Establish identical Values within an index range

Ian Bell
Ian Bell used Ask the Experts™
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Hi,
I would like a formula to establish identical values within an index range.
Please see attached
Many thanks
Ian
identical-values-within-range.xlsx
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Subodh Tiwari (Neeraj)Excel & VBA Expert
Most Valuable Expert 2018
Awarded 2015
Commented:
Please try this...

In C2
=IF(COUNTIF($A$2:$A$15,A2)=COUNTIFS($A$2:$A$15,A2,$B$2:$B$15,B2),0,1)

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and copy it down.
Another formula

=--(SUMPRODUCT(--(A2:A15=A2))<>SUMPRODUCT((A2:A15=A2)*(B2:B15=B2)))
Subodh Tiwari (Neeraj)Excel & VBA Expert
Most Valuable Expert 2018
Awarded 2015

Commented:
To counter the Saqib's formula, here is the more elegant formula... Lol
=--(COUNTIF($A$2:$A$15,A2)<>COUNTIFS($A$2:$A$15,A2,$B$2:$B$15,B2))

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and copy it down. :)
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Ian Bellretired

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Commented:
They all take a long time to calculate
What is the maximum number of entries in any one index set?
Ian Bellretired

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Commented:
40
Subodh Tiwari (Neeraj)Excel & VBA Expert
Most Valuable Expert 2018
Awarded 2015

Commented:
And how many rows of data in there?
After copying the formula down the rows, you can copy the formula range and paste it back as values so that the formulas don't recalculate every time.
Or you can use VBA to place the formula.
In the attached, place the data from your original file in the column A and B starting from Row2 and click the button called "Place Formula" and see if it works faster than manually placing the formula.
identical-values-within-range.xlsm
Try this formula

=--(COUNTIF(A2:A51,A2)<>COUNTIFS(A2:A51,A2,B2:B51,B2))
Subodh Tiwari (Neeraj)Excel & VBA Expert
Most Valuable Expert 2018
Awarded 2015

Commented:
@Saqib
The formula you posted is the same I posted here. What I am missing here?
Subodh, I have removed the absolute marks $

But that formula would fail in some scenarios. So a new formula.

=--(COUNTIF(OFFSET(A2,MAX(-50,-ROW()+2),0,100),A2)<>COUNTIFS(OFFSET(A2,MAX(-50,-ROW()+2),0,100),A2,OFFSET(B2,MAX(-50,-ROW()+2),0,100),B2))
Ian Bellretired

Author

Commented:
Here is a formula using the indirect condition.
In fact the rows and columns are correct so a matter of inserting your new formula.
DC and DD are the F & G columns I referred to in my example.
JR being the Index column

=IF(JR14="","",IF(JR14=JR13,CS13,SUM(INDEX((INDIRECT("$JR"&DC14&":$JR"&DD14)=JR14)*(INDIRECT("$A"&DC14&":$A"&DD14)),0))))
I think this would be even more quicker

=IF(A2=A1,G1,--(COUNTIF(A2:A51,A2)<>COUNTIFS(A2:A51,A2,B2:B51,B2)))
Rob HensonFinance Analyst

Commented:
Applying a bit of logic, if all values in an index range are the same they have to be the same as the Average for the index range.

Therefore, you can use:
=IF(B2=SUMIFS(B:B,A:A,A2)/COUNTIFS(A:A,A2),0,1)

For later versions of Excel:
=IF(B2=AVERAGEIFS(B:B,A:A,A2),0,1)
Rob HensonFinance Analyst
Commented:
However, if one of the values in the range is equal to the average then it will result in 0 on that line. So to allow for that:

=IF(COUNTIFS(A$1:A1,A2,C$1:C1,1),1,IF(B2=SUMIFS(B:B,A:A,A2)/COUNTIFS(A:A,A2),0,1))

One scenario where it gives the wrong result is if the first value for the Index range is equal to the average. To overcome that, I would suggest sorting the values in each range in ascending order. On the same logic, if the values are not equal to the average there has to be some that are less than the average and some that are higher. Sorting in ascending order will ensure that the first value is less than the average; if the first/lowest value is equal to the average then all of them will be.
Ian Bellretired

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Commented:
Thank you guys. I'd be lost without you :)
Ian Bellretired

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Commented:
Hi guys,
I was trying out Rob's simple solution and worked well and calculated in less than a couple of minutes, which is fast
for this 160,000 row spreadsheet

I will work with this one for now but will also run the other formulas and will claim them all as a solution if that is permissible.
Thank You Rob, Saqib and Neeraj
Ian
Rob HensonFinance Analyst

Commented:
Thanks Ian, glad it worked well.
Are you sure that is is Rob's solution which is fastest?

Besides, you accepted mine.
Ian Bellretired

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Commented:
There's not a lot of difference in speed between some of the formulas.
That can't be right. I repeated the given data for more than 160,000 rows and the formula gave the results in only a second or two.
Ian Bellretired

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Commented:
Which one of your formulas did you use ?
The accepted one
Ian Bellretired

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Commented:
I will have to check this later as I need to go out.
Ian Bellretired

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Commented:
Hi Saqib,
Seems I selected wrong solution.
I need the formula to return either 1 or 0
I am leaving for home. I am going to send a file showing comparisons. You can then see the difference.
Ian Bellretired

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Commented:
Thanks Saqib very kind of you..
Ian Bellretired

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Commented:
Saqib,
I have found an alternative method to arrive at the conclusion without the use of
this formula. I'm understanding more and more about this monster of a spreadsheet
Apologies for inconvenience.
Ian
Here is the file. If you see the formulas in the first few columns they take ages to compute. But if you look at the formulas in the last two columns then you would see that the calculations are very much instantaneous.
identical-values-within-range.xlsx
Ian Bellretired

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Commented:
okay thanks Saqib I'll check it out.
Ian Bellretired

Author

Commented:
I finally got onto your formula having been away from here yesterday.
I tried your end col formula and it certainly was fast as only took a few
seconds to calculate. I'm off for a long walk and when I get back will
check out your other formula.
Thanks again Saqib your effort was certainly worthwhile from my point of view
Ian
Ian Bellretired

Author

Commented:
Just back from my usual 2 hour Friday walk with my neighbour and tested your formula in G column.
Both work perfectly and came up with identical results and both calculated at a fast speed of a few seconds for 170k rows.
Thanks a million Saqib.
Ian
Ian Bellretired

Author

Commented:
A big thank you to all you guys who submitted solutions. They worked well, some calculated
quicker than others.

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