Finding the nth smallest number in a range, discounting duplicates

I used to know how to do this I think, but I can't find it anywhere. I need something as simple as possible that when I drag it down, each succeeding row gives me the next smallest unique number in the range [PercentsAll].

In other words if the lowest number is 92.53%, and there are 3 such occurrences, I can't have "=SMALL(PercentsAll,ROW(A2))" return 92.53%. I need it to return the next lowest unique number 93.15%

The method I used to know involved a fairly complicated formula, so I'm hoping there's a simpler one out there.

Thanks!

John
=SMALL(PercentsAll,ROW(A1))
=SMALL(PercentsAll,ROW(A2))
=SMALL(PercentsAll,ROW(A3))

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John CarneyReliability Business Tools Analyst IIAsked:
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barry houdiniConnect With a Mentor Commented:
Sorry, the formula for C3 should refer to the cell above, C2, not C3, like this

=MIN(IF(PercentsAll>C3,PercentsAll))

See attached

regards, barry
27405556.xls
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barry houdiniCommented:
Try using this formula in the first cell, e.g. in C2

=MIN(PercentsAll)

Then in C3 use this array formula confirmed with CTRL+SHIFT+ENTER and copied down

=MIN(IF(PercentsAll>C3,PercentsAll))

regards, barry
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ScriptAddictCommented:
If this is a manual process I would

Take the numbers your interested in an filter the list by unique records.  

Then I would just sort it ascending or decending.  

If you need a complex formula to do it, just let me know.
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ScriptAddictCommented:
You would filter the list by unique records by selecting data, then in the filter section advanced, then check the unique records box.
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barry houdiniCommented:
An improvement to my suggestion...

Use this formula for a count of the different values

=SUMPRODUCT((A2:A50<>"")/COUNTIF(A2:A50,A2:A50&""))

then the formula to list the unique values from smallest to largest can be amended so that it can be copied down further than you need, extra cells will be blank - revised formula in D3 copied down

=IF(C$2>ROWS(D$3:D3),MIN(IF(PercentsAll>D2,PercentsAll)),"")

regards, barry
27405556v2.xls
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John CarneyReliability Business Tools Analyst IIAuthor Commented:
This is perfect, thanks, Barry. And I love how simple it is. :-)

- John
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