Question

我正在尝试编写一个程序，该程序将使用指定的字母和通配符（'*'）并根据列表中的单词检查它们以打印所有可用的匹配项。我编写了下面的代码，当使用两个通配符时，它将起作用：

def wildcard_search(letters, count):
    alpha = 'abcdefghijklmnopqrstuvwxyz'
    words = ['hello', 'hi', 'good', 'help', 'hellos', 'helloing', 'hallow', 'no']
    count = 0
    wild_loc = []

    while count < len(letters):
        for letter in letters:
            if letter == '*':
                wild_loc.append(count)
                count += 1

    for letter in alpha:
        new_letters = letters[:wild_loc[1]].replace('*', letter)

        for each in words:
                    each = each.strip('')

            if new_letters in each:
                holder = new_letters

                for letter in alpha:
                    new_letters = letters[wild_loc[1]:].replace('*', letter)

                    for each in words:
                        each = each.strip('')

                        if holder + new_letters in each:
                             print each

我的问题是，当使用两个以上的通配符时，如何编写此代码以返回结果？我尝试使用下面的while循环，但最终我的索引超出了范围错误：

count = 0
store = ''
while count <= len(wild_loc)-1:
    for letter in alpha:
        if count != len(wild_loc) - 1:
            new_letter = letters[:wild_loc[count]].replace('*', letter)
            for each in words:
                each = each.strip('')
                if new_letter in each:
                    res = store + new_letter
                    store = new_letter
            count += 1

        elif count == len(wild_loc) - 1:
            new_letter = letters[wild_loc[count]:].replace('*', letter)
            for each in words:
                each = each.strip('')
                if (res + new_letter) in each:
                    print each
            count += 1

Answer 1

使用fnmatch.filter。基本上，它使用re module实现类似shell的模式匹配，并且完全符合您的要求。

使用两个以上通配符时的问题

1 个答案: