1. it removes all publications with sited equals to zero
-- 2. for every publication (pub_row_number) it creates a set of all available sited publications, so that it is possible to
-- count the number of publications that has been cited at least same number of times
-- scientist_id pub_row_number publication_cited
-- 1 1 > 0 >> filtered out
-- 1 1 <= 1
-- 1 1 <= 2
-- 1 1 <= 3
-- 1 2 <= 2
-- 1 2 <= 3
-- 1 3 <= 3
group by p.scientist_id, pub_row_number --, p.publication_cited
having pub_row_number <= count(p.publication_cited)
-- {pub_count <= count(p.publication_cited)} this tiangulation creates a final count
-- there are 3 publications sited at least once, 2 - sited at least 2 times, and one sited at least 3 times
-- scientist_id pub_row_number count(publication_cited)
-- 1 1 <= 3
-- 1 2 <= 2
-- 1 3 > 1 >> filtered out via tiangulation
) as final
-- finally, max(pub_count) pulls the answare
-- scientist_id h_index
-- 1 2
group by scientist_id
UNION
-- include scientist without publications
select scientist_id, 0 from scientist where scientist_id not in (select scientist_id from publication)
UNION
-- include scientist with publications but without citation
select scientist_id, 0 from publication group by scientist_id having sum(publication_cited) = 0
答案 3 :(得分：-1)
SELECT F.id,F.name,SUM(H_INDEX) AS H_INDEX
FROM (
SELECT a.id,a.name
,CASE WHEN h.HI IS NULL then 0
WHEN cited>=HI then 1 else 0 end AS H_INDEX
,HI,cited
FROM author a
LEFT JOIN (select publication2.author_id,publication2.title,publication2.cited, count(1) AS HI
from publication publication2
INNER JOIN publication publication1 on publication2.cited >= publication1.cited
and publication2.author_id = publication1.author_id
GROUP BY publication2.author_id,publication2.title,publication2.cited
) h on a.id=h.author_id
) F GROUP BY F.id,F.name ORDER BY H_INDEX DESC