Pointless puzzle:
A. sum &&& genericLength >>> uncurry (/)
Q. ?
HT @dibblego
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Replying to @missingfaktor
@missingfaktor@steshaw with a single list traversal.1 reply 0 retweets 0 likes -
Replying to @TheColonial
@TheColonial@steshaw I think this will traverse the list twice.2 replies 0 retweets 0 likes -
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Replying to @TheColonial
@TheColonial@steshaw Well, that will be very hard to do in Haskell in a point-free way. :)2 replies 0 retweets 0 likes -
Replying to @missingfaktor
@missingfaktor@TheColonial@steshaw Its easy to do point free with one traversal.1 reply 0 retweets 0 likes -
Replying to @dibblego
@dibblego@TheColonial@steshaw I am finding it hard to remove points from `(\(sum, length) cur -> (sum + cur, length + 1))`.2 replies 0 retweets 0 likes -
Replying to @missingfaktor1 reply 0 retweets 0 likes
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Replying to @dibblego
@dibblego@missingfaktor@TheColonial can do sum and length with foldM but not together/composed/parallel. Hint?1 reply 0 retweets 0 likes
Replying to @steshaw
@steshaw @missingfaktor @TheColonial stateSucc f = state (f &&& succ)
6:11 PM - 5 Aug 2013
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