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Bob Phillips Bob Phillips is offline
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Default Count of Unique Values in Table of data

Didn't get how to differentiate between Goods In and Out, but this counts
Goods

=COUNT(1/IF($C$2:$C$20="Goods",MATCH($A$2:$A$20,$A$2:$A$20, 0)=ROW($A$2:$A$10)-ROW($A$2)+1))

which is an array formula, so commit with Ctrl-Shift-Enter, not just Enter

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HTH

Bob

(change the xxxx to gmail if mailing direct)


wrote in message
ups.com...
Have a data table that contains the following information. This table
will eventually contain thousands of records, so I've just included a
sample below.

ID Title Type Value Elapsed
PR583089 PL10 - rec6 Goods £2,640.00 5d0h1m
PR583090 PL151 req1 Goods £3,900.00 0d0h7m
PR583090 PL151 req1 Goods £3,900.00 0d0h7m
PR583090 PL151 req1 Goods £3,900.00 0d0h7m
PR583091 PL151 req2 Service £3,900.00 3d0h3m
PR583091 PL151 req2 Service £3,900.00 3d0h3m

TABLE 1:
Type SLA
Goods 3d
Service 5d

I need to be able to analyse how many 'unique' IDs have an elapsed time
of greater than the entries within Table 1. In the example extract
above PR583090 appears 3 times and PR583091 2 times so these only need
to be counted twice. So in total there would be 3 entries. The table
for the above would then show:

Goods Services
In SLA 1 1
Out SLA 1

1 goods entry in SLA (PR583090), 1 goods out of SLA (PR583089) and one
service entry in SLA (PR583091).

Is this possible @ all?

Thanks in advance as always and Happy New Year.

Cheers, Al.