How can I group this list by second element of tuples:
[(3,2),(17,2),(50,3),(64,3)]
to get something like:
[[(3,2),(17,2)],[(50,3),(64,3)]]
I'm actually a newcomer to Haskell...and seems to be falling in love with it. Hope you would help me find an efficient way.
It sounds like you've already identified that you want Data.List.groupBy. The type of this function is
groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
So it takes a binary predicate, i.e. an equivalence relation determining how to group elements. You want to group elements by equality on the second term of a pair, so you want
groupBy (\x y -> snd x == snd y) myList
Where snd is a built-in function that gets the second element of a pair.
Incidentally, this pattern of "apply a function to two arguments and then apply a binary function to the results" is very common, especially when calling Data.List functions, so Data.Function provides on.
on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
Weird signature, but the use case is just what we want.
((+) `on` f) x y = f x + f y
So your desired groupBy can be written as
groupBy ((==) `on` snd)
Note that groupBy only finds consecutive equal elements. You didn't indicate whether you wanted consecutive equal elements or all equal elements, but if you want the latter, then I don't believe Haskell base provides that function, though you could certainly write it recursively yourself.