On Oct 21, 2009, at 3:16 AM, satorisanitarium wrote:
I just started learning haskell and I just came to my first wtf
moment.
I'm trying to do something like this:
calling:
foo 3 [1,2,3,2,4,1,6,3,6,2,3,5,2,5,2,1,6,4]
returns:
[[1,2,3],[2,4,1,6,3],[2,3]]
but i have no idea how to put something into a sublist.
From the way you talk about the problem, I suspect that you are
trying to understand in a rather complicated way.
Here's how I approach your example:
foo item list =
if there is an occurrence of item in list then
a new list whose first element is everything up to and
including
that occurrence
and whose remaining elements are found by applying foo
item to
everything after that occurrence
else
an empty list
We can do this in just a couple of lines using List.elemIndex and
splitAt,
but let's do it in an elementary way. We'll write a helper function
that
either returns (before-and-including-item, after-item) or nothing.
find_and_split :: Eq t => t -> [t] -> Maybe ([t], [t])
find_and_split item list = loop list []
where loop [] _ = Nothing
loop (x:xs) before | x == item = Just (reverse
(x:before), xs)
loop (x:xs) before = loop xs (x:before)
foo :: Eq t => t -> [t] -> [[t]]
foo item list =
case find_and_split item list of
Nothing -> []
Just (before, after) -> before : foo item after
The answer I get is
*Main> foo 3 [1,2,3,2,4,1,6,3,6,2,3,5,2,5,2,1,6,4]
[[1,2,3],[2,4,1,6,3],[6,2,3]]
^
which differs from yours at the marked place, but which I think is
right.
List.elemIndex tells you where an element occurs.
Just 0 is the first position.
splitAt splits a list into two pieces, the size argument tells you
the length of the first piece. So
foo item list =
case List.elemIndex item list of
Nothing -> []
Just pos -> before : foo item after
where (before, after) = splitAt (pos+1) list
This gives the same answer as the previous version.
In neither case do we do ANYTHING to "put" items into a "sublist",
we just construct a list whose elements happen to be lists. It's
no different from calculating a list of numbers.
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