Hi All,

I have a Named Range called "Data" that spans 8 columns and 7 rows. A Numeric
Value will appear only once in a row. Each row's Numeric Values are in
ascending order. A unique consecutive duplicate MUST be in the SAME column.

I would like to Return across a single row unique consecutive duplicates
(single instance
of a consecutive duplicate value in the SAME column) in ascending order.

Sample Data Layout:

101 102 107 110 145 370 490 501
104 106 107 144 360 430 470 580
125 129 140 150 350 430 460 590
101 102 129 130 149 330 440 578
101 108 120 129 200 280 430 535
100 111 170 175 176 280 420 520
121 189 170 202 229 230 420 521

Expected Results: Unique Duplicate Returned across Single Row
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Thanks,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Sub sam()
Dim s As String
For i = 1 To 8
For j = 2 To 7
v = Cells(j, i).Value
If Cells(j - 1, i).Value = v Then
If s = "" Then
s = v
Else
s = s & "," & v
End If
MsgBox ("column " & i & " " & v)
End If
Next
Next
MsgBox (s)
End Sub

gives result in MSGBOX format. You can modify the code if you need the
results put back into the worksheet.
--
Gary's Student
gsnu200703


"Sam via OfficeKB.com" wrote:

Hi All,

I have a Named Range called "Data" that spans 8 columns and 7 rows. A Numeric
Value will appear only once in a row. Each row's Numeric Values are in
ascending order. A unique consecutive duplicate MUST be in the SAME column.

I would like to Return across a single row unique consecutive duplicates
(single instance
of a consecutive duplicate value in the SAME column) in ascending order.

Sample Data Layout:

101 102 107 110 145 370 490 501
104 106 107 144 360 430 470 580
125 129 140 150 350 430 460 590
101 102 129 130 149 330 440 578
101 108 120 129 200 280 430 535
100 111 170 175 176 280 420 520
121 189 170 202 229 230 420 521

Expected Results: Unique Duplicate Returned across Single Row
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Thanks,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Hi Gary's Student,

Thank you very much for your time and assistance. That's Great! Provides the
desired results.

What do I need to modify to put the results back into the worksheet?

Cheers,
Sam



Gary''s Student wrote:
Sub sam()
Dim s As String
For i = 1 To 8
For j = 2 To 7
v = Cells(j, i).Value
If Cells(j - 1, i).Value = v Then
If s = "" Then
s = v
Else
s = s & "," & v
End If
MsgBox ("column " & i & " " & v)
End If
Next
Next
MsgBox (s)
End Sub

gives result in MSGBOX format. You can modify the code if you need the
results put back into the worksheet.

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Not that this is better than the VBA solution posted.....
but, since you posted in the worksheet functions group,
and I felt like a challenge...
try this:

With your posted list in A1:H7

This formula locates the duplicate items in the grid
A10:
=LARGE(INDEX((FREQUENCY(($A$1:$H$7+COLUMN($A$1:$H$ 7)*1000),($A$1:$H$7+COLUMN($A$1:$H$7)*1000))1)*RO W($A$1:INDEX($A:$A,COUNT($A$1:$H$7)+1)),0),COLUMNS ($A$10:A10))

Copy that formula across to the right, 10 columns or so

This formula translates the location reference to an item in the grid
A11: =INDEX($A$1:$H$7,CEILING(A10/8,1),MOD(A10-1,8)+1)
Copy that formula across to the right also

I hope that helps.
***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi All,

I have a Named Range called "Data" that spans 8 columns and 7 rows. A Numeric
Value will appear only once in a row. Each row's Numeric Values are in
ascending order. A unique consecutive duplicate MUST be in the SAME column.

I would like to Return across a single row unique consecutive duplicates
(single instance
of a consecutive duplicate value in the SAME column) in ascending order.

Sample Data Layout:

101 102 107 110 145 370 490 501
104 106 107 144 360 430 470 580
125 129 140 150 350 430 460 590
101 102 129 130 149 330 440 578
101 108 120 129 200 280 430 535
100 111 170 175 176 280 420 520
121 189 170 202 229 230 420 521

Expected Results: Unique Duplicate Returned across Single Row
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Thanks,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Assuming "data" is in the range A1:H7.

