Tag: multiplication

How to Write a Multiplication Aggregate Function in SQL

Posted on by lukaseder

Everyone knows the SQL SUM() aggregate function (and many people also know its window function variant).

When querying the Sakila database, we can get the daily revenue (using PostgreSQL syntax):

WITH p AS (
  SELECT
    CAST (payment_date AS DATE) AS date,
    amount
  FROM payment
)
SELECT
  date,
  SUM (amount) AS daily_revenue,
  SUM (SUM (amount)) OVER (ORDER BY date) AS cumulative_revenue
FROM p
GROUP BY date
ORDER BY date

The result will look something like this:

date       |daily_revenue |cumulative_revenue 
-----------|--------------|-------------------
2005-05-24 |29.92         |29.92              
2005-05-25 |573.63        |603.55             
2005-05-26 |754.26        |1357.81            
2005-05-27 |685.33        |2043.14            
2005-05-28 |804.04        |2847.18            
2005-05-29 |648.46        |3495.64            
2005-05-30 |628.42        |4124.06            
2005-05-31 |700.37        |4824.43            
2005-06-14 |57.84         |4882.27            
...

Doing the same with multiplication

This is already quite useful. Very occasionally, however, we do not need to aggregate multiple values in a sum (through addition), but in a product (through multiplication). I’ve just stumbled upon such a case on Stack Overflow, recently.

The question wanted to achieve the following result:

date        factor          accumulated
---------------------------------------
1986-01-10  null            1000
1986-01-13  -0.026595745    973.4042548
1986-01-14  0.005464481     978.7234036
1986-01-15  -0.016304348    962.7659569
1986-01-16  0               962.7659569
1986-01-17  0               962.7659569
1986-01-20  0               962.7659569
1986-01-21  0.005524862     968.0851061
1986-01-22  -0.005494506    962.765957
1986-01-23  0               962.765957
1986-01-24  -0.005524862    957.4468078
1986-01-27  0.005555556     962.7659569
1986-01-28  0               962.7659569
1986-01-29  0               962.7659569
1986-01-30  0               962.7659569
1986-01-31  0.027624309     989.3617013
1986-02-03  0.016129032     1005.319148
1986-02-04  0.042328041     1047.872338
1986-02-05  0.04568528      1095.744679

If this were a Microsoft Excel spreadsheet, the ACCUMULATED column would simply start with 1000 and have the following formula in all other rows:

accumulated(i) = accumulated(i - 1) * (1 + factor)

In other words (values truncated for simplicity):

1000.0 = start
 973.4 = 1000.0 * (1 - 0.026)
 978.7 =  973.4 * (1 + 0.005)
 962.7 =  978.7 * (1 - 0.016)
 962.7 =  962.7 * (1 - 0.000)
 962.7 =  962.7 * (1 - 0.000)
 962.7 =  962.7 * (1 - 0.000)
 968.0 =  962.7 * (1 + 0.005)
 ...

This is exciting because we’re not only requiring multiplicative aggregation, but even cumulative multiplicative aggregation. So, another window function.

But regrettably, SQL doesn’t offer a MUL() aggregate function, even if it were relatively simple to implement. We have two options:

  • Implementing a custom aggregate function (stay tuned for a future blog post)
  • Using a trick by summing logarithms, rather than multiplying operands directly

We’re implementing the latter for now. Check out this cool Wikipedia website about logarithmic identities, which we are going to blindly trust. In the middle of it, we have:

bx * by = bx + y

Which leads to:

logb(x * y) = logb(x) + logb(y)

How cool is that? And thus:

x * y = blogb(x) + logb(y)

So, we can define any multiplication in terms of a bunch of exponentiation to some base (say e) and logarithms to some base (say e). Or, in SQL:

x * y = EXP(LN(x) + LN(y))

Or, as an aggregate function:

MUL(x) = EXP(SUM(LN(x)))

Heh!

Our original problem can thus be solved very easily using this, as shown in my stack overflow answer:

SELECT
  date,
  factor,
  EXP(SUM(LN(1000 * (1 + COALESCE(factor, 1)))) 
       OVER (ORDER BY date)) AS accumulated
FROM t

And we get the nice result as previously shown. You may have to replace LN() by LOG() depending on your database.

Caveat: Negative numbers

Try running this:

SELECT LN(-1)

You’ll get:

SQL Error [2201E]: ERROR: cannot take logarithm of a negative number

Logarithms are defined only for strictly positive numbers, unless your database is capable of handling complex numbers as well. In case of which a single zero value would still break the aggregation.

