Python: how to print the first element of a dictionary and position of the second?-CodePudding

I just started dictionaries on my class today, now I'm trying to create a cipher based on the keypad of a phone. I have, for example, this dictionary:

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

Now, if I wanted to encrypt for example, a, the output would be 21, the two from the dictionary and 1 from the position of a inside it, but I don't really know how to do it. Here's what I've tried so far but isn't working.

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

CodePudding user response：

If you really want to do it that way around, then as @Barmar say's when you check if i in keypad its checking the keys of the dictionary, i.e. the number values. So that'll never match. Also seems like you want to get the cipher for each character in the string, thus two nest loops are needed. I think this works how you wanted.

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

Although it terribly inefficient, i'd suggest twisting your dictionary on its head, so it is letters to numbers rather than numbers to letters.

so maybe:

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

CodePudding user response：

There are two basic options you can do. One is to iterate each list in the dict and then iterate the list and see if the value is in there. The other is to have a forward and a reverse dict. This takes up more memory but is much faster.

Option 1 Iterate list in 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

Option 2 Use reverse lookup 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))

CodePudding user response：

Try this,

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")

Output -