Need help for formula to identify the duplicates

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Dear All,

I have entries in excel, all numbers are same but entries in plus and mines.

correct figureĀ  in plus (for reference sheet attached)

please suggest me any formula for identifying the correct entry out of them.

Regards

Amit Varshney

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I have a similar or more complex problem that I need a solution for.

I have data derived from an ACD report, I'm trying to determine how many repeat calls agent x has received within 24 hours, is there any formula I can use to derive the number based on total calls p/agent and repeat instances?
Sam Lee (rep: 2) Jul 12, '19 at 2:30 am
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=MAX(A2:A12)
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Hahaha - I voted this answer up not because I think it solves the OP's problem but because it answers the OP's question so perfectly. I think the OP's real life range should contain other numbers as well, which may have to be disregarded, or the range is difficult to define. Neither of these contingencies is mentioned in the question. Therefore k1 correctly ignores them.
Variatus (rep: 3543) Jun 27, '19 at 9:00 pm
Thank you Variatus. I agree the solution was a little flippant and having exercised my physic powers also deduced there may have been a range of values involved. However being more a macro man rather than function with only a small fraction of you knowledge that was about all the time I could spend on it. K1
k1w1sm (rep: 197) Jun 27, '19 at 10:42 pm
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The correct amount is

=ABS(A2)

When copied down, this will write the correct amount next to each amount in column A. Please read my comment on @k1w1sm's answer above.

A list of distinct amounts, where each amount appears only once, could be created using filters or cell formatting or even by tweaking the formula. One would need to kow the requirement and, most probably, what's in the other columns of the sheet.

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WOW. From the way it's structured I'm amazed anyone could even understand the question. My hat's off to you guys.

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I would post the cry-laughing emoji here if I could for this comment! haha
don (rep: 1745) Jun 30, '19 at 11:33 pm
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