文字列を先頭から見て同じところまで除去

Haskell Python

ふむ。Haskell だとどうなるか。
素直に書くとこうかな。

sameAll :: Eq a => [a] -> Bool
sameAll (x:xs) = all (x==) xs
sameAll []     = True

dropCommonPrefix :: Eq a => [[a]] -> [[a]]
dropCommonPrefix strings | any null strings           = strings
                         | sameAll (map head strings) = dropCommonPrefix (map tail strings)
                         | otherwise                  = strings

こゆのもありか。面倒なので、longestCommonPrefix はソートして先頭と末尾だけ比較。(なので、対象が Ord クラスに限られる)

longestCommonPrefix :: Ord a => [[a]] -> [a]
longestCommonPrefix lst = map fst $ takeWhile eq $ zip (minimum lst) (maximum lst)
  where
    eq (x, y) = x == y

dropCommonPrefix :: Ord a => [[a]] -> [[a]]
dropCommonPrefix lst = map (drop longestCommonPrefixLength) lst
  where
    longestCommonPrefixLength = length $ longestCommonPrefix lst

Python だとこんな感じか。

longestCommonPrefix :: Ord a => [[a]] -> [a]
longestCommonPrefix lst = map fst $ takeWhile eq $ zip (minimum lst) (maximum lst)
  where
    eq (x, y) = x == y

dropCommonPrefix :: Ord a => [[a]] -> [[a]]
dropCommonPrefix lst = map (drop longestCommonPrefixLength) lst
  where
    longestCommonPrefixLength = length $ longestCommonPrefix lst

def dropCommonPrefix(*strings):
    def commonPrefixLength(xs, ys):
        maxLen = min(len(xs), len(ys))
        for i in xrange(maxLen):
            if xs[i] != ys[i]: return i
        return maxLen

    cpl = commonPrefixLength(min(strings), max(strings))

    return tuple(s[cpl:] for s in strings)