Hello All, The standard function groupBy of List.hs doesn't work as I expect in this case: groupBy (\x y -> (last x) == (last y)) ["abc", "bd","cac"] results in: [["abc"],["bd"],["cac"]] where I want: [["abc","cac"], ["bd"]] Am I doing something wrong, or is this a bug? The function is defined (in List.hs in the hugs Data library) as : -- | The 'groupBy' function is the non-overloaded version of 'group'. groupBy :: (a -> a -> Bool) -> [a] -> [[a]] groupBy _ [] = [] groupBy eq (x:xs) = (x:ys) : groupBy eq zs where (ys,zs) = span (eq x) xs Thanks, Hans van Thiel
On Sat, Apr 28, 2007 at 03:49:44PM +0200, Hans van Thiel wrote:
Hello All,
The standard function groupBy of List.hs doesn't work as I expect in this case:
groupBy (\x y -> (last x) == (last y)) ["abc", "bd","cac"]
results in:
[["abc"],["bd"],["cac"]]
where I want:
[["abc","cac"], ["bd"]]
Am I doing something wrong, or is this a bug? The function is defined (in List.hs in the hugs Data library) as :
-- | The 'groupBy' function is the non-overloaded version of 'group'. groupBy :: (a -> a -> Bool) -> [a] -> [[a]] groupBy _ [] = [] groupBy eq (x:xs) = (x:ys) : groupBy eq zs where (ys,zs) = span (eq x) xs
You are doing something wrong. groupBy is specified to never reorder elements. You probably want to use sortBy first. Stefan
Hi Hans,
The standard function groupBy of List.hs doesn't work as I expect in this case:
groupBy (\x y -> (last x) == (last y)) ["abc", "bd","cac"]
You are doing something wrong. groupBy is specified to never reorder elements. You probably want to use sortBy first.
I have a defined groupSetBy which does groupBy working as though the list was in fact a set (which is often what it really is). Its a common thing to want - perhaps it should be a library function. Thanks Neil
On Sat, 2007-04-28 at 15:08 +0100, Neil Mitchell wrote:
Hi Hans,
The standard function groupBy of List.hs doesn't work as I expect in this case:
groupBy (\x y -> (last x) == (last y)) ["abc", "bd","cac"]
You are doing something wrong. groupBy is specified to never reorder elements. You probably want to use sortBy first.
I have a defined groupSetBy which does groupBy working as though the list was in fact a set (which is often what it really is). Its a common thing to want - perhaps it should be a library function. And thanks again. I'll roll my own then..
Hans
Thanks
Neil
On Sat, 2007-04-28 at 06:45 -0700, Stefan O'Rear wrote:
On Sat, Apr 28, 2007 at 03:49:44PM +0200, Hans van Thiel wrote:
Hello All,
The standard function groupBy of List.hs doesn't work as I expect in this case:
groupBy (\x y -> (last x) == (last y)) ["abc", "bd","cac"]
results in:
[["abc"],["bd"],["cac"]]
where I want:
[["abc","cac"], ["bd"]]
Am I doing something wrong, or is this a bug? The function is defined (in List.hs in the hugs Data library) as :
-- | The 'groupBy' function is the non-overloaded version of 'group'. groupBy :: (a -> a -> Bool) -> [a] -> [[a]] groupBy _ [] = [] groupBy eq (x:xs) = (x:ys) : groupBy eq zs where (ys,zs) = span (eq x) xs
You are doing something wrong. groupBy is specified to never reorder elements. You probably want to use sortBy first.
Stefan Ah, many thanks...
Hans
On Sat, 28 Apr 2007, Hans van Thiel wrote:
Hello All,
The standard function groupBy of List.hs doesn't work as I expect in this case:
groupBy (\x y -> (last x) == (last y)) ["abc", "bd","cac"]
results in:
[["abc"],["bd"],["cac"]]
where I want:
[["abc","cac"], ["bd"]]
I think you must roll your own one. How about repeated 'partition' to extract all elements that have the same trailing character like the head word of the list? I have done this in a more complex setting and called it 'slice': http://darcs.haskell.org/haskore/src/Haskore/Basic/TimeOrderedList.lhs
slice :: (Eq a, Num time) => (body -> a) -> T time body -> [(a, T time body)] slice f perf = let splitByHeadKey pf = fmap (\ev -> let i = f (eventBody ev) (pf0, pf1) = partition ((i==) . f) 0 0 pf in ((i,pf0), pf1)) (listToMaybe pf) in List.unfoldr splitByHeadKey perf
Thanks for the help! Strangely, I just now received your messages from April 28, hence the late reply... Hans van Thiel On Sat, 2007-04-28 at 18:09 +0200, Henning Thielemann wrote:
On Sat, 28 Apr 2007, Hans van Thiel wrote:
Hello All,
The standard function groupBy of List.hs doesn't work as I expect in this case:
groupBy (\x y -> (last x) == (last y)) ["abc", "bd","cac"]
results in:
[["abc"],["bd"],["cac"]]
where I want:
[["abc","cac"], ["bd"]]
I think you must roll your own one. How about repeated 'partition' to extract all elements that have the same trailing character like the head word of the list?
I have done this in a more complex setting and called it 'slice': http://darcs.haskell.org/haskore/src/Haskore/Basic/TimeOrderedList.lhs
slice :: (Eq a, Num time) => (body -> a) -> T time body -> [(a, T time body)] slice f perf = let splitByHeadKey pf = fmap (\ev -> let i = f (eventBody ev) (pf0, pf1) = partition ((i==) . f) 0 0 pf in ((i,pf0), pf1)) (listToMaybe pf) in List.unfoldr splitByHeadKey perf
participants (4)
-
Hans van Thiel -
Henning Thielemann -
Neil Mitchell -
Stefan O'Rear