1

我有以下功能，它是填字游戏求解器的一部分：

def CrosswordPossibleWords(p_words, p_cw_words):
    """For each word found in the crossword, find the possible words and keep track of the one with the minimum possible words.

    Keyword arguments:
    p_words    -- The dictionary words.
    p_cw_words -- The crossword word attributes.
    """
    l_min = 999999999
    l_min_index = -1
    l_index = 0
    l_choices = []
    for l_cw_word in p_cw_words:
        if l_cw_word[2] >= l_min_length and '-' in l_cw_word[4]:
            pattern = re.compile('^' + l_cw_word[4].replace('.', '%').replace('-', '.').upper() + '$', re.UNICODE)
            l_choice = []
            for l_word in [w for w in p_words if len(w) == len(l_cw_word[4])]:
                if re.match(pattern, l_word):
                    l_choice.append(l_word)
            l_choices.append(l_choice)
            if len(l_choice) < l_min:
                l_min_index = l_index
                l_min = len(l_choice)
        else:
            l_choices.append([])
        l_index = l_index + 1
    return (l_choices, l_min_index)

填字游戏的形式如下：

[row, col, length, direction, word]

'.'如果我不能解决那个单词，我有一个单词，'-'如果我不知道那个字母。

我怎样才能使这段代码更快？目前运行大约需要 2.5 秒。正在考虑使用 numpy 字符串；因为显然 numpy 快了 10 倍，但我对 numpy 一无所知，也不知道我是否能够使用所有当前的字符串函数。

有任何想法吗？

4

2 回答 2

1

您可以在调用此函数之前按字长对字典进行分区，因此不必在每次调用时都重新执行此操作。

于 2013-02-26T21:54:26.130 回答
1

虽然我同意 Scott Hunter 的观点，但您可能正在寻找这样的东西，其中列表被替换为 dicts：

def CrosswordPossibleWords(p_words, p_cw_words):
    """For each word found in the crossword, find the possible words and keep track of the one with the minimum possible words.

    Keyword arguments:
    p_words    -- The dictionary words.
    p_cw_words -- The crossword word attributes.
    """
    l_min = 999999999
    l_min_index = -1
    l_index = 0
    l_choices = {}    # using dict instead of list
    for l_cw_word in p_cw_words:
        if l_cw_word[2] >= l_min_length and '-' in l_cw_word[4]:
            pattern = re.compile('^' + l_cw_word[4].replace('.', '%').replace('-', '.').upper() + '$', re.UNICODE)
                l_choice = {}  # using dict instead of list

            for l_word in [w for w in p_words if len(w) == len(l_cw_word[4])]:
                if re.match(pattern, l_word):

                    l_choice[l_word]=None

            l_choices[l_choice]=None

            if len(l_choice) < l_min:  ##
                l_min_index = l_index  ## Get rid of this.
                l_min = len(l_choice)  ##
        else:
            l_choices.append([])    # why append empty list?
        l_index = l_index + 1
        l_choices=list(l_choices.keys())   # ...you probably need the list again...
    return (l_choices, l_min_index)
于 2013-02-26T22:03:39.517 回答