我怎樣才能正確地迭代通過這個列表

Question

下面的例子是問題的簡化。 我有一個列表[Either Foo Bar]，另一份[Biz]。 的想法是，我每次迭代元素Biz通過[Either Foo Bar]，從[Either Foo Bar]的開始，直到Biz是空的。其結果將是，有現在在[Either Foo Bar]

的問題是能夠在[Either Foo Bar]開始的時候它的時間來使用下一個元素[Biz]開始更加Bar s以上且Foo秒。

我可以張貼的什麼，我試圖做的，如果它會幫助一個例子。

更新： 好的這裏是我正在使用的實際類型，仍然試圖忽略我認爲可能是無關的信息。請讓我知道，如果我已經離開了一些重要的東西

[Either UnFlaggedDay CalendarDay] [(CalFlag,Product, Day)]

data CalFlag = FirstPass 
       | SecondPass 
       | ThirdPass 
       deriving (Enum,Eq,Show) 

我試圖做的是檢查Day反對[Either UnFlaggedDay CalendarDay]的Left值。當我比賽，我希望做一個新的列表，這是完全一樣的除外以下更改：我會改變UnFlaggedDay，加上未來兩個UnflaggedDay S IN列表中CalendarDay小號. At that point, I want to use the newly built list, that has the same number of elements still, and the [（CalFlag，產品，日） ] minus the（CalFlag，Product，Day）``剛剛被檢查。下面是我在解決這個問題的不同方法之間的一些破解代碼。

flagReserved :: [Either UnFlaggedDay CalendarDay] -> Handler 
                [Either UnFlaggedDay 
                  CalendarDay] 
flagReserved ((Left (MkUFD day)):rest) = do 
    reserved <- runDB $ selectList [TestQueue ==. Scheduled_Q, 
            TestStatus /<-. [Passed,Failed]] [] 

    case (L.null reserved) of 
    True -> do 
      processedDays <- ((Left $ MkUFD day) :) <$> flagReserved rest 
      return processedDays 

    False -> return $ 
      flagReserved' (map prepList ((Left (MkUFD day)):rest)) 
         (flagProductTuple reserved) 

flagReserved ((Right (MkCal day)):rest) = do 
    processedDays <- ((Right $ MkCal day):) <$> flagReserved rest 
    return processedDays 
flagReserved _ = return [] 

flagReserved' :: [Either (UnFlaggedDay) CalendarDay] -> 
       [(CalFlag,Product,Maybe C.Day)] -> 
       [Either UnFlaggedDay CalendarDay] 

flagReserved' ((Left (MkUFD day)):restD) 
       ((calFlag,firmware,Just startDate):restF) = 
    case (startDate == day || not (calFlag == FirstPass)) of 
     True | (calFlag == ThirdPass) -> 
        flagReserved' ((Right $ 
            conScheduled day firmware Reserved) : restD) restF 

      | otherwise -> 
       flagReserved (Right $ 
           consScheduled day firmware Reserved) : 
           flagReserved' restD 
              ((succ calFlag, 
                firmware, 
                Just startDate) : 
                restF) 
     False -> (Left (MkUFD day)) : flagReserved' restD ((calFlag, 
                  firmware, 
                  Just startDate) : restF) 




flagReserved' ((Right (MkCal (Left (MkAD (dayText,day))))):restD) 
       ((calFlag,firmware,Just startDate):restF) = 
     case (startDate == day || not (calFlag == FirstPass)) of 
       True | (calFlag == ThirdPass) -> 
         (Right $ consScheduled day firmware Reserved) : 
         flagReserved' restD restF 
        | otherwise -> 
         (Right $ consScheduled day firmware Reserved) : 
          flagReserved' restD ((succ calFlag, 
                firmware, 
                Just startDate):restF) 
       False -> 
       (Right (MkCal (Left (MkAD (dayText,day))))) : 
       flagReserved' restD ((calFlag,firmware,Just startDate) : 
             restF) 


flagReserved' ((Right (MkCal (Right unAvailable))):restD) 
       ((calFlag,firmware,startDate):restF) = 
       (Right $ 
       MkCal $ 
       Right unAvailable) : 
       flagReserved' restD ((calFlag,firmware,startDate) : restF) 

flagReserved' unprocessed [] = unprocessed 
flagReserved' [] _ = [] 

更新：

我做了一些測試代碼，以制定出我的想法。這裏是我到目前爲止

let reservedDays = [(FirstPass,IM,C.fromGregorian 2012 01 15), 
        (FirstPass,WAF,C.fromGregorian 2012 01 14), 
        (FirstPass,Backup,C.fromGregorian 2012 01 13) 
        ] 

dummyFunc :: [Either UnFlaggedDay CalendarDay] -> (CalFlag,Product,C.Day) 
    dummyFunc dayList (cFlag,product,day) = if day `elem` dayList 
             then dummyFunc' dayList (cFlag,product,day) 
             else dayList 

dummyFunc' dayList (cFlag,product,day) = 
    if (cFlag == ThirdPass) 
    then 

好了，這裏就是我卡住了。我需要能夠將接下來的三個Left值更改爲Right值。我對dummyFunc'意圖是在第一Left價值分裂列表，刪除它，添加新Right值，與之前分拆名單，然後重複兩次。有沒有更好的辦法？如果沒有，是否已經有一個函數根據我提到的標準將列表分成兩半？我可以弄清楚如何手工完成，但我並沒有試圖重新發明輪子。

Answer 1

我認爲在[Biz]每個元件可能調整在[Either Foo Bar]一個或多個元件遠離左邊（Foo）類型和向右（Bar）型。這只是一個方面：

eitherList = [Left(), Left(), Right 5, Right 9, Left()] 
bizList = [4,5,6,7,1] 

func eitherlst biz = if (Left()) `elem` eitherlst 
         then Right biz : delete (Left()) eitherlst 
         else eitherlst 

eitherList' = foldl func eitherList bizList 

以上是未經檢驗的，但你可以看到更新eitherList如何每次調用之間傳遞給func作爲考慮原eitherList和所有Biz元素到這一點的結果。正如你所看到的，你的func的實現將會使這個有用。