此代码段不适用于大写和小写字母

Question

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 scrabble_score(word):
    count=0
    word.lower()
    print word
    for char in word:
        count=count+score[char]
    return count

我基本上必须取输入词并根据字典计算其分数.

Answer 1

此修改后的代码将起作用:

def scrabble_score(word):
    count=0
    word = word.lower() #assign the result of word.lower() to word

word.lower()返回修改后的单词,它不会修改字符串inplace.字符串在Python中是不可变的..lower()返回字符串的事实定义如下:

>>> help(str.lower)
Help on method_descriptor:

lower(...)
    S.lower() -> string

    Return a copy of the string S converted to lowercase.

Answer 2

str.lower()返回字符串的副本 - 它不会更改原始字符串.试试这个:

word = word.lower()