Extracting data in groups of 3's and copying to another worksheet

[Tony Morgan]

Hi All,

New to this forum and Excel/VBA and could really do with some help!

I have the a table set up like the following:

Excel Forum Pic.png

The problem I have is that I need a Macro to extract data in sets of 3's and only if they have the same
concatenated figure i.e. in column E.

For the example attached then I would want Assignment 4,5,6 to be taken out (7,8 left as although these are the same concatenated number there are only two of them) then I'd want 9,10,11 and 12,13,14 (15,16 would be ignored etc).

I'm guessing I might need a loop and an array but not a clue how to go about this?

I have used the following code previous to extract and copy some data but it is limited by a user having to input a variable (not dynamic) and I get
get the list to look at blocks of 3.

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[mrice]

This will delete rows until there are multiples of 3 which you can use on a copy of the data

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[Tony Morgan]

Hi Martin,

Thanks Brilliant!! Thanks for that. Much Appreciated.

Just 2 quick things...

Is there any way to copy keep the deleted rows? I could do with retaining these.

Once I have removed the duplicate rows is there a way to give a each set of 3's a new assignment number common to both?

Thanks,

Superb help

[mrice]

This mod will copy the deleted rows to the second sheet.

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Sorry but didn't understand the assignment number one.

[Cutter]

@ Tony Morgan

Welcome to the forum.

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Thanks.

[Tony Morgan]

Sorry Martin,

My apologies, still getting used to describing some of this stuff - lol

I'll have another go....

Once I have removed the duplicate rows and I'm left with data in blocks of threes, if there a way I can give each block of three
a unique id? i.e. I could rename assignments 4,5,6, to 1 and then the next block in the sequence 9,10,11 to 2 etc

Lastly, is it possible to reverse the way duplicate rows are removed i.e. the duplicates are left on the original page and the non-duplicates (blocks of 3) are removed to another page instead?

Thanks again for all your help!!