Python：如何列印字典的第一個元素和第二個的位置？

我今天剛開始在課堂上使用字典，現在我正在嘗試根據手機的鍵盤創建一個密碼。例如，我有這本字典：

2:['a','b','c']

現在，如果我想加密例如 a，輸出將是 21，其中兩個來自字典，1 來自其中 a 的位置，但我真的不知道該怎么做。這是我到目前為止嘗試過的，但沒有奏效。

def phoneCipher(string):
    cipher = ''
    keypad = {'2':['a','b','c'],'3':['d','e','f'],'4':['g','h','i'],'5':['j','k','l'],'6' : ['m','n','o'],
              '7':['p','q','r','s'],'8':['t','u','v'],'9':['w','x','y','z']}
    for i in string:
        if i in keypad:
            position = keypad.find(i)
            cipher = keypad[i]   position
    return cipher

uj5u.com熱心網友回復：

如果您真的想這樣做，那么正如@Barmar所說，當您檢查i in keypad它是否檢查字典的鍵時，即數字值。所以這永遠不會匹配。似乎您還想獲取字串中每個字符的密碼，因此需要兩個嵌套回圈。我認為這可以按照您的意愿進行。

def phoneCipher(string):
    cipher = ''
    keypad = {'2':['a','b','c'],'3':['d','e','f'],'4':['g','h','i'],'5':['j','k','l'],'6' : ['m','n','o'],
              '7':['p','q','r','s'],'8':['t','u','v'],'9':['w','x','y','z']}
    for i in string:
        for key in keypad:
            if i in keypad[key]:
                #position = keypad[key].index(i) # note its index not find, but we don't actually need it
                cipher  = key
    return cipher

盡管效率極低，但我建議您將字典顛倒過來，因此它是字母對數字而不是數字對字母。

所以也許：

def phoneCipher2(string):
    cipher = ''
    keypad = {'a':'2','b':'2','c':'2',
    'd':'3','e':'3','f':'3',
    'g':'4','h':'4','i':'4',
    'j':'5','k':'5','l':'5',
    'm':'6','n':'6','o':'6',
    'p':'7','q':'7','r':'7','s':'7',
    't':'8','u':'8','v':'8',
    'w':'9','x':'9','y':'9','z':'9',}
    for i in string:
        cipher  = keypad[i]
    return cipher

uj5u.com熱心網友回復：

您可以執行兩個基本選項。一種是迭代其中的每個list，dict然后迭代list并查看值是否在其中。另一種是有正向和反向dict。這會占用更多記憶體，但速度更快。

選項list1迭代dict

def phoneCipher(string):
    cipher = ''
    keypad = {'2': ['a', 'b', 'c'], '3': ['d', 'e', 'f'], '4': ['g', 'h', 'i'],
              '5': ['j', 'k', 'l'], '6': ['m', 'n', 'o'], '7': ['p', 'q', 'r', 's'],
              '8': ['t', 'u', 'v'], '9': ['w', 'x', 'y', 'z']}

    for c in string:
        for number, letters in keypad.items():
            if c in letters:
                cipher  = number   str(letters.index(c))
    return cipher

選項 2使用反向查找dict

keypad = {'2': ['a', 'b', 'c'], '3': ['d', 'e', 'f'], '4': ['g', 'h', 'i'],
          '5': ['j', 'k', 'l'], '6': ['m', 'n', 'o'], '7': ['p', 'q', 'r', 's'],
          '8': ['t', 'u', 'v'], '9': ['w', 'x', 'y', 'z']}

# Creating a reverse dict should be done only once and saved for future/repeated use
reverse_lookup = dict()

for number, letters in keypad.items():
    for index, letter in enumerate(letters):
        reverse_lookup[letter] = number   str(index)


def phoneCipher(string):
    return ''.join(map(lambda char: reverse_lookup[char], string))

uj5u.com熱心網友回復：

試試這個，

def phoneCipher(string):
    cipher = ''
    keypad = {'2':['a','b','c'],'3':['d','e','f'],'4':['g','h','i'],'5':['j','k','l'],'6' : ['m','n','o'],
              '7':['p','q','r','s'],'8':['t','u','v'],'9':['w','x','y','z']}
    
    for i in string:
      for key in keypad:
        if i in keypad[key]:
          cipher  = key
          continue
    
    return cipher

phoneCipher("ishan")

輸出 -

47426

標籤：Python 字典

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