是否有替代此代碼評估

Question

下面的處理是爲了解析變量dealList填充4字典SHD和C.我認爲問題是在我使用eval(suit)代碼的倒數第二行，西裝是循環通過。

所有的西裝SHD，和C初始化爲空的字典，但是是爲了最終包含13項字典，一個在其訴訟每張卡。

我懷疑所發生的事情是，每次的eval（套裝）執行它找到一本字典來處理，但不會保留字典的名稱，因此所需的命名字典不會被更新。換句話說，它看起來像通過使用從列表中取出的字母S循環遍歷，但該字母也命名字典。但我不知道如何告訴蟒蛇這兩者是相關的。

我需要使用什麼來代替eval（），或者除了eval（）來完成我的目標？

keys = list('23456789TJQKA') 
values = range (13) 
suitDict = {} 
for key,value in zip(keys, values): 
    suitDict[key] = value 
dealList = 'AQJT5.KQ.8.KQT95 3.A765.QT743.843 974.T93.J92.AJ62 K862.J842.AK65.7'.split() 
players = list('NESW') 
suits = list('SHDC') 
S = H = D = C = {} 
playerHand ={} 
for player,hand in zip(players,dealList): 
    playerHand[player]=hand 
    print player,hand 
    for suit,cards in zip(suits,playerHand[player].split('.')): 
     print "SC:",suit,cards 
     for card in cards: 
      eval(suit)[suitDict[card]]= player 
      print "card",card,"Suit",suit,"X",eval(suit),"card",suitDict[card],"player",player  

也許輸出的列表將顯示問題。請注意，在下面的示例輸出列表中，前5張是黑桃，接下來的2張是心，K和Q都是特定的。但黑桃詞典正在改變，而不是心靈詞典。正在更新的陣列的這種缺乏改變會一直持續下去。

N AQJT5.KQ.8.KQT95 
SC: S AQJT5 
card A Suit S X [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 'N'] card 12 player N 
card Q Suit S X [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 'N', 11, 'N'] card 10 player N 
card J Suit S X [0, 1, 2, 3, 4, 5, 6, 7, 8, 'N', 'N', 11, 'N'] card 9 player N 
card T Suit S X [0, 1, 2, 3, 4, 5, 6, 7, 'N', 'N', 'N', 11, 'N'] card 8 player N 
card 5 Suit S X [0, 1, 2, 'N', 4, 5, 6, 7, 'N', 'N', 'N', 11, 'N'] card 3 player N 
SC: H KQ 
card K Suit H X [0, 1, 2, 'N', 4, 5, 6, 7, 'N', 'N', 'N', 'N', 'N'] card 11 player N 
card Q Suit H X [0, 1, 2, 'N', 4, 5, 6, 7, 'N', 'N', 'N', 'N', 'N'] card 10 player N 
SC: D 8 
card 8 Suit D X [0, 1, 2, 'N', 4, 5, 'N', 7, 'N', 'N', 'N', 'N', 'N'] card 6 player N 
SC: C KQT95 
card K Suit C X [0, 1, 2, 'N', 4, 5, 'N', 7, 'N', 'N', 'N', 'N', 'N'] card 11 player N 
card Q Suit C X [0, 1, 2, 'N', 4, 5, 'N', 7, 'N', 'N', 'N', 'N', 'N'] card 10 player N 
card T Suit C X [0, 1, 2, 'N', 4, 5, 'N', 7, 'N', 'N', 'N', 'N', 'N'] card 8 player N 
card 9 Suit C X [0, 1, 2, 'N', 4, 5, 'N', 'N', 'N', 'N', 'N', 'N', 'N'] card 7 player N 
card 5 Suit C X [0, 1, 2, 'N', 4, 5, 'N', 'N', 'N', 'N', 'N', 'N', 'N'] card 3 player N 
E 3.A765.QT743.843 
SC: S 3 
card 3 Suit S X [0, 'E', 2, 'N', 4, 5, 'N', 'N', 'N', 'N', 'N', 'N', 'N'] card 1 player E 
SC: H A765 
card A Suit H X [0, 'E', 2, 'N', 4, 5, 'N', 'N', 'N', 'N', 'N', 'N', 'E'] card 12 player E 
card 7 Suit H X [0, 'E', 2, 'N', 4, 'E', 'N', 'N', 'N', 'N', 'N', 'N', 'E'] card 5 player E 
card 6 Suit H X [0, 'E', 2, 'N', 'E', 'E', 'N', 'N', 'N', 'N', 'N', 'N', 'E'] card 4 player E 
card 5 Suit H X [0, 'E', 2, 'E', 'E', 'E', 'N', 'N', 'N', 'N', 'N', 'N', 'E'] card 3 player E 
SC: D QT743 
card Q Suit D X [0, 'E', 2, 'E', 'E', 'E', 'N', 'N', 'N', 'N', 'E', 'N', 'E'] card 10 player E 
card T Suit D X [0, 'E', 2, 'E', 'E', 'E', 'N', 'N', 'E', 'N', 'E', 'N', 'E'] card 8 player E 
card 7 Suit D X [0, 'E', 2, 'E', 'E', 'E', 'N', 'N', 'E', 'N', 'E', 'N', 'E'] card 5 player E 
card 4 Suit D X [0, 'E', 'E', 'E', 'E', 'E', 'N', 'N', 'E', 'N', 'E', 'N', 'E'] card 2 player E 
card 3 Suit D X [0, 'E', 'E', 'E', 'E', 'E', 'N', 'N', 'E', 'N', 'E', 'N', 'E'] card 1 player E 
SC: C 843 
card 8 Suit C X [0, 'E', 'E', 'E', 'E', 'E', 'E', 'N', 'E', 'N', 'E', 'N', 'E'] card 6 player E 
card 4 Suit C X [0, 'E', 'E', 'E', 'E', 'E', 'E', 'N', 'E', 'N', 'E', 'N', 'E'] card 2 player E 
card 3 Suit C X [0, 'E', 'E', 'E', 'E', 'E', 'E', 'N', 'E', 'N', 'E', 'N', 'E'] card 1 player E 

Answer 1

您的代碼古怪的主要原因是事實，你的所有衣服詞典（S，H，D和C）引用的是相同的字典。如果您希望它們指向不同的參考（如S, H, D, C = {}, {}, {}, {}），則需要分別初始化它們。

這就是說，eval是（幾乎）從來沒有解決方案。在你的情況，我會做一些事情，如：

keys = list('23456789TJQKA') 
values = range (13) 
suitDict = {} 
for key,value in zip(keys, values): 
    suitDict[key] = value 
dealList = 'AQJT5.KQ.8.KQT95 3.A765.QT743.843 974.T93.J92.AJ62 K862.J842.AK65.7'.split() 
players = list('NESW') 
suits = {"S": {}, "H": {}, "D": {}, "C": {}} 
playerHand = {} 
for player,hand in zip(players,dealList): 
    playerHand[player] = hand 
    print player,hand 
     for suit,cards in zip(suits,playerHand[player].split('.')): 
      print "SC:",suit,cards 
      for card in cards: 
       suits[suit][suitDict[card]] = player 
       print "card",card,"Suit",suit,"X",suits[suit],"card",suitDict[card],"player",player 

話雖這麼說，我覺得你過於複雜你的情況......