Array entered:

=IF(ISERROR(SMALL(IF($A1:$H6=$A2:$H7,$A1:$H6),COLU MNS($A:A))),"",SMALL(IF($A1:$H6=$A2:$H7,$A1:$H6),C OLUMNS($A:A)))

Copy across until you get blanks.

Biff

"Sam via OfficeKB.com" <u4102@uwe wrote in message
news:6d50abbcea48c@uwe...
Hi All,

I have a Named Range called "Data" that spans 8 columns and 7 rows. A
Numeric
Value will appear only once in a row. Each row's Numeric Values are in
ascending order. A unique consecutive duplicate MUST be in the SAME
column.

I would like to Return across a single row unique consecutive duplicates
(single instance
of a consecutive duplicate value in the SAME column) in ascending order.

Sample Data Layout:

101 102 107 110 145 370 490 501
104 106 107 144 360 430 470 580
125 129 140 150 350 430 460 590
101 102 129 130 149 330 440 578
101 108 120 129 200 280 430 535
100 111 170 175 176 280 420 520
121 189 170 202 229 230 420 521

Expected Results: Unique Duplicate Returned across Single Row
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Thanks,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Well, maybe not!

That formula works ok on the sample you posted but when I put it through the
"wringer" it chokes!

For example: (I guess this is possible?):

100...101
100...101
101...110
101...120
101...107

Biff

"T. Valko" wrote in message
...
Assuming "data" is in the range A1:H7.

Array entered:

=IF(ISERROR(SMALL(IF($A1:$H6=$A2:$H7,$A1:$H6),COLU MNS($A:A))),"",SMALL(IF($A1:$H6=$A2:$H7,$A1:$H6),C OLUMNS($A:A)))

Copy across until you get blanks.

Biff

"Sam via OfficeKB.com" <u4102@uwe wrote in message
news:6d50abbcea48c@uwe...
Hi All,

I have a Named Range called "Data" that spans 8 columns and 7 rows. A
Numeric
Value will appear only once in a row. Each row's Numeric Values are in
ascending order. A unique consecutive duplicate MUST be in the SAME
column.

I would like to Return across a single row unique consecutive duplicates
(single instance
of a consecutive duplicate value in the SAME column) in ascending order.

Sample Data Layout:

101 102 107 110 145 370 490 501
104 106 107 144 360 430 470 580
125 129 140 150 350 430 460 590
101 102 129 130 149 330 440 578
101 108 120 129 200 280 430 535
100 111 170 175 176 280 420 520
121 189 170 202 229 230 420 521

Expected Results: Unique Duplicate Returned across Single Row
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Thanks,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Hi Ron,

Thank you for your time and assistance. Unfortunately, I am not getting the
expected results. I receive #N/A in the cells using this Formula:

=LARGE(INDEX((FREQUENCY(($A$1:$H$7+COLUMN($A$1:$H $7)*1000),($A$1:$H$7+COLUMN($A$1:$H$7)*1000))1)*R OW($A$1:INDEX($A:$A,COUNT($A$1:$H$7)+1)),0),COLUMN S($A$10:A10))

This formula locates the duplicate items in the grid A10:

Copy that formula across to the right, 10 columns or so

My Data starts in Column "D", Row "2". Row "1" has Text Labels. I think I've
made the necessary adjustments for that but I'm still getting #N/A.

Further help appreciated.

Cheers,
Sam

Ron Coderre wrote:
Not that this is better than the VBA solution posted.....
but, since you posted in the worksheet functions group,
and I felt like a challenge...
try this:

With your posted list in A1:H7

This formula locates the duplicate items in the grid
A10:
=LARGE(INDEX((FREQUENCY(($A$1:$H$7+COLUMN($A$1:$H $7)*1000),($A$1:$H$7+COLUMN($A$1:$H$7)*1000))1)*R OW($A$1:INDEX($A:$A,COUNT($A$1:$H$7)+1)),0),COLUMN S($A$10:A10))

Copy that formula across to the right, 10 columns or so

This formula translates the location reference to an item in the grid
A11: =INDEX($A$1:$H$7,CEILING(A10/8,1),MOD(A10-1,8)+1)
Copy that formula across to the right also

I hope that helps.
***********
Regards,
Ron

XL2002, WinXP

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Hi Biff,

Thank you for your time and assistance.


