[Haskell-cafe] Separate a string into a list of strings
clifford.beshers at linspire.com
Tue Jun 13 02:36:30 EDT 2006
Donn Cave wrote:
> Quoth Clifford Beshers <clifford.beshers at linspire.com>:
> | Well, I couldn't resist the puzzle. Here are solutions using foldr and
> | unfoldr. Don't know if they are cunning or not, but they were kind of fun.
> | splitByElem1 e xs =
> | foldr f [] xs
> | where f a b = if a == e then  : b else (a : head b) : (tail b)
> This does the right thing with trailing separators, which is not to be
> taken for granted among Haskell split implementations. The splits I have
> been seeing in this thread are conservative, so if the separator is ':',
> then "::a" splits to ["", "", "a"]. Frequently however the implementation
> fails to deal with the trailing separator, so "a:" is ["a"], where it
> should be ["a", ""]. It's not something you run into right away.
> In a liberal split, "a " should indeed be ["a"], but that's a different
> matter. Neither of the two I've looked at seems to be shooting for a
> liberal "words" white space split.
Good point. My solutions are inconsistent on white space, which I don't
Your criterion eliminates this solution as well:
filter (/= ",") $ groupBy (\x y -> x /= ',' && y /= ',') "Haskell,
Haskell, and Haskell,"
["Haskell"," Haskell"," and Haskell"]
More information about the Haskell-Cafe