Just curious:
Which of your 9 suggested solutions use the least computing power and
computes the fastest?

"ryguy7272" wrote:

I can come up with at least 9 ways to do this; any more is moot:
=SUMPRODUCT((A1:A78<"")/(COUNTIF(A1:A78,A1:A78)+(A1:A78="")))
=SUM(IF(FREQUENCY(IF(LEN(A1:A971)0,MATCH(A1:A971, A1:A971,0),""),IF(LEN(A1:A971)0,MATCH(A1:A971,A1: A971,0),""))0,1))
(this is a CSE function; you must hit Ctrl + Shift + Enter at the same time)
=SUM(IF(A1:A400<"",1/COUNTIF(A1:A400,A1:A400)))
(this is a CSE function; you must hit Ctrl + Shift + Enter at the same time)
=SUMPRODUCT((A1:A78<"")/(COUNTIF(A1:A78,A1:A78&"")))
(this is a CSE function; you must hit Ctrl + Shift + Enter at the same time)
=SUM(--(FREQUENCY(IF(A1:A2676<"",MATCH(A1:A2676,A1:A2676 ,0)),ROW(INDIRECT("1:"&ROWS(A1:A2676))))0))
(this is a CSE function; you must hit Ctrl + Shift + Enter at the same time)

Regards,
Ryan---

--
RyGuy


"Lars-Ã…ke Aspelin" wrote:

On Mon, 14 Jul 2008 04:17:02 -0700, matt3542
wrote:

Hi there,

I am trying to count the number of values in a column (A2:A217) ignoring
duplicated values. As an eg, applying this to the data below I would expect
the count value to be 3. Please can anyone help, many thanks, Matt

00013270
00018038
00018038
00018038
00018038
00018038
00018038
00018038
00018038
00018038
00018038
00023049
00023049
00023049


Try this formula:

=SUM(1/COUNTIF(A2:A217,A2:A217))

Hope this helps / Lars-Ã…ke