Combining Rank and Unique

[alibini]

Is there a way I can combine Rank and Unique to give me ranking of unique values from a column.

Tried this formula:

=RANK(U9,UNIQUE(FILTER($U$9:$U$18,U9:U18>0)))

but RANK doesn't seem to accept Unique as the range.
Thanks.

 

[JEC]

Book1
	A	B	C	D
1	absolute
2	1000		1000	2
3	50045		50045	1
4	1000		455	4
5	455		500	3
6	0
7	500
8	0
9	0
10	500
Sheet1

Cell Formulas
Range	Formula
C2:D5	C2	=LET(z,A2:A10,r,UNIQUE(FILTER(z,z>0)),HSTACK(r,MATCH(r,SORT(r,,-1),0)))
Dynamic array formulas.

 

[Fluff]

How about

=LET(f,UNIQUE(FILTER(U9:U18,U9:U18>0)),XMATCH(U9,SORT(f,,-1)))

 

[Eric W]

Another:

=SUM(--(U9<UNIQUE(FILTER($U$9:$U$18,$U$9:$U$18<>0))))+1

Just FYI, RANK requires the second parameter to be a range, not an array. That's why it didn't work. Also, your profile shows you use Excel 2016, which doesn't have UNIQUE or FILTER. You might wish to update your profile.

 

[alibini]

Another:

=SUM(--(U9<UNIQUE(FILTER($U$9:$U$18,$U$9:$U$18<>0))))+1

Just FYI, RANK requires the second parameter to be a range, not an array. That's why it didn't work. Also, your profile shows you use Excel 2016, which doesn't have UNIQUE or FILTER. You might wish to update your profile.

Thanks for the solution and profile updated.

 

[Eric W]

Glad we could help, and thanks for updating your profile!

 

[alibini]

Another:

=SUM(--(U9<UNIQUE(FILTER($U$9:$U$18,$U$9:$U$18<>0))))+1

Just FYI, RANK requires the second parameter to be a range, not an array. That's why it didn't work. Also, your profile shows you use Excel 2016, which doesn't have UNIQUE or FILTER. You might wish to update your profile.

Hi, just looking back on this solution, which works great for me. I was wondering since I apply this formula to a very large data set - approx 80,000 rows..... Is there a faster solution, possibly by doing the calculation in vba rather than with actual formula?

Thanks

 

[Peter_SSs]

I apply this formula to a very large data set - approx 80,000 rows..... Is there a faster solution, possibly by doing the calculation in vba rather than with actual formula?

My test was with nearly 8,000 rows only but these approximate calculation times on my (old) machine may give you something to consider.
From your attempted formula in post 1 I wasn't certain what you were trying to do with zero values or whether you may have blank cells in the range. For the moment I have assumed no blanks and 0 values to be ranked at the bottom like Eric's formula does.

Formula from post 4 (adjusted to row 8,000)
About 30 seconds

Formula from post 3 (adjusted to row 8,000)
About 30 seconds

This formula (only needs to be placed in the top cell and not copied down)
0.04 seconds

=MATCH(U9:U8000,SORT(UNIQUE(U9:U8000),,-1),0)

The vba code below (which places the results in Y9 and below but can be varied in the second last line of the code)
0.09 seconds

Sub Rank_Unique()
  Dim SL As Object
  Dim a As Variant, b As Variant
  Dim i As Long, SLC As Long

  Set SL = CreateObject("System.Collections.Sortedlist")
  a = Range("U9", Range("U" & Rows.Count).End(xlUp)).Value
  ReDim b(1 To UBound(a), 1 To 1)
  For i = 1 To UBound(a)
    If Not SL.Contains(a(i, 1)) Then SL.Add a(i, 1), 1
  Next i
  SLC = SL.Count
  For i = 1 To UBound(a)
    b(i, 1) = SLC - SL.IndexOfKey(a(i, 1))
  Next i
  Range("Y9").Resize(UBound(b)).Value = b
End Sub

 

[alibini]

My test was with nearly 8,000 rows only but these approximate calculation times on my (old) machine may give you something to consider.
From your attempted formula in post 1 I wasn't certain what you were trying to do with zero values or whether you may have blank cells in the range. For the moment I have assumed no blanks and 0 values to be ranked at the bottom like Eric's formula does.

Formula from post 4 (adjusted to row 8,000)
About 30 seconds

Formula from post 3 (adjusted to row 8,000)
About 30 seconds

This formula (only needs to be placed in the top cell and not copied down)
0.04 seconds

=MATCH(U9:U8000,SORT(UNIQUE(U9:U8000),,-1),0)

The vba code below (which places the results in Y9 and below but can be varied in the second last line of the code)
0.09 seconds

Sub Rank_Unique()
  Dim SL As Object
  Dim a As Variant, b As Variant
  Dim i As Long, SLC As Long

  Set SL = CreateObject("System.Collections.Sortedlist")
  a = Range("U9", Range("U" & Rows.Count).End(xlUp)).Value
  ReDim b(1 To UBound(a), 1 To 1)
  For i = 1 To UBound(a)
    If Not SL.Contains(a(i, 1)) Then SL.Add a(i, 1), 1
  Next i
  SLC = SL.Count
  For i = 1 To UBound(a)
    b(i, 1) = SLC - SL.IndexOfKey(a(i, 1))
  Next i
  Range("Y9").Resize(UBound(b)).Value = b
End Sub

Thanks Steve, I apologise but I think I was not explicit enough in what I'm trying to do.

I am trying to rank rows, based on a value in a column, where values in other columns are unique.

In the below table for example, I want to rank the Value (without duplicates) where the Identifier AND Date are the same. I have manually added the desired result for example. As mentioned my challenge is a large data set.

Currently I am using something like this:

=SUM(--(c3<UNIQUE(FILTER(Value_Range,(Identifier=A2)*(Date = B2)))))+1

Identifier	Date	Value	Desired Result
1	1-May	1.5	1
1	1-May	1.5	1
1	1-May	1.6	2
1	1-May	1.7	3
1	5-Jun	1.8	3
1	5-Jun	1.8	3
1	5-Jun	1.6	2
1	5-Jun	1.6	2
1	5-Jun	1.9	4
1	5-Jun	1.5	1
2	3-Nov	1.5	1
2	3-Nov	1.5	1
2	3-Nov	1.6	2
2	3-Nov	1.7	3
2	5-Jun	1.8	4
2	5-Jun	1.8	4
2	5-Jun	1.6	3
2	5-Jun	1.6	3
2	5-Jun	1.1	1
2	5-Jun	1.5	2

 

[Peter_SSs]

Thanks Steve, ..

Peter?
Edit: Name correction noted as per your direct message.

looking back on this solution, which works great for me. I was wondering since I apply this formula to a very large data set - approx 80,000 rows..... Is there a faster solution,

I was answering the question that you asked - a faster method for a solution that "works great" for you.

I want to rank the Value (without duplicates) where the Identifier AND Date are the same. I have manually added the desired result for example.

This is a very different question. Not only have you introduced other columns that were not mentioned in post 1 or shown in your post 1 formula, but you also now seem to be ranking low numbers first rather than high numbers as in your post 1 formula and the suggestion that you said "works great".