根据维基百科：
一个科学家的索引为h，如果他/她的Np篇论文中的h每篇至少有h次引用，并且其他(Np - h)篇论文每篇的引用不超过h。
想象一下，我们有 SCIENTISTS、PAPERS、CITATIONS 表，SCIENTISTS 和 PAPERS 之间的关系为 1-n，PAPERS 和 CITATION TABLES 之间的关系为 1-n。如何编写一个 SQL 语句来计算 SCIENTISTS 表中每个科学家的 h 分数？
为了展示我在这里所做的一些研究工作，是对每篇论文的引用次数进行 SQL 计算：
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;
h 值的作用是以两种方式计算引用次数。假设一位科学家的引用次数如下：
10
8
5
5
2
1
让我们看看有那么多或更多引用的数字，以及两者之间的区别：
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 的事实使这变得困难，因为它不在原始数据中。这使计算变得更加困难，因为您需要生成一个数字表。
下面使用 SQL Server 语法来生成包含 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做到这一点。h-score 是引用次数大于或等于引用次数的案例计数。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;
“也许”晚了，但这是另一种不依赖分析功能的解决方案。希望它对其他有类似问题的人有用。
（我之前的回答没有考虑到您的数据库设计）
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);
我已经使用 SQLite3.8 运行它。源代码和 SQLite 示例数据库可以在我的 github存储库中找到。
这是一个 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
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