How to compare a tsvector against another tsvector? - sql

I'm trying to get "possibly related" records of a given record.
There's a tsvector (tsv) on the table, so I'm thinking how to convert the source tsv to a tsquery format to then find the most closely related matches like a normal ranked search.
SELECT title,
link,
IMAGE,
intro,
created_at,
updated_at,
ts_rank_cd(tsv, q.match::tsquery) AS rank
FROM items,
(
SELECT tsv AS match
FROM items
WHERE id = 1234
) AS q
WHERE id <> 1234
ORDER BY rank DESC LIMIT 10;
Is there a nice way to achieve this?


I did some poking around and it didn't seem like there was an easy way of doing this. I think to do it effectively you would probably need your own C functions which could provide a distance from one tsvector to another (then you could use KNN searches).
Either way there is not a very easy way to do this and it is likely a significant amount of work, but it seems like it should be a generally applicable problem so the general community might be interested in a solution.
Note this is not as trivial as it sounds. Suppose I write a book about Albert Lord's the Singer of Tales and his emphasis on poetic formulas. Suppose I call it "Albert Lord and the Ring of Words." This would create a tsvector of Albert:1 Lord:2 Ring:5 Words:7, The Lord of the Rings is Lord:2 Ring:5 which would create a very false sense of similarity. If you have any categorization involved, you would want to leverage that as well.


You could perhaps compare tsvector with similarity from the pg_trgm extension.
Something like this:
SELECT title, similarity(STRIP(to_tsvector('english', title))::text, STRIP(to_tsvector('english', 'The Lord of the Rings'))::text) sim
FROM (VALUES
('Albert Lord and the Ring of Words'),
('The Ring of Words'),
('Albert Lord')
) t(title)
ORDER BY sim DESC

Related

Sorting with many to many relationship

I have a 3 tables person, person_speaks_language and language.
person has 80 records
language has 2 records
I have the following records
the first 10 persons speaks one language
the first 70 persons (include the first group) speaks 2 languages
the last 10 persons dont speaks any language
Following with the example I want sort the persons by language, How I can do it correctly.
I'm trying to use the the following SQL but seems quite strange
SELECT "person".*
FROM "person"
LEFT JOIN "person_speaks_language" ON "person"."id" = "person_speaks_language"."person_id"
LEFT JOIN "language" ON "person_speaks_language"."language_id" = "language"."id"
ORDER BY "language"."name"
ASC
dataset
71,Catherine,Porter,male,NULL
72,Isabelle,Sharp,male,NULL
73,Scott,Chandler,male,NULL
74,Jean,Graham,male,NULL
75,Marc,Kennedy,male,NULL
76,Marion,Weaver,male,NULL
77,Melvin,Fitzgerald,male,NULL
78,Catherine,Guerrero,male,NULL
79,Linnie,Strickland,male,NULL
80,Ann,Henderson,male,NULL
11,Daniel,Boyd,female,English
12,Ora,Beck,female,English
13,Hulda,Lloyd,female,English
14,Jessie,McBride,female,English
15,Marguerite,Andrews,female,English
16,Maurice,Hamilton,female,English
17,Cecilia,Rhodes,female,English
18,Owen,Powers,female,English
19,Ivan,Butler,female,English
20,Rose,Bishop,female,English
21,Franklin,Mann,female,English
22,Martha,Hogan,female,English
23,Francis,Oliver,female,English
24,Catherine,Carlson,female,English
25,Rose,Sanchez,female,English
26,Danny,Bryant,female,English
27,Jim,Christensen,female,English
28,Eric,Banks,female,English
29,Tony,Dennis,female,English
30,Roy,Hoffman,female,English
31,Edgar,Hunter,female,English
32,Matilda,Gordon,female,English
33,Randall,Cruz,female,English
34,Allen,Brewer,female,English
35,Iva,Pittman,female,English
36,Garrett,Holland,female,English
37,Johnny,Russell,female,English
38,Nina,Richards,female,English
39,Mary,Ballard,female,English
40,Adrian,Sparks,female,English
41,Evelyn,Santos,female,English
42,Bess,Jackson,female,English
43,Nicholas,Love,female,English
44,Fred,Perkins,female,English
45,Cynthia,Dunn,female,English
46,Alan,Lamb,female,English
47,Ricardo,Sims,female,English
48,Rosie,Rogers,female,English
49,Susan,Sutton,female,English
50,Mary,Boone,female,English
51,Francis,Marshall,male,English
52,Carl,Olson,male,English
53,Mario,Becker,male,English
54,May,Hunt,male,English
55,Sophie,Neal,male,English
56,Frederick,Houston,male,English
57,Edwin,Allison,male,English
58,Florence,Wheeler,male,English
59,Julia,Rogers,male,English
60,Janie,Morgan,male,English
61,Louis,Hubbard,male,English
62,Lida,Wolfe,male,English
63,Alfred,Summers,male,English
64,Lina,Shaw,male,English
65,Landon,Carroll,male,English
66,Lilly,Harper,male,English
67,Lela,Gordon,male,English
68,Nina,Perry,male,English
69,Dean,Perez,male,English
70,Bertie,Hill,male,English
1,Nelle,Gill,female,Spanish
2,Lula,Wright,female,Spanish
3,Anthony,Jensen,female,Spanish
4,Rodney,Alvarez,female,Spanish
5,Scott,Holmes,female,Spanish
6,Daisy,Aguilar,female,Spanish
7,Elijah,Olson,female,Spanish
8,Alma,Henderson,female,Spanish
9,Willie,Barrett,female,Spanish
10,Ada,Huff,female,Spanish
11,Daniel,Boyd,female,Spanish
12,Ora,Beck,female,Spanish
13,Hulda,Lloyd,female,Spanish
14,Jessie,McBride,female,Spanish
15,Marguerite,Andrews,female,Spanish
16,Maurice,Hamilton,female,Spanish
17,Cecilia,Rhodes,female,Spanish
18,Owen,Powers,female,Spanish
19,Ivan,Butler,female,Spanish
20,Rose,Bishop,female,Spanish
21,Franklin,Mann,female,Spanish
22,Martha,Hogan,female,Spanish
23,Francis,Oliver,female,Spanish
24,Catherine,Carlson,female,Spanish
25,Rose,Sanchez,female,Spanish
26,Danny,Bryant,female,Spanish
27,Jim,Christensen,female,Spanish
28,Eric,Banks,female,Spanish
29,Tony,Dennis,female,Spanish
30,Roy,Hoffman,female,Spanish
31,Edgar,Hunter,female,Spanish
32,Matilda,Gordon,female,Spanish
33,Randall,Cruz,female,Spanish
34,Allen,Brewer,female,Spanish
35,Iva,Pittman,female,Spanish
36,Garrett,Holland,female,Spanish
37,Johnny,Russell,female,Spanish
38,Nina,Richards,female,Spanish
39,Mary,Ballard,female,Spanish
40,Adrian,Sparks,female,Spanish
41,Evelyn,Santos,female,Spanish
42,Bess,Jackson,female,Spanish
43,Nicholas,Love,female,Spanish
44,Fred,Perkins,female,Spanish
45,Cynthia,Dunn,female,Spanish
46,Alan,Lamb,female,Spanish
47,Ricardo,Sims,female,Spanish
48,Rosie,Rogers,female,Spanish
49,Susan,Sutton,female,Spanish
50,Mary,Boone,female,Spanish
51,Francis,Marshall,male,Spanish
52,Carl,Olson,male,Spanish
53,Mario,Becker,male,Spanish
54,May,Hunt,male,Spanish
55,Sophie,Neal,male,Spanish
56,Frederick,Houston,male,Spanish
57,Edwin,Allison,male,Spanish
58,Florence,Wheeler,male,Spanish
59,Julia,Rogers,male,Spanish
60,Janie,Morgan,male,Spanish
61,Louis,Hubbard,male,Spanish
62,Lida,Wolfe,male,Spanish
63,Alfred,Summers,male,Spanish
64,Lina,Shaw,male,Spanish
65,Landon,Carroll,male,Spanish
66,Lilly,Harper,male,Spanish
67,Lela,Gordon,male,Spanish
68,Nina,Perry,male,Spanish
69,Dean,Perez,male,Spanish
70,Bertie,Hill,male,Spanish
Update
the expect results are: each person must be appears only one time using the language order
For explain the case further, I'll take a new and small dataset, using only the person id and the language name
1,English
2,English
3,English
4,English
19,English
1,Spanish
2,Spanish
3,Spanish
4,Spanish
5,Spanish
14,Spanish
15,Spanish
16,Spanish
19,Spanish
21,Spanish
25,Spanish
I'm using the same order but if I use a limit for example LIMIT 8 the results will be
1,English
2,English
3,English
4,English
19,English
1,Spanish
2,Spanish
3,Spanish
And the expected result is
1,English
2,English
3,English
4,English
19,English
5,Spanish
14,Spanish
15,Spanish
What I'm trying to do
What I'm trying to do is sorting, paginating and filtering a list of X that may have a many-to-many relationship with Y, in this case X is a person and Y is the language. I need do it in a general way. I found a trouble if I want ordering the list by some Y properties.
The list will show in this way:
firstname, lastname, gender , languages
Daniel , Boyd , female , English Spanish
Ora , Beck , female , English
Anthony , Jensen , female , Spanish
....
I only need return a array with the IDs in the correct order
this is the main reason I need that the results only appears the person one time is because the ORM (that I'm using) try to hydrate each result and if I paginate the results using offset and limit. the results maybe aren't the expected. I'm doing assumptions many to many relationships
I can't use the string_agg or group_concat because I dont know the real data, I dont know if are integers or strings

If you want each person to appear only once, then you need to aggregate by that person. If you then want the list of languages, you need to combine them in some way, concatenation comes to mind.
The use of double quotes suggests Postgres or Oracle to me. Here is Postgres syntax for this:
SELECT p.id, string_agg(l.name) as languages
FROM person p LEFT JOIN
person_speaks_language psl
ON p.id = psl.person_id LEFT JOIN
language l
ON psl.language_id = l.id
GROUP BY p.id
ORDER BY COUNT(l.name) DESC, languages;
Similar functionality to string_agg() exists in most databases.

There is nothing wrong with Bertie Hill appearing in two rows, with one language each, that is the Tabular View of Data per the Relational Model. There are no dependencies on data values or number of data values. It is completely correct and un-confused.
But here, the requirement is confused, because you really want three separate lists:
speaks one language
speaks two languages [or the number of languages currently in the language file]
speaks no language [on file] ) ...
But you want those three lists in one list.
Concatenating data values is never, ever a good idea. It is a breach of rudimentary standards, specifically 1NF. It may be common, but it is a gross error. It may be taught by the so-called "theoreticians", but it remains a gross error. Even in a result set, yes.
It creates confusion, such as I have detailed at the top.
With concatenated strings, as the number of languages changes, the width of that concatenated field will grow, and eventually exceed space, wherever it appears (eg. the width of the field on the screen).
Just two of the many reasons why it is incorrect, not expandable, sub-standard.
By the way, in your "dataset" (it isn't the result set produced by your code), the sexes appear to be nicely mixed up.
Therefore the answer, and the only correct one, even if it isn't popular, is that your code is correct (it can be cleaned it up, sure), and you have to educate the user re the dangers of sub-standard code or reports.
You can sort by person.name (rather than by language.name) and then write smarter SQL such that (eg) the person.name is not repeated on the second and subsequent row for persons who speak more than one language, etc. That is just pretty printing.
The non-answer, for those who insist on sub-standard code that will break one day when, is Gordon's response.
Response to Comments
In the Relational Model:
There is no order to the rows, that is deemed a physical or implementation aspect, which we have no control over, and which changes anyway, and which we are warned not to rely upon. If order is sought in the output result set, then we must us ORDER BY, that is its purpose in life.
The data has meaning, and that meaning is carried in Relational Keys. Meaning cannot be carried in surrogates (ie. ID columns).
Limiting myself to the files (they are not tables) that you have given, there is no such thing in the data as:
the first 10 persons who speaks one language
Obtaining persons who speak one language is simple, I believe you already understand that:
SELECT person.first_name,
person.last_name
FROM person P,
(SELECT person_id
FROM person_speaks_language
GROUP BY person_id
HAVING COUNT(*) = 1 -- change this for 2 languages, etc
) AS PL
WHERE P.person_id = PL.person_id
But "first" ? "first" by what criteria ? Record creation date ?
ORDER BY date_created -- if it exists in the data
Record ID does not give first anything: as records are added and deleted, any "order" that may exist initially is completely lost.
You cannot extract meaning out of, or assign meaning to something that, by definition, has no meaning. If the Record ID is relevant, ie. you are going to use it for some purpose, then it is not a Record ID, name the field for what it actually is.
I fail to see, I do not understand, the relevance of the difference between the "dataset" and the updated "small dataset". The "dataset" size is irrelevant, the field headings are irrelevant, what the result set means, is relevant.
The problem is not some "limitation" in the Relational Model, the problem is (a) your fixed view of data values, and (b) your lack of understanding about what the Relational Model is, what it does, understanding of which makes this whole question disappear, and we are left with a simple SQL (as tagged) "how to" question. Eg. If I had a Relational Database, with persons and languages, with no ID columns, there is nothing that I cannot do with it, no report that I cannot produce from it, from the data.
Please try to use an example that conveys the meaning in the data, in what you are trying to do.
the expect results are: each person must be appear only one time
They already appear only once (for each language)
using the language order
Well, there is no order in the language file. We can give it some order, whatever order is meaning-ful, to you, in the result set, based on the data. Eg. language.name. Of course, many persons speak each language, so what order would you like within language.name? How about last_name, first_name. The Record IDs are meaningless to the user, so I won't display them in the result set. NULL is also meaningless, and ambiguous, so I will make the meaning here explicit. This is pretty much what you have, tidied up:
SELECT [language] = CASE name
WHEN NULL THEN "[None]"
ELSE name
END,
last_name,
first_name
FROM person P
LEFT JOIN person_speaks_language PL
ON P.id = PL.person_id
LEFT JOIN language L
ON PL.language_id = L.id
ORDER BY name,
last_name,
first_name
But then you have:
And the expected result is
The example data of which contradicts your textual descriptions:
the expect results are: each person must be appear only one time using the language order
So now, if I ignore the text, and examine the example data re what you want
(which is a horrible thing to do, because I am joining you in the incorrect activity of focussing on the data values, rather than understanding the meaning),
it appears you want the person to appear only once, full stop, regardless of how many languages they speak. Your example data is meaningless, so I cannot be asked to reproduce it. See if this has some meaning.
SELECT last_name,
first_name,
[language] = ( -- correlated subquery
SELECT TOP 1 -- get the "first" language
CASE name -- make meaning of null explicit
WHEN NULL THEN "[None]"
ELSE name
END
FROM person_speaks_language PL
JOIN language L
ON PL.language_id = L.id
WHERE P.id = PL.person_id -- the subject person
ORDER BY name -- id would be meaningless
)
FROM person P -- vector for person, once
ORDER BY last_name,
first_name
Now if you wanted only persons who speak a language (on file):
SELECT last_name,
first_name,
[language] = ( -- correlated subquery
SELECT TOP 1 -- get the "first" language
name
FROM person_speaks_language PL
JOIN language L
ON PL.language_id = L.id
WHERE P.id = PL.person_id -- the subject person
ORDER BY name -- id would be meaningless
)
FROM person P,
(
SELECT DISTINCT person_id -- just one occ, thanks
FROM person_speaks_language PL -- vector for speakers
) AS PL_1
WHERE P.id = PL_1.person_id -- join them to person fields
There, not an outer join anywhere to be seen, in either solution. LEFT or RIGHT will confuse you. Do not attempt to "get everything", so that you can "see" the data values, and then mangle, hack and chop away at the result set, in order to get what you want from that. No, forget about the data values and get only what you want from the record filing system.
Response to Update
I was trying to explain the case with a data set, I think I made things tougher than they actually were
Yes, you did. Reviewing the update then ...
The short answer is, get rid of the ORM. There is nothing in it of value:
you can access the RDB from the queries that populate your objects directly. The way we did for decades before the flatulent beast came along. Especially if you understand and implement Open Architecture Standards.
Further, as evidenced, it creates masses of problems. Here, you are trying to work around the insane restrictions of the ORM.
Pagination is a straight-forward issue, if you have your data Normalised, and Relational Keys.
The long answer is ... please read this Answer. I trust you will understand that the approach you take to designing your app components, your design of windows, will change. All your queries will be simplified, you get only what you require for the specific window or object.
The problem may well disappear entirely (except for possibly the pagination, you might need a method).
Then please think about those architectural issues carefully, and make specific comments of questions.

display the order in relational algebra

Suppose we have this relational schema
homebuilder(hID, hName, hStreet, hCity, hZip, hPhone)
model(hID, mID, mName, sqft, story) subdivision(sName,
sCity, sZip) offered(sName, hID, mID, price) lot(sName,
lotNum, lStAddr, lSize, lPremium) sold(sName, lotNum, hID, mID,
status)
I have problem by doing relational algebra for each subdivision , find the number of models offered and the average, minimum and maximum price of the models offered at that subdivision. Also display the result in descending order on the average price of a home.
I am done with SQL formula, but it hard for me to translate this SQL to relational algebra. Can someone help me?
Here is what I got so far:
SQL:=
SELECT S, avg (O.price), min (O.price), max (O.price), count(*)
FROM offered O, subdivision S
WHERE O.sName = S.sName
GROUP BY S.sName
ORDER BY 4 desc;

+1 to DPenner's comment: quite true that you can't do ordering in RA. (Although those q's and a's referenced seem to have some 'difficulties'.)
Another thing you can't do in RA (contra the SQL that JaveLeave shows) is to have anonymous columns referenced by position. If SQL were a sensible language (or indeed any sort of language at all), you could name the column in the SELECT clause ..., max (O.price) AS maxPrice, ... then ORDER BY maxPrice desc. But no, you can't do that. In SQL you have to repeat ORDER BY max (O.price) desc. (By the way, the question asked for ordering by average price, not max(?) That's column 2.)
Contrast that the RA Group operation returns a relation. And being a relation it must have attributes only addressable by name.
Back to the question as asked. The nearest you can get to an ordering is to put a column on each row with the ordinal position of this row relative to the overall table. Since the question asks for descending sequence, the first step is to find the subdivision with minimum average price and tag it with ordinal 1. Then select all but that one, get the minmium of those, tag it with 2. And in general: take all not tagged so far; get the minmum; tag it with highest tag so far +1; recurse. So you need the transitive closure operation (which is another 'missing' feature of standard RA). You can find some SQL code to achieve this sort of thing in comp.database.theory -- from memory Joe Celko gives examples.
Off-topic: I'm puzzled why courses/professors/textbooks in SQL also ask you to do impossible things in RA. Certainly it's good to have a grounding in RA. It's a powerful mental model to understand data structures that SQL only obscures. RA (as an algebra) underpins most SQL engines. But then why leave the impression that RA is some sort of 'poor cousin' to SQL? There are no commercial implementations of RA; there are no job advertisements for RA programmers. Why try to make it what it isn't?

Is there away in SQL Server to sort by the number of matched words in a contains function on a full text index

I have a table in a database that has a description of an item. I want to be able to have the user type a search term and return the rows that had at least one match, sorted by the number of matches they had, descending.
I don't know if this is possible, I haven't been able to find an answer googling so I'm coming here.
Basically if the user enters "truck blue with gold two tone", this will be generated:
SELECT * FROM MyItemsTable
WHERE contains(Description, 'truck or blue or with or gold or two or tone')
and have that return sorted by the number of words that matched.
Any advice would be greatly appreciated. This table will become very large in time so efficiency is also in the back of my mind as well.

This seems to have worked very well, thanks very much to Gordon Linoff.
SELECT * FROM MyItemsTable m
INNER JOIN
CONTAINSTABLE(MyItemsTable, Description, 'truck or blue or with or gold or two or tone') AS l ON m.MyItemsTable=l.[KEY]

Reference
In case you have a record like "truck blue with gold two tone". You can use below query.
SELECT * FROM
MyItemsTable as t
JOIN CONTAINSTABLE(MyItemsTable , Description,'"truck"') fulltextSearch
ON
t.[Id] = fulltextSearch.[KEY]
This will also bring this record.

Custom SQL sort by

Use:
The user searches for a partial postcode such as 'RG20' which should then be displayed in a specific order. The query uses the MATCH AGAINST method in boolean mode where an example of the postcode in the database would be 'RG20 7TT' so it is able to find it.
At the same time it also matches against a list of other postcodes which are in it's radius (which is a separate query).
I can't seem to find a way to order by a partial match, e.g.:
ORDER BY FIELD(postcode, 'RG20', 'RG14', 'RG18','RG17','RG28','OX12','OX11')
DESC, city DESC
Because it's not specifically looking for RG20 7TT, I don't think it can make a partial match.
I have tried SUBSTR (postcode, -4) and looked into left and right, but I haven't had any success using 'by field' and could not find another route...
Sorry this is a bit long winded, but I'm in a bit of a bind.
A UK postcode splits into 2 parts, the last section always being 3 characters and within my database there is a space between the two if that helps at all.
Although there is a DESC after the postcodes, I do need them to display in THAT particular order (RG20, RG14 then RG18 etc..) I'm unsure if specifying descending will remove the ordering or not

Order By Case
When postcode Like 'RG20%' Then 1
When postcode Like 'RG14%' Then 2
When postcode Like 'RG18%' Then 3
When postcode Like 'RG17%' Then 4
When postcode Like 'RG28%' Then 5
When postcode Like 'OX12%' Then 6
When postcode Like 'OX11%' Then 7
Else 99
End Asc
, City Desc

You're on the right track, trimming the field down to its first four characters:
ORDER BY FIELD(LEFT(postcode, 4), 'RG20', 'RG14', ...),
-- or SUBSTRING(postcode FROM 1 FOR 4)
-- or SUBSTR(postcode, 1, 4)
Here you don't want DESC.
(If your result set contains postcodes whose prefixes do not appear in your FIELD() ordering list, you'll have a bit more work to do, since those records will otherwise appear before any explicitly ordered records you specify. Before 'RG20' in the example above.)

If you want a completely custom sorting scheme, then I only see one way to do it...
Create a table to hold the values upon which to sort, and include a "sequence" or "sort_order" field. You can then join to this table and sort by the sequence field.
One note on the sequence field. It makes sense to create it as an int as... well, sequences are often ints :)
If there is any possibility of changing the sort order, you may want to consider making it alpha numeric... It is a lot easier to insert "5A" between "5 and "6" than it is to insert a number into a sequence of integers.

Another method I use is utilising the charindex function:
order by charindex(substr(postcode,4,1),"RG20RG14RG18...",1)
I think that's the syntax anyway, I'm just doing this in SAS at the moment so I've had to adapt from memory!
But essentially the sooner you hit your desired part of the string, the higher the rank.
If you're trying to rank on a large variety of postcodes then a case statement gets pretty hefty.

Suggestion For Finding Similar Rows In Mysql

i want select similar rows accourding to row's title columun. Title columun has mostly have 5 or 6 six keywords. Which algorithm do you recommend ? Soundex Maybe ?
P.S: Title columun has unicode chracters like Ç, Ö, Ş...

My question's answer mysql full text search. Also it supports unicode.
SELECT *, match(project_title) against('sample project 55') as similarity
FROM projects
WHERE status IN(1, 2, 3, 4, 5, 6) AND id != ? AND match('sample project 55') against(?)
ORDER BY similarity DESC

If you mean similar in spelling and pronunciation, I'd look into using the SOUNDEX function.

Honestly, I'd create a table for keywords(id, external_id, keyword), and then I would join the table against itself, order by how many matches there are, and then grab the rows back out.
If you're matching against a single row, you can select only that one, for much better efficiency with the join.
This could be combined with SOUNDEX to match together things that are close