While browsing the forum for useful tips I spotted this formula posted
by Bob Phillips On Sept 21. It counts the number of unique records in a
list (A2:A20 in the example):
=SUMPRODUCT((A2:A20<"")/COUNTIF(A2:A20,A2:A20&""))

I found this formula very interesting and I just had to try it out. It
works, but I can't for the life of me understand how it works.
I've read the SUMPRODUCT Help file but that hasn't helped me.
COUNTIF Help file didn't help either. I can't see how A2:A20&"" works
as the Criterion for the COUNTIF part of the formula.
Can anybody help me understand Bob's incredible formula?

Ken,

Spookily, Richard Buttrey asked the same question this morning.

Here is my reply re-posted



Let's start by defining the range A1:A20 to talk specifics.

Bob,John,Bob,Bob,John,John,Bob,Bill,Bill,Max

or data in just A1:A10

The basic formula to count unique items is
=SUMPRODUCT(1/COUNTIF($A$1:$A$10,$A$1:$A$10))
The COUNTIF($A$1:$A$10,$A$1:$A$10) part of the formula builds an array of
the number of occurrences of each item, in this case{4;3;4;4;3;3;4;2;2;1}.
As can be seen, each occurrence of the repeated value is counted, so there
are four occurrences of Bob in the array. There will always be the same
number of occurrences of value as the count of that value, unless two or
more items are repeated the same number of times, in which case it will be
some multiple of that count.
Thus the item that is repeated 4 times has 4 instances of that count,
dividing 1 by the count of 4, gives 0.25 4 times. The full array of values
is
{0.25;0.333333333333333;0.25;0.25;0.33333333333333 3;0.333333333333333;0.25;0
..5;0.5;1}.
The item that repeats 4 times sums to 1. The item that repeats 3 times also
sums to 1. It should be clear from this that every value works in the same
way and sums to 1. In other words, 1 is returned for every unique item. The
sum of these values becomes the count of unique items.
As our test range is A1:A20, and some of the items in A1:A20 are blank,
extending this formula to A1:A20 would return a #DIV/0! Error.
The reason for the error is blank cells in the full range A1:A20. Each blank
cell returns a 0 value from the COUNTIF formula, which gives the #DIV/0!
Error when divided into 1.
The solution to this is to force it to count the empty cells as well, and
not return a zero. Adding &"" to the end of the COUNTIF formula forces a
count of the blanks.
This addition on its own removes the #DIV/0! error, but will cause the
blanks to be counted as a unique item. A further addition to the formula
resolves this by testing for those blanks. Instead of dividing the array of
counts into 1 each time, adding the test creates an array of TRUE/FALSE
values to be divided by the equivalent element in the counts array. Each
blank will resolve to FALSE in the dividend array, and the count of the
blanks in the divisor array. The result of this will be 0, so the blanks do
not get counted.


--
HTH

Bob Phillips

"Ken Johnson" wrote in message
oups.com...
While browsing the forum for useful tips I spotted this formula posted
by Bob Phillips On Sept 21. It counts the number of unique records in a
list (A2:A20 in the example):
=SUMPRODUCT((A2:A20<"")/COUNTIF(A2:A20,A2:A20&""))

I found this formula very interesting and I just had to try it out. It
works, but I can't for the life of me understand how it works.
I've read the SUMPRODUCT Help file but that hasn't helped me.
COUNTIF Help file didn't help either. I can't see how A2:A20&"" works
as the Criterion for the COUNTIF part of the formula.
Can anybody help me understand Bob's incredible formula?

Bob,
Thanks for that, it's going to take me a while to work through it. It's
easily the most amazing bit of excel wizardry I've ever seen. Is it
your own invention?

I wish I could lay some claim to that but I cannot. It is doubtful that
anyone can claim these things absolutely as so many contribute to their
evolution. I first saw it in a more simplistic form

=SUMPRODUCT(1/COUNTIF(A1:A20,A1:A20)

and it has evolved over the years to what I showed that day.

There are better ones out there IMO, although that does have the appeal of
brevity. I saw one from Domenic a few weeks back that sums in non-contiguous
cells that I really liked

=SUMPRODUCT(--ISNUMBER(MATCH(COLUMN(C1:N1)-COLUMN(C1)+1,{1,3,6,8,12},0)),--(
C1:N1=1))

This sums C1,E1,H1,J1,N1 only. I had a solution to this problem, but not as
elegant.

Here is one of mine for extracting the numbers from a delimited string and
sum them. So if A1 contains 1,12,123 it will get the 1 and 123 out and sum
them

=SUMPRODUCT(--MID(A1,FIND("~",SUBSTITUTE(B1&A1&B1,B1,"~",ROW(IND IRECT("1:"&L
EN(A1)-LEN(SUBSTITUTE(A1,B1,""))+1)))),
FIND("~",SUBSTITUTE(B1&A1&B1,B1,"~",ROW(INDIRECT(" 2:"&LEN(A1)-LEN(SUBSTITUTE
(A1,B1,""))+2))))-
FIND("~",SUBSTITUTE(B1&A1&B1,B1,"~",ROW(INDIRECT(" 1:"&LEN(A1)-LEN(SUBSTITUTE
(A1,B1,""))+1))))-1))

If you think that all this is useful, take a look at
http://www.xldynamic.com/source/xld.SUMPRODUCT.html


--
HTH

Bob Phillips

"Ken Johnson" wrote in message
ups.com...
Bob,
Thanks for that, it's going to take me a while to work through it. It's
easily the most amazing bit of excel wizardry I've ever seen. Is it
your own invention?

Bob,
thanks for that Bob!
I'm just starting a two week break from teaching (High school science)
and you've given me plenty to think about and keep me off the streets.
Ken Johnson