python - これをリストとして返すにはどうすればよいですか

Question

私はこのデータを持っています：

inventory = { HAMMER: (10,100),
              SCREW: (1, 1000),
              NAIL: (1, 1000),
              SCREWDRIVER: (8, 100),
              DRILL: (50, 20),
              WORKBENCH: (150, 5),
              HANDSAW: (15, 50),
              CHAINSAW: (80, 30)
            }

私は次の関数を書きました：

def get_items(cheapness):
    """ Return a list of (item, (price, count)) tuples that are the given
    cheapness. Note that the second element of the tuple is another tuple. """

    if cheapness == 'CHEAP':
        return [(key, value) for (key,value) in inventory.items() if value in ((cost,quantity) for (cost, quantity) in inventory.values() if cost < 20)]
    elif cheapness == 'MODERATE':
        return [(key, value) for (key,value) in inventory.items() if value in ((cost,quantity) for (cost, quantity) in inventory.values() if 20 < cost < 100)]
    elif cheapness == 'EXPENSIVE':
        return [(key, value) for (key,value) in inventory.items() if value in ((cost,quantity) for (cost, quantity) in inventory.values() if cost > 100)]

私の問題は、 を実行するprint type(get_items(CHEAP))と、 が得られること<type 'NoneType'>です。結果をリストとして返すにはどうすればよいですか?

Answer 1

このコードはより機能的で、保守が容易になる可能性があります。

def get_items(cheapness):
    def a(value):
        if cheapness == 'CHEAP':
            return value < 20
        if cheapness == 'MODERATE':
            return value >= 20 and value < 100
        if cheapness == 'EXPENSIVE':
            return value >= 100
        return False
    return filter(lambda x: a(x[1][0]), inventory.items())

Answer 2

あなたの言いたいことを言ってください：

def get_items(cheapness):
    """ Return a list of (item, (price, count)) tuples that are the given
    cheapness. Note that the second element of the tuple is another tuple. """

    if cheapness == 'CHEAP':
        return [(item, (price,count)) for (item, (price,count)) in inventory.items() if price < 20]
    elif cheapness == 'MODERATE':
        return [(item, (price,count)) for (item, (price,count)) in inventory.items() if price > 20 and price < 100]
    elif cheapness == 'EXPENSIVE':
        return [(item, (price,count)) for (item, (price,count)) in inventory.items() if price > 100]

Answer 3

print type(get_items('CHEAP')) はどうですか? （引用符に注意してください）