T. Valko wrote:
Well, maybe not!

That formula works ok on the sample you posted but when I put it through the
"wringer" it chokes!

For example: (I guess this is possible?):

Yes, it is possible.

100...101
100...101
101...110
101...120
101...107

Biff

Cheers,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Not sure why you're getting errors.....

I used your posted data table, beginning in A1
Maybe you have column headings in Row_1?

If that's the case and the data range is in A2:H8
then these are the formulas

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H$ 8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*RO W($A$1:INDEX($A:$A,COUNT($A$2:$H$8)+1)),0),COLUMNS ($A$11:A11))

A12: =INDEX($A$2:$H$8,CEILING(A11/8,1),MOD(A11-1,8)+1)

Does that help?
***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi Ron,

Thank you for your time and assistance. Unfortunately, I am not getting the
expected results. I receive #N/A in the cells using this Formula:

=LARGE(INDEX((FREQUENCY(($A$1:$H$7+COLUMN($A$1:$H $7)*1000),($A$1:$H$7+COLUMN($A$1:$H$7)*1000))1)*R OW($A$1:INDEX($A:$A,COUNT($A$1:$H$7)+1)),0),COLUMN S($A$10:A10))

This formula locates the duplicate items in the grid A10:

Copy that formula across to the right, 10 columns or so

My Data starts in Column "D", Row "2". Row "1" has Text Labels. I think I've
made the necessary adjustments for that but I'm still getting #N/A.

Further help appreciated.

Cheers,
Sam

Ron Coderre wrote:
Not that this is better than the VBA solution posted.....
but, since you posted in the worksheet functions group,
and I felt like a challenge...
try this:

With your posted list in A1:H7

This formula locates the duplicate items in the grid
A10:
=LARGE(INDEX((FREQUENCY(($A$1:$H$7+COLUMN($A$1:$H $7)*1000),($A$1:$H$7+COLUMN($A$1:$H$7)*1000))1)*R OW($A$1:INDEX($A:$A,COUNT($A$1:$H$7)+1)),0),COLUMN S($A$10:A10))

Copy that formula across to the right, 10 columns or so

This formula translates the location reference to an item in the grid
A11: =INDEX($A$1:$H$7,CEILING(A10/8,1),MOD(A10-1,8)+1)
Copy that formula across to the right also

I hope that helps.
***********
Regards,
Ron

XL2002, WinXP

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Hi Ron,

Thanks for reply. Even with Sample Data starting in cell A1 and using your
original version of the Formula I still get #N/A?

Ron Coderre wrote:
Not sure why you're getting errors.....

If anything comes to mind would appreciate further help.

I used your posted data table, beginning in A1
Maybe you have column headings in Row_1?

Yes, I just mentioned in my previous post my Data starts in Column "D", Row
"2". Row "1" has Text Labels. I think I've made the necessary adjustments for
that but I'm still getting #N/A.

If that's the case and the data range is in A2:H8
then these are the formulas

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H $8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*R OW($A$1:INDEX($A:$A,COUNT($A$2:$H$8)+1)),0),COLUMN S($A$11:A11))

A12: =INDEX($A$2:$H$8,CEILING(A11/8,1),MOD(A11-1,8)+1)

Does that help?
***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

The issue must be data related. Are there Blanks? Text?
Those are the only exceptions that throw errors in my testing.

With the data in A2:H11, this formula is durable against blanks, but not text

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H$ 8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*RO W($A$1:INDEX($A:$A,ROWS($A$2:$H$8)*COLUMNS($A$2:$H $8)+1)),0),COLUMNS($A$11:A11))

Does that help?
or.... do you see anything else that may be an issue?

***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi Ron,

Thanks for reply. Even with Sample Data starting in cell A1 and using your
original version of the Formula I still get #N/A?

Ron Coderre wrote:
Not sure why you're getting errors.....

