问题
I have a list of dictionaries that I would like to filter based on multiple criteria. A shortened version of the list looks like so:
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"}]
I am looking to end up with a list that contains the dictionaries with the lowest price for each dictionary with the same "name" key. For example, the above would become:
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"}]
I have accomplished this with an abomination of nested if-statements and for-loops, however I was hoping there was a more "Pythonic" way of achieving things.
Either reusing the same list or creating a new one is fine.
Thank you for the help.
EDIT: Thank you for the answers, I tried timing each of them with the following code
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))
And the results were:
Number of dictionaries in orders: 20867
listcomp_wikiben: 39.78123s
itertools_MrGeek: 0.01562s
itertools_CoryKramer: 0.01565s
pandas_Trenton_M: 0.29685s
回答1:
If you first sort your list by "name"
, you can use itertools.groupby
to group them, then use min
with a lambda to find the minimum "price"
in each group.
>>> from itertools import groupby
>>> sorted_orders = sorted(orders, key=lambda i: i["name"])
>>> [min(g, key=lambda j: j["price"]) for k,g in groupby(sorted_orders , key=lambda i: i["name"])]
[{'name': 'v', 'price': 123, 'location': 'Mars'},
{'name': 'x', 'price': 124, 'location': 'Mars'},
{'name': 'y', 'price': 456, 'location': 'Mars'},
{'name': 'z', 'price': 123, 'location': 'Mars'}]
回答2:
You can use itertools.groupby
:
from itertools import groupby
print([min(g[1], key = lambda x : x['price']) for g in groupby(orders, lambda o : o['name'])])
Output:
[
{'name': 'v', 'price': 123, 'location': 'Mars'},
{'name': 'x', 'price': 124, 'location': 'Mars'},
{'name': 'y', 'price': 456, 'location': 'Mars'},
{'name': 'z', 'price': 123, 'location': 'Mars'}
]
回答3:
Solution without itertools
[x for x in orders if all(x["price"] <= y["price"] for y in orders if x["name"] == y["name"])]
回答4:
Use pandas
:
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": "Pluto"},
{"name": "z", "price": 5623, "location": "Mars"}]
import pandas as pd
df = pd.DataFrame(orders)
df.groupby(['name', 'location'])['price'].min()
来源:https://stackoverflow.com/questions/57597433/filtering-a-list-of-dictionaries-based-on-multiple-values