Question

我有一个清单：

scrabble_scores = [(1, "E A O I N R T L S U"), (2, "D G"), (3, "B C M P"),
               (4, "F H V W Y"), (5, "K"), (8, "J X"), (10, "Q Z")]

我必须写一个函数，它给出一个dict，其中包含一个得分映射的字母，只要上面的列表被传递。请帮助我编写函数... !!

Answer 1

用两个for循环来拯救Dict：...

letter_score = {letter: score for score, letters in scrabble_scores
                              for letter in letters.split()}

E.g。对于字符串中的每个字母（由空格分隔），在输出字典中生成一个键和值对;关键是字母，分数值。

演示：

>>> scrabble_scores = [(1, "E A O I N R T L S U"), (2, "D G"), (3, "B C M P"),
...                (4, "F H V W Y"), (5, "K"), (8, "J X"), (10, "Q Z")]
>>> letter_score = {letter: score for score, letters in scrabble_scores
...                               for letter in letters.split()}
>>> letter_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}
>>> letter_score['Q']
10

奖金单词分数计算器：

>>> word = 'QUICK'
>>> sum(letter_score[c] for c in word)
20

其中word是一个仅包含（拼字）字母的大写字符串，忽略了双字母和三字母的评分。

Answer 2

另一种更冗长的方式：

def makeScoreDict(scrabble_scores):
        score_dict = {}
        for row in scrabble_scores:
            for letter in row[1].split():
                score_dict[letter] = row[0]
        return score_dict

Answer 3

在python中，你可以循环遍历这样的对：

for score, letters in scrabble_scores:
    print("{}: {}".format(letters, score))

这将给出结果：

"E A O I N R T L S U": 1
"D G": 2
...

在python中，您可以拆分包含空格的字符串：

>>> "E A O I N R T L S U".split()
['E', 'A', 'O', 'I', 'N', 'R', 'T', 'L', 'S', 'U']

现在只需将两者结合起来：

for score, letters in scrabble_scores:
    for letter in letters.split():
        # voila, you've got score and letter.

Answer 4

这是你想要的吗？似乎是你所描述的。

scrabble_scores = [(1, "E A O I N R T L S U"), (2, "D G"), (3, "B C M P"),
               (4, "F H V W Y"), (5, "K"), (8, "J X"), (10, "Q Z")]
out_dict = {} 
for score in scrabble_scores:
    val = score[0]
    for letter in score[1].split():
        out_dict[letter] = val
print out_dict

Answer 5

非字典理解答案。

scrabble_scores = [(1, "E A O I N R T L S U"), (2, "D G"), (3, "B C M P"),
           (4, "F H V W Y"), (5, "K"), (8, "J X"), (10, "Q Z")]

d = dict()
for value,letters in scrabble_scores:
    d.update(dict(zip(letters.split(),[value]*len(letters.split())))

我肯定会把它作为一个dict comp，我确信这是OP的目的，但dict(zip(iter,iter))是值得一提的有用表达。

用于映射拼字游戏字母分布的字典理解

5 个答案: