python-如何将字典的键与字符串字符进行比较

Question

I am new to python and I need to compare a character in string with a key in a dictionary. But I am not able to figure out a way to compare that character with a key. I am only able to compare it with the value at dict[key]

I am trying to implement something like this:

score = {"a": 1, "c": 3, "b": 3, "e": 1, "d": 2, "g": 2, 
     "f": 4, "i": 1, "h": 4, "k": 5, "j": 8, "m": 3, 
     "l": 1, "o": 1, "n": 1, "q": 10, "p": 3, "s": 1, 
     "r": 1, "u": 1, "t": 1, "w": 4, "v": 4, "y": 4, 
     "x": 8, "z": 10}

def compare(word):
   res = 0
   for letter in word:
       if score[**What should i put in here**] == letter:
          res += score[letter]
   return res

where score[key] represents the value at that particular key as a whole. Is there a way to compare a key to the letter, instead of the value?

My aim is to compare the "letter" in "word" with the keys in dictionary and add the values against the characters and return the result.

Answer 1

Looks like you're thinking about this strangely. All you need to do is check if the letter is in your score dict, and if it is, to add that number to your total.

def compare(word):
    res = 0
    for letter in word:
        if letter in score:
            res += score[letter]
    return res

However there's an easier way to do this. Since you're just using res as an accumulator, you can add score[letter] if it exists or 0 if it doesn't. This is easy using the dict.get method.

def compare(word):
    res = 0
    for letter in word:
        res += score.get(letter, 0)
        # dict.get(key, defaultvalue)
    return res

In fact you can even make it into an ugly lambda .

compare = lambda word: sum([scores.get(letter,0) for letter in word])

Answer 2

score = {"a": 1, ...}

def compare(word):
   res = 0
   for letter in word:
       if letter in score:
           res += score[letter]
   return res

That's probably what you want. You can even omit if letter in score if you're sure all your letters will exist in score . You don't really need to compare anything.

Answer 3

score = {"a": 1, "c": 3, "b": 3, "e": 1, "d": 2, "g": 2, 
 "f": 4, "i": 1, "h": 4, "k": 5, "j": 8, "m": 3, 
 "l": 1, "o": 1, "n": 1, "q": 10, "p": 3, "s": 1, 
 "r": 1, "u": 1, "t": 1, "w": 4, "v": 4, "y": 4, 
 "x": 8, "z": 10}

def compare(word):
    res = 0
    for letter in word:
       if letter in score:
          res += score[letter]
    return res

Answer 4

As the other answer by Adam has pointed out you are thinking about this in a strange way. There is NO way to get the key of the dictionary score using something like score['*get_me_the_key*'] .

If you really want the keys. score.keys() will give you a list of keys in the dictionary. Something like this:

keys = score.keys()
print keys
['a', 'c', 'b', 'e', 'd', 'g', 'f', 'i', 'h', 'k', 'j', 'm', 'l', 'o', 'n', 'q', 'p', 's', 'r', 'u', 't', 'w', 'v', 'y', 'x', 'z']

Answer 5

dict = {"a": 1, "c": 3, "b": 3, "e": 1, "d": 2, "g": 2, 
 "f": 4, "i": 1, "h": 4, "k": 5, "j": 8, "m": 3, 
 "l": 1, "o": 1, "n": 1, "q": 10, "p": 3, "s": 1, 
 "r": 1, "u": 1, "t": 1, "w": 4, "v": 4, "y": 4, 
 "x": 8, "z": 10}

def compare(word):

res = 0

for letter in word:

   if letter in dict:

      res += dict['letter']

return res

example:

dict['c']

get corresponding value 3

Answer 6

I recognize this problem in the Codecademy: Python course. In the problem you are doing, you must add word = word.lower() at the start of the function. Otherwise, if you check for words containing capital letters, the program will get errors. In other words, "D" is not in the dictionary, but "d" is. Hope this helps.