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First solution in Clear category for Bigger Price by vlad.bezden
"""
Bigger Price.
You have a table with all available goods in the store. The data is represented as a list of dicts
Your mission here is to find the TOP most expensive goods. The amount we are looking for will be
given as a first argument and the whole data as the second one
Input: int and list of dicts. Each dicts has two keys "name" and "price"
Output: the same as the second Input argument.
"""
def bigger_price(limit: int, data: list) -> list:
"""
TOP most expensive goods
"""
return sorted(data, key=lambda x: x["price"], reverse=True)[:limit]
if __name__ == "__main__":
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"
Sept. 29, 2018
