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:

b^{x}* b^{y}= b^{x + y}

Which leads to:

log_{b}(x * y) = log_{b}(x) + log_{b}(y)

How cool is that? And thus:

x * y = b^{logb(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.