輸入：word_types 列表，例如 ["verb", "adjective", "name", "noun]期望的輸出：按 word_type 排列的每個字母-word_type 組合的三個單詞的字典。IE{“動詞”：[“Averb1”，“Averb2”，“Averb3”]，[Bverb1，...]，...，“形容詞”：[“Aadjective1”，“Aadjectives2”，“Aadjectives3”] .. 。 ETC這是我用來生成數據的內容。import stringA_to_Z = list(string.ascii_uppercase) # list of letters a-zlettered_word_types = ["verb", "adjective", "name", "noun"]class Lexicon: def __init__(self, testing = True): if testing == True: for TYPE in lettered_word_types: setattr(self, TYPE + "s", [letter + TYPE for letter in A_to_Z]) self.__dict__ = {key: [[element + "1", element + "2", element + "3"] for element in value] for key, value in self.__dict__.items()}data = Lexicon().__dict__
1 回答
冉冉說
TA貢獻1877條經驗 獲得超1個贊
該類沒有任何作用，它只是混淆了你正在做的事情。要將代碼音譯為dict直接使用 a，您只需要：
result = {}
for TYPE in lettered_word_types:
result[TYPE + 's'] = [letter + TYPE for letter in A_to_Z]
result = {key: [[element + "1", element + "2", element + "3"] for element in value] for key, value in result.items()}
所以，只是為了證明這確實是等價的：
In [1]: import string
...: A_to_Z = list(string.ascii_uppercase) # list of letters a-z
...: lettered_word_types = ["verb", "adjective", "name", "noun"]
...:
...: class Lexicon:
...: def __init__(self, testing = True):
...: if testing == True:
...: for TYPE in lettered_word_types:
...: setattr(self, TYPE + "s", [letter + TYPE for letter in A_to_Z])
...: self.__dict__ = {key: [[element + "1", element + "2", element + "3"] for element in value] for key, value in self.__dict__.items()}
...:
...:
...: data = Lexicon().__dict__
In [2]: result = {}
In [3]: for TYPE in lettered_word_types:
...: result[TYPE + 's'] = [letter + TYPE for letter in A_to_Z]
...:
In [4]: result = {key: [[element + "1", element + "2", element + "3"] for element in value] for key, value in result.items()}
In [5]: result == data
Out[5]: True
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