Question

我有一个sideH函数，冒着Prelude.head []的风险。因此，我使用Maybe编写它，以避免这种情况：

sideH :: Residue -> Maybe (Atom, Atom)
sideH res
    -- Make sure the elements exist
    | nits /= [] && cars /= [] && oxys /= [] = Just (newH1, newH2)
    | otherwise = Nothing where
    ...

以上工作完全符合预期，没有错误。现在，在调用sideH（不是do构造）的函数中，我必须处理sideH返回Nothing的情况：

callerFunc :: [Residue] -> Aromatic -> [(Double, Double)]
callerFunc [] _ = []
callerFunc (r:rs) aro
    -- Evaluate only if there is something to evaluate
    | newHs /= Nothing = (newH1Pos, newH2Pos)
    | otherwise = callerFunc rs aro where
    newHs = sideH r
    newH1Pos = atomPos $ fst $ fromJust newHs
    newH2Pos = atomPos $ snd $ fromJust newHs

如果我在newH1Pos时尝试评估newH2Pos或newH = Nothing，则会失败，因为fromJust Nothing是错误的。但是，我希望这永远不会发生。我希望callerFunc评估newHs，Just something或Nothing。如果是Nothing，则callerFunc将进入下一步，而不会评估newH1Pos或newH2Pos。情况似乎并非如此。我收到了*** Exception: Maybe.fromJust: Nothing错误，我希望newHs能够返回Nothing。

我被要求提供更多代码。我试图提出一个重现错误的最小情况，但同时，这是完整有问题的callerFunc代码。

-- Given a list of residues and an aromatic, find instances where there
--  is a Hydrogen bond between the aromatic and the Hydrogens on Gln or Asn
callerFunc :: [Residue] -> Aromatic -> [(Double, Double)]
callerFunc [] _ = []
callerFunc (r:rs) aro
    -- GLN or ASN case
    | fst delR <= 7.0 && (resName r == gln || resName r == asn) &&
        newHs /= Nothing && snd delR <= 6.0 = 
        [(snd delR, fst delR)] ++ hBondSFinder rs aro
    | otherwise = hBondSFinder rs aro where
    -- Sidechain identifying strings
    gln = B.pack [71, 76, 78]
    asn = B.pack [65, 83, 78]
    -- Get the location of the Hydrogens on the residue's sidechain
    newHs = sideH r
    newH1Pos = atomPos $ fst $ fromJust newHs
    newH2Pos = atomPos $ snd $ fromJust newHs
    -- Get the location of the Nitrogen on the mainchain of the Residue
    ats = resAtoms r
    backboneNPos = atomPos $ head $ getAtomName ats "N"
    hNVect1 = Line2P {lp1 = newH1Pos, lp2 = backboneNPos}
    hNVect2 = Line2P {lp1 = newH2Pos, lp2 = backboneNPos}
    interPoint1 = linePlaneInter (aroPlane aro) hNVect1
    interPoint2 = linePlaneInter (aroPlane aro) hNVect2
    delR = minimum [(interPoint1 `dist` newH1Pos, delr1), 
        (interPoint2 `dist` newH2Pos, delr2)]
    delr1 = interPoint1 `dist` (aroCenter aro)
    delr2 = interPoint2 `dist` (aroCenter aro)

我知道这是一个痛苦的代码转储。我试图减少它。

Answer 1

这个问题的答案（在评论中提到）不符合评论：“我不确定如何使用模式匹配来删除这些if语句。”。

就像这样，例如，虽然仍有一些代码气味可能会通过一些额外的重构得到改善：

sideH :: Residue -> Maybe (Atom, Atom)
sideH res = case (nits, cars, oxys) of
    (_:_, _:_, _:_) -> Just (newH1, newH2)
    _ -> Nothing
    where
    ...

如果你有灵活的道德，你可以尝试这样的事情：

sideH :: Residue -> Maybe (Atom, Atom)
sideH res = do
    _:_ <- return nits
    _:_ <- return cars
    _:_ <- return oxys
    return (newH1, newH2)
    where
    ...

同样，如果有更多的上下文和代码可用于推荐，这两个代码示例都可能会改进大约十倍。

Haskell：预期的懒惰，为什么评估这个？

1 个答案: