Calculating Descriptive Statistics

13.2.1 Problem

You want to characterize a dataset by computing general descriptive or summary statistics.

13.2.2 Solution

Many common descriptive statistics, such as mean and standard deviation, can be obtained by applying aggregate functions to your data. Others, such as median or mode, can be calculated based on counting queries.

13.2.3 Discussion

Suppose you have a table testscore containing observations representing subject ID, age, sex, and test score:

mysql> SELECT subject, age, sex, score FROM testscore ORDER BY subject; +---------+-----+-----+-------+ | subject | age | sex | score | +---------+-----+-----+-------+ | 1 | 5 | M | 5 | | 2 | 5 | M | 4 | | 3 | 5 | F | 6 | | 4 | 5 | F | 7 | | 5 | 6 | M | 8 | | 6 | 6 | M | 9 | | 7 | 6 | F | 4 | | 8 | 6 | F | 6 | | 9 | 7 | M | 8 | | 10 | 7 | M | 6 | | 11 | 7 | F | 9 | | 12 | 7 | F | 7 | | 13 | 8 | M | 9 | | 14 | 8 | M | 6 | | 15 | 8 | F | 7 | | 16 | 8 | F | 10 | | 17 | 9 | M | 9 | | 18 | 9 | M | 7 | | 19 | 9 | F | 10 | | 20 | 9 | F | 9 | +---------+-----+-----+-------+

A good first step in analyzing a set of observations is to generate some descriptive statistics that summarize their general characteristics as a whole. Common statistical values of this kind include:

Aside from the median and mode, all of these can be calculated easily by invoking aggregate functions:

mysql> SELECT COUNT(score) AS n, -> SUM(score) AS sum, -> MIN(score) AS minimum, -> MAX(score) AS maximum, -> AVG(score) AS mean, -> STD(score) AS 'std. dev.' -> FROM testscore; +----+------+---------+---------+--------+-----------+ | n | sum | minimum | maximum | mean | std. dev. | +----+------+---------+---------+--------+-----------+ | 20 | 146 | 4 | 10 | 7.3000 | 1.7916 | +----+------+---------+---------+--------+-----------+

The aggregate functions as used in this query count only non-NULL observations. If you use NULL to represent missing values,you may want to perform an additional characterization to assess the extent to which values are missing. (See Recipe 13.5.)

Variance is not shown in the query, and MySQL has no function for calculating it. However, variance is just the square of the standard deviation, so it's easily computed like this:

STD(score) * STD(score)

STDDEV( ) is a synonym for STD( ).

Standard deviation can be used to identify outliersvalues that are uncharacteristically far from the mean. For example, to select values that lie more than three standard deviations from the mean, you can do something like this:

SELECT @mean := AVG(score), @std := STD(score) FROM testscore; SELECT score FROM testscore WHERE ABS(score-@mean) > @std * 3;

For a set of n values, the standard deviation produced by STD( ) is based on n degrees of freedom. This is equivalent to computing the standard deviation "by hand" as follows (@ss represents the sum of squares):

mysql> SELECT -> @n := COUNT(score), -> @sum := SUM(score), -> @ss := SUM(score*score) -> FROM testscore; mysql> SELECT @var := ((@n * @ss) - (@sum * @sum)) / (@n * @n); mysql> SELECT SQRT(@var); +------------+ | SQRT(@var) | +------------+ | 1.791647 | +------------+

To calculate a standard deviation based on n-1 degrees of freedom instead, do it like this:

mysql> SELECT -> @n := COUNT(score), -> @sum := SUM(score), -> @ss := SUM(score*score) -> FROM testscore; mysql> SELECT @var := ((@n * @ss) - (@sum * @sum)) / (@n * (@n - 1)); mysql> SELECT SQRT(@var); +------------+ | SQRT(@var) | +------------+ | 1.838191 | +------------+

Or, more simply, like this:

mysql> SELECT @n := COUNT(score) FROM testscore; mysql> SELECT STD(score)*SQRT(@n/(@n-1)) FROM testscore; +----------------------------+ | STD(score)*SQRT(@n/(@n-1)) | +----------------------------+ | 1.838191 | +----------------------------+

MySQL has no built-in function for computing the mode or median of a set of values, but you can compute them yourself. The mode is the value that occurs most frequently. To determine what it is, count each value and see which one is most common:

mysql> SELECT score, COUNT(score) AS count -> FROM testscore GROUP BY score ORDER BY count DESC; +-------+-------+ | score | count | +-------+-------+ | 9 | 5 | | 6 | 4 | | 7 | 4 | | 4 | 2 | | 8 | 2 | | 10 | 2 | | 5 | 1 | +-------+-------+

In this case, 9 is the modal score value.

The median of a set of ordered values can be calculated like this:[1]

[1] Note that the definition of median given here isn't fully general; it doesn't address what to do if there's duplication of the middle values in the dataset.

Based on that definition, use the following procedure to determine the median of a set of observations stored in the database:

For example, if a table t contains a score column with 37 values (an odd number), you need to select a single value, using a query like this:

SELECT score FROM t ORDER BY 1 LIMIT 18,1

If the column contains 38 values (an even number), the query becomes:

SELECT score FROM t ORDER BY 1 LIMIT 18,2

Then you can select the value or values returned by the query and compute the median from their average.

The following Perl function implements a median calculation. It takes a database handle and the names of the table and column that contain the set of observations, then generates the query that retrieves the relevant values, and returns their average:

sub median { my ($dbh, $tbl_name, $col_name) = @_; my ($count, $limit); $count = $dbh->selectrow_array ("SELECT COUNT($col_name) FROM $tbl_name"); return undef unless $count > 0; if ($count % 2 == 1) # odd number of values; select middle value { $limit = sprintf ("LIMIT %d,1", ($count-1)/2); } else # even number of values; select middle two values { $limit = sprintf ("LIMIT %d,2", $count/2 - 1); } my $sth = $dbh->prepare ( "SELECT $col_name FROM $tbl_name ORDER BY 1 $limit"); $sth->execute ( ); my ($n, $sum) = (0, 0); while (my $ref = $sth->fetchrow_arrayref ( )) { ++$n; $sum += $ref->[0]; } return ($sum / $n); }

The preceding technique works for a set of values stored in the database. If you happen to have already fetched an ordered set of values into an array @val, you can compute the median like this instead:

if (@val == 0) # if array is empty, median is undefined { $median = undef; } elsif (@val % 2 == 1) # if array size is odd, median is middle number { $median = $val[(@val-1)/2]; } else # array size is even; median is average { # of two middle numbers $median = ($val[@val/2 - 1] + $val[@val/2]) / 2; }

The code works for arrays that have an initial subscript of 0; for languages that used 1-based array indexes, adjust the algorithm accordingly.

Related

Категории