If anything comes to mind would appreciate further help.

I used your posted data table, beginning in A1
Maybe you have column headings in Row_1?

Yes, I just mentioned in my previous post my Data starts in Column "D", Row
"2". Row "1" has Text Labels. I think I've made the necessary adjustments for
that but I'm still getting #N/A.

If that's the case and the data range is in A2:H8
then these are the formulas

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H $8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*R OW($A$1:INDEX($A:$A,COUNT($A$2:$H$8)+1)),0),COLUMN S($A$11:A11))

A12: =INDEX($A$2:$H$8,CEILING(A11/8,1),MOD(A11-1,8)+1)

Does that help?
***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Hi Ron,

Thanks very much for reply and further input.

Ron Coderre wrote:
The issue must be data related. Are there Blanks? Text?
No Blanks, text only in Row "1" .

Those are the only exceptions that throw errors in my testing.

With the data in A2:H11, this formula is durable against blanks, but not text

No text, although I did have a problem copying back my original Sample Data
using Data Text to Columns which created the #N/A errors but sorted now.

However, regarding my Sample Data and the Expected Results, I've noticed that
your Formula picks out Numeric Value 102 as a consecutive duplicate value in
the SAME column. It is a duplicate in the same column but NOT a consecutive
duplicate as described in my original post. It should not be included in the
results.

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H $8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*R OW($A$1:INDEX($A:$A,ROWS($A$2:$H$8)*COLUMNS($A$2:$ H$8)+1)),0),COLUMNS($A$11:A11))

Does that help?
or.... do you see anything else that may be an issue?

Numeric Value 102 is NOT a consecutive duplicate as described in my original
post. However, the Formula does incorrectly return Numeric Value 102. I think
this could be part of the problem and solution.

***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

--
Message posted via http://www.officekb.com

Well, the formulas didn't get any prettier, but....I *think* this works.....

A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$8=$A$3:$H$9)*($A $2:$H$8+COLUMN($A$2:$H$8)*1000)),($A$2:$H$8+COLUMN ($A$2:$H$8)*1000))=1)*ROW($A$1:INDEX($A:$A,ROWS($ A$2:$H$8)*COLUMNS($A$2:$H$8)+1)),0),COLUMNS($A$11: A11))

A12: =IF(A11,INDEX($A$2:$H$8,CEILING(A11/8,1),MOD(A11-1,8)+1),"")

Are we done yet?

***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi Ron,

Thanks very much for reply and further input.

Ron Coderre wrote:
The issue must be data related. Are there Blanks? Text?
No Blanks, text only in Row "1" .

Those are the only exceptions that throw errors in my testing.

With the data in A2:H11, this formula is durable against blanks, but not text

No text, although I did have a problem copying back my original Sample Data
using Data Text to Columns which created the #N/A errors but sorted now.

However, regarding my Sample Data and the Expected Results, I've noticed that
your Formula picks out Numeric Value 102 as a consecutive duplicate value in
the SAME column. It is a duplicate in the same column but NOT a consecutive
duplicate as described in my original post. It should not be included in the
results.

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H $8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*R OW($A$1:INDEX($A:$A,ROWS($A$2:$H$8)*COLUMNS($A$2:$ H$8)+1)),0),COLUMNS($A$11:A11))

Does that help?
or.... do you see anything else that may be an issue?

Numeric Value 102 is NOT a consecutive duplicate as described in my original
post. However, the Formula does incorrectly return Numeric Value 102. I think
this could be part of the problem and solution.

***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

--
Message posted via http://www.officekb.com

Done?....not quite. The previous formula calc'd consecutive blanks as dupes.

This fixes that:
A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$7<0)*($A$2:$H$7 =$A$3:$H$8)*($A$2:$H$7+COLUMN($A$2:$H$7)*1000)),($ A$2:$H$7+COLUMN($A$2:$H$7)*1000))=1)*ROW($A$1:IND EX($A:$A,ROWS($A$2:$H$7)*COLUMNS($A$2:$H$7)+1)),0) ,COLUMNS($A$11:A11))

How're we doing?

***********
Regards,
Ron

XL2002, WinXP


"Ron Coderre" wrote:

Well, the formulas didn't get any prettier, but....I *think* this works.....

A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$8=$A$3:$H$9)*($A $2:$H$8+COLUMN($A$2:$H$8)*1000)),($A$2:$H$8+COLUMN ($A$2:$H$8)*1000))=1)*ROW($A$1:INDEX($A:$A,ROWS($ A$2:$H$8)*COLUMNS($A$2:$H$8)+1)),0),COLUMNS($A$11: A11))

A12: =IF(A11,INDEX($A$2:$H$8,CEILING(A11/8,1),MOD(A11-1,8)+1),"")

Are we done yet?

***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi Ron,

Thanks very much for reply and further input.

Ron Coderre wrote:
The issue must be data related. Are there Blanks? Text?
No Blanks, text only in Row "1" .

Those are the only exceptions that throw errors in my testing.

With the data in A2:H11, this formula is durable against blanks, but not text

No text, although I did have a problem copying back my original Sample Data
using Data Text to Columns which created the #N/A errors but sorted now.

However, regarding my Sample Data and the Expected Results, I've noticed that
your Formula picks out Numeric Value 102 as a consecutive duplicate value in
the SAME column. It is a duplicate in the same column but NOT a consecutive
duplicate as described in my original post. It should not be included in the
results.

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H $8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*R OW($A$1:INDEX($A:$A,ROWS($A$2:$H$8)*COLUMNS($A$2:$ H$8)+1)),0),COLUMNS($A$11:A11))

Does that help?
or.... do you see anything else that may be an issue?

Numeric Value 102 is NOT a consecutive duplicate as described in my original
post. However, the Formula does incorrectly return Numeric Value 102. I think
this could be part of the problem and solution.

***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

--
Message posted via http://www.officekb.com

It's picking up the 100 in A7.

Biff

"Ron Coderre" wrote in message
...
Done?....not quite. The previous formula calc'd consecutive blanks as
dupes.

This fixes that:
A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$7<0)*($A$2:$H$7 =$A$3:$H$8)*($A$2:$H$7+COLUMN($A$2:$H$7)*1000)),($ A$2:$H$7+COLUMN($A$2:$H$7)*1000))=1)*ROW($A$1:IND EX($A:$A,ROWS($A$2:$H$7)*COLUMNS($A$2:$H$7)+1)),0) ,COLUMNS($A$11:A11))

How're we doing?

***********
Regards,
Ron

XL2002, WinXP


"Ron Coderre" wrote:

Well, the formulas didn't get any prettier, but....I *think* this
works.....

A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$8=$A$3:$H$9)*($A $2:$H$8+COLUMN($A$2:$H$8)*1000)),($A$2:$H$8+COLUMN ($A$2:$H$8)*1000))=1)*ROW($A$1:INDEX($A:$A,ROWS($ A$2:$H$8)*COLUMNS($A$2:$H$8)+1)),0),COLUMNS($A$11: A11))

A12: =IF(A11,INDEX($A$2:$H$8,CEILING(A11/8,1),MOD(A11-1,8)+1),"")

Are we done yet?

***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi Ron,

Thanks very much for reply and further input.

Ron Coderre wrote:
The issue must be data related. Are there Blanks? Text?
No Blanks, text only in Row "1" .

Those are the only exceptions that throw errors in my testing.

With the data in A2:H11, this formula is durable against blanks, but
not text

No text, although I did have a problem copying back my original Sample
Data
using Data Text to Columns which created the #N/A errors but sorted
now.

However, regarding my Sample Data and the Expected Results, I've
noticed that
your Formula picks out Numeric Value 102 as a consecutive duplicate
value in
the SAME column. It is a duplicate in the same column but NOT a
consecutive
duplicate as described in my original post. It should not be included
in the
results.

A11:
=LARGE(INDEX((FREQUENCY(($A$2:$H$8+COLUMN($A$2:$H $8)*1000),($A$2:$H$8+COLUMN($A$2:$H$8)*1000))1)*R OW($A$1:INDEX($A:$A,ROWS($A$2:$H$8)*COLUMNS($A$2:$ H$8)+1)),0),COLUMNS($A$11:A11))

Does that help?
or.... do you see anything else that may be an issue?

Numeric Value 102 is NOT a consecutive duplicate as described in my
original
post. However, the Formula does incorrectly return Numeric Value 102. I
think
this could be part of the problem and solution.

***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

--
Message posted via http://www.officekb.com

Hi Ron,

Using the Sample Data the results do not tie up with the Expected Results -
Now Numeric Value 100 is being listed.

Expected Results: Unique Consecutive Duplicate Returned across Single Row in
ascending order.
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Ron Coderre wrote:
Done?....not quite. The previous formula calc'd consecutive blanks as dupes.

This fixes that:
A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$7<0)*($A$2:$H$ 7=$A$3:$H$8)*($A$2:$H$7+COLUMN($A$2:$H$7)*1000)),( $A$2:$H$7+COLUMN($A$2:$H$7)*1000))=1)*ROW($A$1:IN DEX($A:$A,ROWS($A$2:$H$7)*COLUMNS($A$2:$H$7)+1)),0 ),COLUMNS($A$11:A11))

How're we doing?

Not quite there, yet.

***********
Regards,
Ron

XL2002, WinXP

Well, the formulas didn't get any prettier, but....I *think* this works.....

Cheers,
Sam

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Hi Biff,

Thanks, bit slow in seeing your post.

Cheers,
Sam

T. Valko wrote:
It's picking up the 100 in A7.

Biff

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OK....Here's the latest in a series of final formulas : \

A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$7<0)*($A$2:$H$7 =$A$3:$H$8)*($A$2:$H$7+COLUMN($A$2:$H$7)*1000)),($ A$2:$H$7+COLUMN($A$2:$H$7)*1000)+($A$2:$H$7<$A$3: $H$8)*9999)=1)*ROW($A$1:INDEX($A:$A,ROWS($A$2:$H$ 7)*COLUMNS($A$2:$H$7)+1)),0),COLUMNS($A$11:A11))

Copied across yields these results:
420 170 280 101 430 107

In ascending order that would be:
101 107 170 280 420 430

Dare I ask?

***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi Ron,

Using the Sample Data the results do not tie up with the Expected Results -
Now Numeric Value 100 is being listed.

Expected Results: Unique Consecutive Duplicate Returned across Single Row in
ascending order.
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Ron Coderre wrote:
Done?....not quite. The previous formula calc'd consecutive blanks as dupes.

This fixes that:
A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$7<0)*($A$2:$H$ 7=$A$3:$H$8)*($A$2:$H$7+COLUMN($A$2:$H$7)*1000)),( $A$2:$H$7+COLUMN($A$2:$H$7)*1000))=1)*ROW($A$1:IN DEX($A:$A,ROWS($A$2:$H$7)*COLUMNS($A$2:$H$7)+1)),0 ),COLUMNS($A$11:A11))

How're we doing?

Not quite there, yet.

***********
Regards,
Ron

XL2002, WinXP

Well, the formulas didn't get any prettier, but....I *think* this works.....

Cheers,
Sam

--
Message posted via OfficeKB.com
http://www.officekb.com/Uwe/Forums.a...tions/200702/1

Hi Ron,

Thank you very much for persevering.

Ron Coderre wrote:
OK....Here's the latest in a series of final formulas : \

A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$7<0)*($A$2:$H$ 7=$A$3:$H$8)*($A$2:$H$7+COLUMN($A$2:$H$7)*1000)),( $A$2:$H$7+COLUMN($A$2:$H$7)*1000)+($A$2:$H$7<$A$3 :$H$8)*9999)=1)*ROW($A$1:INDEX($A:$A,ROWS($A$2:$H $7)*COLUMNS($A$2:$H$7)+1)),0),COLUMNS($A$11:A11))

Copied across yields these results:
420 170 280 101 430 107

This is Great!

In ascending order that would be:
101 107 170 280 420 430

Yes,

Dare I ask?

Can the Formula actually list them in ascending order?

***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

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Message posted via http://www.officekb.com

Assuming that A2:H8 contains the data, try the following...

