我有一个字典列表,我想根据多个条件进行过滤。该列表的简化版本如下所示:
orders = [{"name": "v", "price": 123, "location": "Mars"},
{"name": "x", "price": 223, "location": "Mars"},
{"name": "x", "price": 124, "location": "Mars"},
{"name": "y", "price": 456, "location": "Mars"},
{"name": "z", "price": 123, "location": "Mars"},
{"name": "z", "price": 5623, "location": "Mars"}]
我希望最终得到一个列表,其中包含具有相同“名称”键的每本词典的最低价格的词典。
例如,上面的内容将变为:
minimums = [{"name": "v", "price": 123, "location": "Mars"},
{"name": "x", "price": 124, "location": "Mars"},
{"name": "y", "price": 456, "location": "Mars"},
{"name": "z", "price": 123, "location": "Mars"}]
我通过令人厌恶的嵌套 if 语句和 for 循环实现了这一点,但是我希望有一种更“Pythonic”的方式来实现目标。
重复使用相同的列表或创建一个新列表都可以。
感谢您的帮助。
编辑:
谢谢您的回答,我尝试用以下代码对每个问题进行计时
print("Number of dictionaries in orders: " + str(len(orders)))
t0 = time.time()
sorted_orders = sorted(orders, key=lambda i: i["name"])
t1 = time.time()
sorting_time = (t1 - t0)
t0 = time.time()
listcomp_wikiben = [x for x in orders if all(x["price"] <= y["price"] for y in orders if x["name"] == y["name"])]
t1 = time.time()
print("listcomp_wikiben: " + str(t1 - t0))
t0 = time.time()
itertools_MrGeek = [min(g[1], key=lambda x: x['price']) for g in groupby(sorted_orders, lambda o: o['name'])]
t1 = time.time()
print("itertools_MrGeek: " + str(t1 - t0 + sorting_time))
t0 = time.time()
itertools_Cory = [min(g, key=lambda j: j["price"]) for k,g in groupby(sorted_orders, key=lambda i: i["name"])]
t1 = time.time()
print("itertools_CoryKramer: " + str(t1 - t0 + sorting_time))
t0 = time.time()
pandas_Trenton = pd.DataFrame(orders)
pandas_Trenton.groupby(['name'])['price'].min()
t1 = time.time()
print("pandas_Trenton_M: " + str(t1 - t0))
结果是:
Number of dictionaries in orders: 20867
listcomp_wikiben: 39.78123s
itertools_MrGeek: 0.01562s
itertools_CoryKramer: 0.01565s
pandas_Trenton_M: 0.29685s