Question

有没有办法过滤这样的列表？

movies = [
   [["t", "transformers"],["g", "animation"],["d", "2005"]],
   [["t", "ted"],["g", "comedy"],["d", "2008"]],
   [["t", "starwars"],["g", "action"],["d", "2000"]]
]

有这样的过滤器列表

filter = [["g", "animation"], ["d", "2005"]]

并收到

["transformers"]

我认为由于电影列表的结构不可能，但我不知道Python向导是否有答案。

如果您对此有任何建议，谢谢。

Answer 1

类似的东西：

movies = [
   [["t", "transformers"],["g", "animation"],["d", "2005"]],
   [["t", "ted"],["g", "comedy"],["d", "2008"]],
   [["t", "starwars"],["g", "action"],["d", "2000"]]
]

fil = [["g", "animation"], ["d", "2005"]]


print [i[0][1] for i in movies if fil[0] in i and fil[1] in i]

这将打印：

['transformers']

Answer 2

另一种方法是使用面向对象编程（OPP）。你可以实现一个非常简单的类，它将为你节省大量的工作。

class Movie:

    def __init__(self, title, genre, year):
        self.title = title
        self.genre = genre
        self.year = year


    def filter(self, filter_):
        """ filter_ here is a dict."""
        for feature, value in filter_.items():
            if getattr(self, feature) != value:
                return False
        return True

    def __repr__(self):
        return self.title


movies = [
    Movie("transformers", "animation", 2005),
    Movie("ted", "comedy", 2008),
    Movie("starwars", "action", 2000)
]

filter_ = { "genre": "animation", "year": 2005}

movies = [movie for movie in movies if movie.filter(filter_)]

print(movies)

这产生了输出：

['transformers']

你也可以这样做：

print(movies[0].title)

将产生输出：

'transformers'

Answer 3

第一步，在dicts中进行重组（我将filter重命名为filtr，因为filter是内置函数）：

movies = [dict(item) for item in movies]
filtr = dict(filtr)

然后，根据您的标准进行过滤：

filtered = [item for item in movies if all(item[k] == v for k,v in filtr.iteritems())]

在这里，您循环播放所有电影，然后查找与所有滤镜元素匹配的电影。

然后，您打印所有匹配的标题

print [item['t'] for item in filtered]
# ['transformers']

具有过滤器列表python的理解列表

3 个答案: