Count Of Distinct Characters In Column - sql

Question

Say I have the following data set
Column1 (VarChar(50 or something))
Elias
Sails
Pails
Plane
Games
What I'd like to produce from this column is the following set:
LETTER COUNT
E 3
L 4
I 3
A 5
S 5
And So On...
One solution I thought of was combining all strings into a single string, and then count each instance of the letter in that string, but that feels sloppy.
This is more an exercise of curiosity than anything else, but, is there a way to get a count of all distinct letters in a dataset with SQL?

Answer 1

I would do this by creating a table of your letters similar to:
CREATE TABLE tblLetter
(
letter varchar(1)
);
INSERT INTO tblLetter ([letter])
VALUES
('a'),
('b'),
('c'),
('d'); -- etc
Then you could join the letters to your table where your data is like the letter:
select l.letter, count(n.col) Total
from tblLetter l
inner join names n
on n.col like '%'+l.letter+'%'
group by l.letter;
See SQL Fiddle with Demo. This would give a result:
| LETTER | TOTAL |
|--------|-------|
| a | 5 |
| e | 3 |
| g | 1 |
| i | 3 |
| l | 4 |
| m | 1 |
| p | 2 |
| s | 4 |

Answer 2

If you create a table of letters, like this:
create table letter (ch char(1));
insert into letter(ch) values ('A'),('B'),('C'),('D'),('E'),('F'),('G'),('H')
,('I'),('J'),('K'),('L'),('M'),('N'),('O'),('P')
,('Q'),('R'),('S'),('T'),('U'),('V'),('W'),('X'),('Y'),('Z');
you could do it with a cross join, like this:
select ch, SUM(len(str) - len(replace(str,ch,'')))
from letter
cross join test -- <<== test is the name of the table with the string
group by ch
having SUM(len(str) - len(replace(str,ch,''))) <> 0
Here is a running demo on sqlfiddle.
You can do it without defining a table by embedding a list of letters into a query itself, but the idea of cross-joining and grouping by the letter would remain the same.
Note: see this answer for the explanation of the expression inside the SUM.

Answer 3

To me, this is a problem almost tailored for a CTE (Thanks, Nicholas Carey, for the original, my fiddle here: http://sqlfiddle.com/#!3/44f77/8):
WITH cteLetters
AS
(
SELECT
1 AS CharPos,
str,
MAX(LEN(str)) AS MaxLen,
SUBSTRING(str, 1, 1) AS Letter
FROM
test
GROUP BY
str,
SUBSTRING(str, 1, 1)
UNION ALL
SELECT
CharPos + 1,
str,
MaxLen,
SUBSTRING(str, CharPos + 1, 1) AS Letter
FROM
cteLetters
WHERE
CharPos + 1 <= MaxLen
)
SELECT
UPPER(Letter) AS Letter,
COUNT(*) CountOfLetters
FROM
cteLetters
GROUP BY
Letter
ORDER BY
Letter;
Use the CTE to calculate character positions and deconstruct each string. Then you can just aggregate from the CTE itself. No need for additional tables or anything.

Answer 4

This should work even if you have case sensitivity turned on.
The setup:
CREATE TABLE _test ( Column1 VARCHAR (50) )
INSERT _test (Column1) VALUES ('Elias'),('Sails'),('Pails'),('Plane'),('Games')
The work:
DECLARE #counter AS INT
DECLARE #results TABLE (LETTER VARCHAR(1),[COUNT] INT)
SET #counter=65 --ascii value for 'A'
WHILE ( #counter <=90 ) -- ascii value for 'Z'
BEGIN
INSERT #results (LETTER,[COUNT])
SELECT CHAR(#counter),SUM(LEN(UPPER(Column1)) - LEN(REPLACE(UPPER(Column1), CHAR(#counter),''))) FROM _test
SET #counter=#counter+1
END
SELECT * FROM #results WHERE [Count]>0

Answer 5

It's often useful to have a range or sequence table that gives you a source of large runs of contiguous sequential numbers, like this one covering the range -100,000–+100,000.
drop table dbo.range
go
create table dbo.range
(
id int not null primary key clustered ,
)
go
set nocount on
go
declare #i int = -100000
while ( #i <= +100000 )
begin
if ( #i > 0 and #i % 1000 = 0 ) print convert(varchar,#i) + ' rows'
insert dbo.range values ( #i )
set #i = #i + 1
end
go
set nocount off
go
Once you have such a table, you can do something like this:
select character = substring( t.some_column , r.id , 1 ) ,
frequency = count(*)
from dbo.some_table t
join dbo.range r on r.id between 1 and len( t.some_column )
group by substring( t.some_column , r.id , 1 )
order by 1
If you want to ensure case-insensitivity, just mix in the desired upper() or lower():
select character = upper( substring( t.some_column , r.id , 1 ) ) ,
frequency = count(*)
from dbo.some_table t
join dbo.range r on r.id between 1 and len( t.some_column )
group by upper( substring( t.some_column , r.id , 1 ) )
order by 1
Given your sample data:
create table dbo.some_table
(
some_column varchar(50) not null
)
go
insert dbo.some_table values ( 'Elias' )
insert dbo.some_table values ( 'Sails' )
insert dbo.some_table values ( 'Pails' )
insert dbo.some_table values ( 'Plane' )
insert dbo.some_table values ( 'Games' )
go
The latter query above produces the following results:
character frequency
A 5
E 3
G 1
I 3
L 4
M 1
N 1
P 2
S 5