English 中文(简体)
Getting data for histogram plot
原标题:

Is there a way to specify bin sizes in MySQL? Right now, I am trying the following SQL query:

select total, count(total) from faults GROUP BY total;

The data that is being generated is good enough but there are just too many rows. What I need is a way to group the data into predefined bins. I can do this from a scripting language, but is there a way to do it directly in SQL?

Example:

+-------+--------------+
| total | count(total) |
+-------+--------------+
|    30 |            1 | 
|    31 |            2 | 
|    33 |            1 | 
|    34 |            3 | 
|    35 |            2 | 
|    36 |            6 | 
|    37 |            3 | 
|    38 |            2 | 
|    41 |            1 | 
|    42 |            5 | 
|    43 |            1 | 
|    44 |            7 | 
|    45 |            4 | 
|    46 |            3 | 
|    47 |            2 | 
|    49 |            3 | 
|    50 |            2 | 
|    51 |            3 | 
|    52 |            4 | 
|    53 |            2 | 
|    54 |            1 | 
|    55 |            3 | 
|    56 |            4 | 
|    57 |            4 | 
|    58 |            2 | 
|    59 |            2 | 
|    60 |            4 | 
|    61 |            1 | 
|    63 |            2 | 
|    64 |            5 | 
|    65 |            2 | 
|    66 |            3 | 
|    67 |            5 | 
|    68 |            5 | 
------------------------

What I am looking for:

+------------+---------------+
| total      | count(total)  |
+------------+---------------+
|    30 - 40 |            23 | 
|    40 - 50 |            15 | 
|    50 - 60 |            51 | 
|    60 - 70 |            45 | 
------------------------------

I guess this cannot be achieved in a straight forward manner but a reference to any related stored procedure would be fine as well.

最佳回答

This is a post about a super quick-and-dirty way to create a histogram in MySQL for numeric values.

There are multiple other ways to create histograms that are better and more flexible, using CASE statements and other types of complex logic. This method wins me over time and time again since it s just so easy to modify for each use case, and so short and concise. This is how you do it:

SELECT ROUND(numeric_value, -2)    AS bucket,
       COUNT(*)                    AS COUNT,
       RPAD(  , LN(COUNT(*)),  * ) AS bar
FROM   my_table
GROUP  BY bucket;

Just change numeric_value to whatever your column is, change the rounding increment, and that s it. I ve made the bars to be in logarithmic scale, so that they don t grow too much when you have large values.

numeric_value should be offset in the ROUNDing operation, based on the rounding increment, in order to ensure the first bucket contains as many elements as the following buckets.

e.g. with ROUND(numeric_value,-1), numeric_value in range [0,4] (5 elements) will be placed in first bucket, while [5,14] (10 elements) in second, [15,24] in third, unless numeric_value is offset appropriately via ROUND(numeric_value - 5, -1).

This is an example of such query on some random data that looks pretty sweet. Good enough for a quick evaluation of the data.

+--------+----------+-----------------+
| bucket | count    | bar             |
+--------+----------+-----------------+
|   -500 |        1 |                 |
|   -400 |        2 | *               |
|   -300 |        2 | *               |
|   -200 |        9 | **              |
|   -100 |       52 | ****            |
|      0 |  5310766 | *************** |
|    100 |    20779 | **********      |
|    200 |     1865 | ********        |
|    300 |      527 | ******          |
|    400 |      170 | *****           |
|    500 |       79 | ****            |
|    600 |       63 | ****            |
|    700 |       35 | ****            |
|    800 |       14 | ***             |
|    900 |       15 | ***             |
|   1000 |        6 | **              |
|   1100 |        7 | **              |
|   1200 |        8 | **              |
|   1300 |        5 | **              |
|   1400 |        2 | *               |
|   1500 |        4 | *               |
+--------+----------+-----------------+

Some notes: Ranges that have no match will not appear in the count - you will not have a zero in the count column. Also, I m using the ROUND function here. You can just as easily replace it with TRUNCATE if you feel it makes more sense to you.

I found it here http://blog.shlomoid.com/2011/08/how-to-quickly-create-histogram-in.html

问题回答

Mike DelGaudio s answer is the way I do it, but with a slight change:

select floor(mycol/10)*10 as bin_floor, count(*)
from mytable
group by 1
order by 1

The advantage? You can make the bins as large or as small as you want. Bins of size 100? floor(mycol/100)*100. Bins of size 5? floor(mycol/5)*5.

Bernardo.

SELECT b.*,count(*) as total FROM bins b 
left outer join table1 a on a.value between b.min_value and b.max_value 
group by b.min_value

The table bins contains columns min_value and max_value which define the bins. note that the operator "join... on x BETWEEN y and z" is inclusive.

table1 is the name of the data table

Ofri Raviv s answer is very close but incorrect. The count(*) will be 1 even if there are zero results in a histogram interval. The query needs to be modified to use a conditional sum:

SELECT b.*, SUM(a.value IS NOT NULL) AS total FROM bins b
  LEFT JOIN a ON a.value BETWEEN b.min_value AND b.max_value
GROUP BY b.min_value;
select "30-34" as TotalRange,count(total) as Count from table_name
   where total between 30 and 34
union (
select "35-39" as TotalRange,count(total) as Count from table_name 
   where total between 35 and 39)
union (
select "40-44" as TotalRange,count(total) as Count from table_name
   where total between 40 and 44)
union (
select "45-49" as TotalRange,count(total) as Count from table_name
   where total between 45 and 49)
etc ....

As long as there are not too many intervals, this is a pretty good solution.

I made a procedure that can be used to automatically generate a temporary table for bins according to a specified number or size, for later use with Ofri Raviv s solution.

CREATE PROCEDURE makebins(numbins INT, binsize FLOAT) # binsize may be NULL for auto-size
BEGIN
 SELECT FLOOR(MIN(colval)) INTO @binmin FROM yourtable;
 SELECT CEIL(MAX(colval)) INTO @binmax FROM yourtable;
 IF binsize IS NULL 
  THEN SET binsize = CEIL((@binmax-@binmin)/numbins); # CEIL here may prevent the potential creation a very small extra bin due to rounding errors, but no good where floats are needed.
 END IF;
 SET @currlim = @binmin;
 WHILE @currlim + binsize < @binmax DO
  INSERT INTO bins VALUES (@currlim, @currlim+binsize);
  SET @currlim = @currlim + binsize;
 END WHILE;
 INSERT INTO bins VALUES (@currlim, @maxbin);
END;

DROP TABLE IF EXISTS bins; # be careful if you have a bins table of your own.
CREATE TEMPORARY TABLE bins (
minval INT, maxval INT, # or FLOAT, if needed
KEY (minval), KEY (maxval) );# keys could perhaps help if using a lot of bins; normally negligible

CALL makebins(20, NULL);  # Using 20 bins of automatic size here. 

SELECT bins.*, count(*) AS total FROM bins
LEFT JOIN yourtable ON yourtable.value BETWEEN bins.minval AND bins.maxval
GROUP BY bins.minval

This will generate the histogram count only for the bins that are populated. David West ought to be right in his correction, but for some reason, unpopulated bins do not appear in the result for me (despite the use of a LEFT JOIN — I do not understand why).

That should work. Not so elegant but still:

select count(mycol - (mycol mod 10)) as freq, mycol - (mycol mod 10) as label
from mytable
group by mycol - (mycol mod 10)
order by mycol - (mycol mod 10) ASC

via Mike DelGaudio

SELECT
    CASE
        WHEN total <= 30 THEN "0-30"
        WHEN total <= 40 THEN "31-40"       
        WHEN total <= 50 THEN "41-50"
        ELSE "50-"
    END as Total,
    count(*) as count
GROUP BY Total 
ORDER BY Total;

Equal width binning into a given count of bins:

WITH bins AS(
   SELECT min(col) AS min_value
        , ((max(col)-min(col)) / 10.0) + 0.0000001 AS bin_width
   FROM cars
)
SELECT tab.*,
   floor((col-bins.min_value) / bins.bin_width ) AS bin
FROM tab, bins;

Note that the 0.0000001 is there to make sure that the records with the value equal to max(col) do not make it s own bin just by itself. Also, the additive constant is there to make sure the query does not fail on division by zero when all the values in the column are identical.

Also note that the count of bins (10 in the example) should be written with a decimal mark to avoid integer division (the unadjusted bin_width can be decimal).

In addition to great answer https://stackoverflow.com/a/10363145/916682, you can use phpmyadmin chart tool for a nice result:

enter image description here

enter image description here





相关问题
SQL SubQuery getting particular column

I noticed that there were some threads with similar questions, and I did look through them but did not really get a convincing answer. Here s my question: The subquery below returns a Table with 3 ...

please can anyone check this while loop and if condition

<?php $con=mysql_connect("localhost","mts","mts"); if(!con) { die( unable to connect . mysql_error()); } mysql_select_db("mts",$con); /* date_default_timezone_set ("Asia/Calcutta"); $date = ...

php return a specific row from query

Is it possible in php to return a specific row of data from a mysql query? None of the fetch statements that I ve found return a 2 dimensional array to access specific rows. I want to be able to ...

Character Encodings in PHP and MySQL

Our website was developed with a meta tag set to... <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> This works fine for M-dashes and special quotes, etc. However, I ...

Pagination Strategies for Complex (slow) Datasets

What are some of the strategies being used for pagination of data sets that involve complex queries? count(*) takes ~1.5 sec so we don t want to hit the DB for every page view. Currently there are ~...

Averaging a total in mySQL

My table looks like person_id | car_id | miles ------------------------------ 1 | 1 | 100 1 | 2 | 200 2 | 3 | 1000 2 | 4 | 500 I need to ...

热门标签