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.

I discussed this in detail in two of my books many years ago. Here are the references:

Joe Celko’s SQL for Smarties: Advanced SQL Programming – Page 468

https://books.google.com/books?isbn=012800830X

Chapter 23.8.2 The PRD() Aggregate Function by Logarithms Roy Harvey, another SQL guru, found a different solution, which only someone old enough to remember slide rules and that we can multiply by …

Joe Celko’s Thinking in Sets: Auxiliary, Temporal, and Virtual …

https://books.google.com/books?isbn=008055752X

Doesn’t it blow on zero? Isn’t it a bit slow?

Yes and yes

My version covers zeros and negative numbers in the group. You need a simple CASE expression.

Without having tried too hard, I couldn’t get PostgreSQL to avoid the LN(..) call with a zero argument using a CASE expression. How did you do it?

UPDATE:Never mind. Of course, a`NULLIF()`

call or similar expression prior to passing the argument to`LN()`

will suffice. I’ve updated the blog post. Thanks for your pointers!Here is a “cut & paste” from one of my books:

29.06.02. The PRD() Aggregate Function by Logarithms

Roy Harvey, another SQL guru who answered questions on CompuServe, found a different solution, which only someone old enough to remember slide rules and that we can multiply by adding logs. The nice part of this solution is that you can also use the DISTINCT option in the SUM() function.

But there are a lot of warnings about this approach. The Standard allows only natural logarithms shown as LN(), but you will see LOG10() for the logarithm base ten and perhaps LOG(, ) for a general logarithm function. Since the logarithm of zero or less is undefined, the Standard requires an exception to be raised. But some older SQL might return a zero or a NULL. Likewise, the Standard also defines the exponential function as EXP() as its inverse.

The expression for the product of a column from logarithm and exponential functions is:

The nice part of this is that you can also use the SUM (DISTINCT ) option to get the equivalent of PRD (DISTINCT ).

You should watch the data type of the column involved and use either integer 0 and 1 or decimal 0.00 and 1.00 as is appropriate in the CASE statements. It is worth studying the three CASE expressions that make up the terms of the Prod calculation.

The first CASE expression is to ensure that all zeros and negative numbers are converted to a non-negative or NULL for the SUM() function, just in case your SQL raises an exception.

The second CASE expression will return zero as the answer if there was a zero in the nbr column of any selected row. The MIN(ABS(nbr)) is a handy trick for detecting the existence of a zero in a list of both positive and negative numbers with an aggregate function.

The third CASE expression will return minus one if there was an odd number of negative numbers in the nbr column. The innermost CASE expression uses a SIGN() function which returns + 1 for a positive number, -1 for a negative number and 0 for a zero. The SUM() counts the -1 results then the MOD() functions determines if the count was odd or even.

I present this version of the query first because this is how I developed the answer. We can do a much better job with a little algebra and logic:

The idea is that there are three special cases – all positive numbers, one or more zeros, and some negative numbers in the set. You can find out what your situation is with a quick test on the SIGN() of the minimum value in the set.

In the case where you have negative numbers, there are two sub-cases: (1) an even number of negatives or (2) an odd number of negatives. You then need to apply some High School algebra to determine the sign of the final result.

Itzak Ben-Gan had problems in implementing this in an older version of SQL Server which is worth passing along in case your SQL product also has them. The query as written returns a domain error in SQL Server, even though it should not have the result expressions in the CASE expression been evaluated after the conditional flow had performed a short-circuit evaluation. Examining the execution plan of the above query, it looks like the optimizer evaluates all of the possible result expressions in a step prior to handling the flow of the CASE expression.

This means that in the expression after WHEN 1 … the LN() function is also invoked in an intermediate phase for zeros and negative numbers, and in the expression after WHEN -1 … the LN(ABS()) is also invoked in an intermediate phase for zeroes. This explains the domain error.

To handle this, I had to use the ABS() and NULLIF() functions in the positive numbers WHEN clause, and the NULLIF() function in the negative numbers WHEN clause:

If you are sure that you will have only positive values in the column being computed, then you can use

As an aside, the book BYPASSES: A SIMPLE APPROACH TO COMPLEXITY by Z. A. Melzak (Wiley-Interscience, 1983, ISBN 0-471-86854-X), is a short mathematical book on the general principle of conjugacy. This is the method of using a transform and its inverse to reduce the complexity of a calculation.

I’m having trouble figuring out how to calculate the daily compound interest for an initial amount, over various rates periods, producing a new total that includes the interest amounts from each rate period. The challenge is that for each subsequent rate period you have to calculate the interest on the amount plus the previous interest!!! So it’s not a simple running total.

For example, using the following rates table.

rate from date rate to date rate

2013-07-15 2013-09-30 3

2013-10-01 2013-12-31 4

2014-01-01 2014-03-31 3

Using an initial amount of $32,550.37, I have to traverse each rate period with an interest calculation, producing the final amount of $33,337.34.

rate from rate to rate daysx amount interest

2013-07-15 2013-09-30 .03 78 32,550.37 209.34

2013-10-01 2013-12-31 .04 92 32,759.71 331.94

2014-01-01 2014-03-31 .03 90 33,091.65 245.69

Final Amount 33,337.34

For example, the initial amount of $32,550.37 has interest of $209.34 at 3%. For the second rate period, I add that interest to the amount, which is $32,759.71 and then calculate the interest on $32,759.71 at 4%. Etc.

I’m using Netezza which does not allow recursive SQL, so I have been trying to use windowed functions, but not with any success yet …

DROP TABLE TRATES;

CREATE TABLE TRATES (RATE_FROM_DATE DATE, RATE_TO_DATE DATE, RATE DECIMAL(10,2));

INSERT INTO TRATES VALUES (‘2013-07-15′,’2013-09-30’,.03);

INSERT INTO TRATES VALUES (‘2013-10-01′,’2013-12-31’,.04);

INSERT INTO TRATES VALUES (‘2014-01-01′,’2014-03-31’,.03);

SELECT TRATES.*

, DAYS_BETWEEN(RATE_FROM_DATE, RATE_TO_DATE)+1 AS DAYSX

, (AMOUNT * POW(1+(RATE)/365,(DAYS_BETWEEN(RATE_FROM_DATE, RATE_TO_DATE)+1)))) – AMOUNT

AS INTEREST

, FIRST_VALUE(AMOUNT) OVER(ORDER BY RATE_FROM_DATE)

*(POW(1+(RATE/100)/365,(DAYS_BETWEEN(RATE_FROM_DATE, RATE_TO_DATE)+1)))

AS NEW_AMOUNT

FROM TRATES

JOIN (SELECT 32550.37 AS AMOUNT) AS TPARMS ON 1=1

;

Any help would be greatly appreciated.

Mark, have you considered asking your question on http://stackoverflow.com