[學習] Check.io - Bigger Price

check.io 上的題目：求出 list 裡 topN 貴的物品儲存方式是dicionary，每個物品各有兩個 key 分別為 name & price
此系列文章主要是自己在解題時覺得有趣的題目，記錄下來讓自己以後好複習，有興趣的人可以參考看看

先來看看題目

看完題目可能會覺得小複雜，一般題目出有關 dictionary 的題目大多都只要找出一個最大或最小的 key 值，很直覺就會想到要用max/min的函式來解決(詳細用法在此篇文章有提到)，這裡還要找不只一個，但認真想想其實也沒很難，用sort就可以一行輕鬆解決了。

sorted 的用法

語法：sorted(list, key=parameters, reverse=True/False)

基本

1
2
3

x = [3, 2, 6, 1, 9]
y = sorted(x)
print(y)

output:

1	[1, 2, 3, 6, 9]

反向排序(大到小)

1	y = sorted(x, reverse=True)

output:

1	[9, 6, 3, 2, 1]

進階

example 1:

x = [("Amy", 150, 60),
     ("Emily", 154, 55),
     ("Wen", 145, 70)]
print(sorted(x, key=lambda a: a[1]))

output:

1	[('Wen', 145, 70), ('Amy', 150, 60), ('Emily', 154, 55)]

example 2:

x = [{"fruit": "apple", "price": 20},
     {"fruit": "banana", "price": 18},
     {"fruit": "peach", "price": 25}] 
print(sorted(x, key=lambda x: x["price"]))

output:

1	[{'fruit': 'banana', 'price': 18}, {'fruit': 'apple', 'price': 20}, {'fruit': 'peach', 'price': 25}]

解答：

def bigger_price(limit: int, data: list) -> list:
    """
        TOP most expensive goods
    """
    return sorted(data, key=lambda d: d['price'], reverse=True)[:limit]


if __name__ == '__main__':
    from pprint import pprint
    print('Example:')
    pprint(bigger_price(2, [
        {"name": "bread", "price": 100},
        {"name": "wine", "price": 138},
        {"name": "meat", "price": 15},
        {"name": "water", "price": 1}
    ]))

    # These "asserts" using for self-checking and not for auto-testing
    assert bigger_price(2, [
        {"name": "bread", "price": 100},
        {"name": "wine", "price": 138},
        {"name": "meat", "price": 15},
        {"name": "water", "price": 1}
    ]) == [
        {"name": "wine", "price": 138},
        {"name": "bread", "price": 100}
    ], "First"

    assert bigger_price(1, [
        {"name": "pen", "price": 5},
        {"name": "whiteboard", "price": 170}
    ]) == [{"name": "whiteboard", "price": 170}], "Second"

    print('Done! Looks like it is fine. Go and check it')