Donn Cave wrote:
Quoth Clifford Beshers <[EMAIL PROTECTED]>:
| 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
like.
Your criterion eliminates this solution as well:
filter (/= ",") $ groupBy (\x y -> x /= ',' && y /= ',') "Haskell,
Haskell, and Haskell,"
["Haskell"," Haskell"," and Haskell"]
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