Tile tool to split 10 rows into two equal groups based on numeric 'Rating' value.

Question

I have a list of 10 players

I tried using the Tile tool with equal sum method to split the rows into 2 groups (as close to even as possible)

This has not worked as expected. The tile tool has just categorised the rows in the order they appear, with tile 1 have a much higher rating.

I may be using the Tile application incorrectly, or my understanding of how it works is incorrect. Alternatively can I achieve 2 even groups with an alternative tool? Perhaps by listing out all possible team combinations and summarizing to get average rating across the teams.

Present_guy · Answer

So..... I conclude that the tile tool will not do an equal sum grouping and eagerly await the new tool that will be added to Alteryx that will actually do this. :(

DataNath · Answer

@GarethLiggett I've just quickly put together an iterative macro that splits the data in 2, sums both teams and then creates a % difference of this split. Upon outputting from the macros, the iterations are then sorted by % diff and the top one taken. At the moment, I've just set a hard limit of 50 iterations (I've ran it a few times and usually the best result in this case - 0.43% - is found within ~5 iterations anyway, though there were a few runs that went into the 20s etc). However, you could set a threshold for the % diff as a point for the macro to end if there was a result that you'd be happy with without finding the optimal. Hope this helps and please let me know if you have any questions:

(The error is just the macro hitting the iteration limit I set - don't worry!)

Equal sum finder.yxzp

DenisZ · Answer

I am not so sure how simple it is using Alteryx. You could try to match all possible team combinations, but this would be a 10 factorial number of combinations. However, if you have experience with Python, then that tool can be leveraged as there appears to be multiple codes online for that.

Some examples:

python - Splitting list into 2 parts, as equal to sum as possible - Stack Overflow