On Feb 11, 2014, at 9:08 AM, Frerich Raabe
After grouping the given list, so that you have
[[1,1,1,1],[2,2,2,2],[3,3],[2,2,2],[5],[6],[2,2],[7]]
Sort the list by comparing the first element of each list
ghci> sortBy (comparing head) [[1,1,1,1],[2,2,2,2],[3,3],[2,2,2],[5],[6],[2,2],[7]] [[1,1,1,1],[2,2,2,2],[2,2,2],[2,2],[3,3],[5],[6],[7]]
Then, group that again such that lists with equal elements get put into one list:
ghci> groupBy ((==) `on` head) [[1,1,1,1],[2,2,2,2],[2,2,2],[2,2],[3,3],[5],[6],[7]] [[[1,1,1,1]],[[2,2,2,2],[2,2,2],[2,2]],[[3,3]],[[5]],[[6]],[[7]]]
Finally, select the "maximum" of each inner list by comparing the length of the sub-sub-lists:
ghci> map (maximumBy (comparing length)) [[[1,1,1,1]],[[2,2,2,2],[2,2,2],[2,2]],[[3,3]],[[5]],[[6]],[[7]]]
Thank you for the nicely detailed explanation. In my frustration, I was fearing I was going to have to revert to a more imperative approach to solving this, and I was hoping someone would demonstrate a more functional solution. In the various iterations that I unsuccessfully tried, I think the major idea I was lacking was your second step, the groupBy ((==) `on` head). That’s really the part I just wasn’t able to come up with on my own. Again, thanks so much for the help. Best, James