But if your data set is defined to contain only strictly positive numbers, you’ll be fine – give or take some floating point rounding errors. Or, you’ll do some sign handling, which looks like this:

WITH v(i) AS (VALUES (-2), (-3), (-4))
SELECT 
  CASE 
    WHEN SUM (CASE WHEN i < 0 THEN -1 END) % 2 < 0 
    THEN -1 
    ELSE 1 
  END * EXP(SUM(LN(ABS(i)))) multiplication1
FROM v;

WITH v(i) AS (VALUES (-2), (-3), (-4), (-5))
SELECT 
  CASE 
    WHEN SUM (CASE WHEN i < 0 THEN -1 END) % 2 < 0 
    THEN -1 
    ELSE 1 
  END * EXP(SUM(LN(ABS(i)))) multiplication2
FROM v;

The above yielding

multiplication1      
--------------------
-23.999999999999993 


multiplication2     
-------------------
119.99999999999997

Close enough.

Caveat: Zero

Try running this:

SELECT LN(0)

You’ll get:

SQL Error [2201E]: ERROR: cannot take logarithm of zero

Zero is different from negative numbers. A product that has a zero operand is always zero, so we should be able to handle this. We’ll do it in two steps:

  • Exclude zero values from the actual aggregation that uses EXP() and LN()
  • Add an additional CASE expression that checks if any of the operands is zero

The first step might not be necessary depending on how your database optimiser executes the second step.

WITH v(i) AS (VALUES (2), (3), (0))
SELECT 
  CASE 
    WHEN SUM (CASE WHEN i = 0 THEN 1 END) > 0
    THEN 0
    WHEN SUM (CASE WHEN i < 0 THEN -1 END) % 2 < 0 
    THEN -1 
    ELSE 1 
  END * EXP(SUM(LN(ABS(NULLIF(i, 0))))) multiplication
FROM v;

Extension: DISTINCT

Calculating the product of all DISTINCT values requires to repeat the DISTINCT keyword in 2 out of the above 3 sums:

WITH v(i) AS (VALUES (2), (3), (3))
SELECT 
  CASE 
    WHEN SUM (CASE WHEN i = 0 THEN 1 END) > 0
    THEN 0
    WHEN SUM (DISTINCT CASE WHEN i < 0 THEN -1 END) % 2 < 0 
    THEN -1 
    ELSE 1 
  END * EXP(SUM(DISTINCT LN(ABS(NULLIF(i, 0))))) multiplication
FROM v;

The result is now:

multiplication |
---------------|
6              |

Notice that the first SUM() that checks for the presence of NULL values doesn’t require a DISTINCT keyword, so we omit it to improve performance.

Extension: Window functions

Of course, if we are able to emulate a PRODUCT() aggregate function, we’d love to turn it into a window function as well. This can be done simply by transforming each individual SUM() into a window function:

WITH v(i, j) AS (
  VALUES (1, 2), (2, -3), (3, 4), 
         (4, -5), (5, 0), (6, 0)
)
SELECT i, j, 
  CASE 
    WHEN SUM (CASE WHEN j = 0 THEN 1 END) 
      OVER (ORDER BY i) > 0
    THEN 0
    WHEN SUM (CASE WHEN j < 0 THEN -1 END) 
      OVER (ORDER BY i) % 2 < 0 
    THEN -1 
    ELSE 1 
  END * EXP(SUM(LN(ABS(NULLIF(j, 0)))) 
    OVER (ORDER BY i)) multiplication
FROM v;

The result is now:

i |j  |multiplication      |
--|---|--------------------|
1 | 2 |2                   |
2 |-3 |-6                  |
3 | 4 |-23.999999999999993 |
4 |-5 |119.99999999999997  |
5 | 0 |0                   |
6 | 1 |0                   |

So cool! The cumulative product gets bigger and bigger until it hits he first zero, from then on it stays zero.

jOOQ support

jOOQ 3.12 will support this as well and emulate it correctly on all databases:
https://github.com/jOOQ/jOOQ/issues/5939

A note on Oracle performance

Do note that Oracle is very slow to calculate LN(number_type). It can be MUCH faster to calculate, instead LN(binary_double_type). An explicit type cast produced a 100x performance improvement in a simple test, documented here.