J2:

=SUM(IF(FREQUENCY(IF(A2:H7=A3:H8,A2:H7),IF(A2:H7=A 3:H8,A2:H7)),1))

....confirmed with CONTROL+SHIFT+ENTER

K2:

leave empty

L2, copied across:

=IF(COLUMNS($L$2:L2)<=$J$2,MIN(IF(ISNA(MATCH($A$2: $H$7,$K$2:K2,0)),IF($A$
2:$H$7=$A$3:$H$8,$A$2:$H$7))),"")

....confirmed with CONTROL+SHIFT+ENTER

Hope this helps!

In article <6d50abbcea48c@uwe, "Sam via OfficeKB.com" <u4102@uwe
wrote:

Hi All,

I have a Named Range called "Data" that spans 8 columns and 7 rows. A Numeric
Value will appear only once in a row. Each row's Numeric Values are in
ascending order. A unique consecutive duplicate MUST be in the SAME column.

I would like to Return across a single row unique consecutive duplicates
(single instance
of a consecutive duplicate value in the SAME column) in ascending order.

Sample Data Layout:

101 102 107 110 145 370 490 501
104 106 107 144 360 430 470 580
125 129 140 150 350 430 460 590
101 102 129 130 149 330 440 578
101 108 120 129 200 280 430 535
100 111 170 175 176 280 420 520
121 189 170 202 229 230 420 521

Expected Results: Unique Duplicate Returned across Single Row
101 107 170 280 420 430

Column 1 = 101
Column 3 = 107, 170
Column 6 = 280, 430
Column 7 = 420


Thanks,
Sam

Definitely look at Domenic's response.
Where I misinterpreted your criteria, built a formula that listed
same-column multiples, then fiddled with it to list consecutive dupes....he
paid attention. His formula does what you want without all the side trips.

***********
Regards,
Ron

XL2002, WinXP


"Sam via OfficeKB.com" wrote:

Hi Ron,

Thank you very much for persevering.

Ron Coderre wrote:
OK....Here's the latest in a series of final formulas : \

A11:
=LARGE(INDEX((FREQUENCY((($A$2:$H$7<0)*($A$2:$H$ 7=$A$3:$H$8)*($A$2:$H$7+COLUMN($A$2:$H$7)*1000)),( $A$2:$H$7+COLUMN($A$2:$H$7)*1000)+($A$2:$H$7<$A$3 :$H$8)*9999)=1)*ROW($A$1:INDEX($A:$A,ROWS($A$2:$H $7)*COLUMNS($A$2:$H$7)+1)),0),COLUMNS($A$11:A11))

Copied across yields these results:
420 170 280 101 430 107

This is Great!

In ascending order that would be:
101 107 170 280 420 430

Yes,

Dare I ask?

Can the Formula actually list them in ascending order?

***********
Regards,
Ron

XL2002, WinXP

Cheers,
Sam

--
Message posted via http://www.officekb.com

Hi Ron,

Thank you for all your help.

Will see Domenic's post, thanks.

Cheers,
Sam

Ron Coderre wrote:
Definitely look at Domenic's response.
Where I misinterpreted your criteria, built a formula that listed
same-column multiples, then fiddled with it to list consecutive dupes....he
paid attention. His formula does what you want without all the side trips.

***********
Regards,
Ron

XL2002, WinXP

--
Message posted via http://www.officekb.com

Hi Domenic,

Thank you very much for your assistance. Does the job great. Brilliant!

Cheers,
Sam

Domenic wrote:
Assuming that A2:H8 contains the data, try the following...

J2:

=SUM(IF(FREQUENCY(IF(A2:H7=A3:H8,A2:H7),IF(A2:H7= A3:H8,A2:H7)),1))

...confirmed with CONTROL+SHIFT+ENTER

K2:

leave empty

L2, copied across:

=IF(COLUMNS($L$2:L2)<=$J$2,MIN(IF(ISNA(MATCH($A$2 :$H$7,$K$2:K2,0)),IF($A$
2:$H$7=$A$3:$H$8,$A$2:$H$7))),"")

...confirmed with CONTROL+SHIFT+ENTER

Hope this helps!

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