[英]How to filter list of dictionaries in python?
我有一個字典列表,如下所示-
VehicleList = [
{
'id': '1',
'VehicleType': 'Car',
'CreationDate': datetime.datetime(2021, 12, 10, 16, 9, 44, 872000)
},
{
'id': '2',
'VehicleType': 'Bike',
'CreationDate': datetime.datetime(2021, 12, 15, 11, 8, 21, 612000)
},
{
'id': '3',
'VehicleType': 'Truck',
'CreationDate': datetime.datetime(2021, 9, 13, 10, 1, 50, 350095)
},
{
'id': '4',
'VehicleType': 'Bike',
'CreationDate': datetime.datetime(2021, 12, 10, 21, 1, 00, 300012)
},
{
'id': '5',
'VehicleType': 'Car',
'CreationDate': datetime.datetime(2021, 12, 21, 10, 1, 50, 600095)
}
]
如何根據“CreationDate”獲取每個“VehicleType”的最新車輛列表?
我期待這樣的事情-
latestVehicles = [
{
'id': '5',
'VehicleType': 'Car',
'CreationDate': datetime.datetime(2021, 12, 21, 10, 1, 50, 600095)
},
{
'id': '2',
'VehicleType': 'Bike',
'CreationDate': datetime.datetime(2021, 12, 15, 11, 8, 21, 612000)
},
{
'id': '3',
'VehicleType': 'Truck',
'CreationDate': datetime.datetime(2021, 9, 13, 10, 1, 50, 350095)
}
]
我嘗試根據它們的“VehicleType”將每個字典分離到不同的列表中,然后根據它們的“CreationDate”對它們進行排序,然后選擇最新的。
我相信可能有更優化的方法來做到這一點。
使用從VehicleType
值到最終列表中所需的字典的字典映射。 將輸入列表中每個項目的日期與您的字典進行比較,並保留后一個。
latest_dict = {}
for vehicle in VehicleList:
t = vehicle['VehicleType']
if t not in latest_dict or vehicle['CreationDate'] > latest_dict[t]['CreationDate']:
latest_dict[t] = vehicle
latestVehicles = list(latest_dict.values())
這是使用max
和filter
的解決方案:
VehicleLatest = [
max(
filter(lambda _: _["VehicleType"] == t, VehicleList),
key=lambda _: _["CreationDate"]
) for t in {_["VehicleType"] for _ in VehicleList}
]
結果
print(VehicleLatest)
# [{'id': '2', 'VehicleType': 'Bike', 'CreationDate': datetime.datetime(2021, 12, 15, 11, 8, 21, 612000)}, {'id': '3', 'VehicleType': 'Truck', 'CreationDate': datetime.datetime(2021, 9, 13, 10, 1, 50, 350095)}, {'id': '5', 'VehicleType': 'Car', 'CreationDate': datetime.datetime(2021, 12, 21, 10, 1, 50, 600095)}]
我認為您可以使用來自 itertools 的 groupby function 來實現您想要的。
from itertools import groupby
# entries sorted according to the key we wish to groupby: 'VehicleType'
VehicleList = sorted(VehicleList, key=lambda x: x["VehicleType"])
latestVehicles = []
# Then the elements are grouped.
for k, v in groupby(VehicleList, lambda x: x["VehicleType"]):
# We then append to latestVehicles the 0th entry of the
# grouped elements after sorting according to the 'CreationDate'
latestVehicles.append(sorted(list(v), key=lambda x: x["CreationDate"], reverse=True)[0])
按'VehicleType'
和'CreationDate'
排序,然后從'VehicleType'
和車輛創建字典以獲取每種類型的最新車輛:
VehicleList.sort(key=lambda x: (x.get('VehicleType'), x.get('CreationDate')))
out = list(dict(zip([item.get('VehicleType') for item in VehicleList], VehicleList)).values())
Output:
[{'id': '2',
'VehicleType': 'Bike',
'CreationDate': datetime.datetime(2021, 12, 15, 11, 8, 21, 612000)},
{'id': '5',
'VehicleType': 'Car',
'CreationDate': datetime.datetime(2021, 12, 21, 10, 1, 50, 600095)},
{'id': '3',
'VehicleType': 'Truck',
'CreationDate': datetime.datetime(2021, 9, 13, 10, 1, 50, 350095)}]
這在pandas
中非常簡單。 首先將字典列表加載為 pandas dataframe,然后按日期對值進行排序,取前 n 項(下例中為 3),然后導出到字典。
import pandas as pd
df = pd.DataFrame(VehicleList)
df.sort_values('CreationDate', ascending=False).head(3).to_dict(orient='records')
您可以使用運算符來實現該目標:
import operator
my_sorted_list_by_type_and_date = sorted(VehicleList, key=operator.itemgetter('VehicleType', 'CreationDate'))
一個更易讀的代碼的小請求:
from operator import itemgetter
from itertools import groupby
vtkey = itemgetter('VehicleType')
cdkey = itemgetter('CreationDate')
latest = [
# Get latest from each group.
max(vs, key = cdkey)
# Sort and group by VehicleType.
for g, vs in groupby(sorted(vehicles, key = vtkey), vtkey)
]
Blckknght 的答案的一個變體,使用defaultdict
來避免長if
條件:
from collections import defaultdict
import datetime
from operator import itemgetter
latest_dict = defaultdict(lambda: {'CreationDate': datetime.datetime.min})
for vehicle in VehicleList:
t = vehicle['VehicleType']
latest_dict[t] = max(vehicle, latest_dict[t], key=itemgetter('CreationDate'))
latestVehicles = list(latest_dict.values())
最新車輛:
[{'id': '5', 'VehicleType': 'Car', 'CreationDate': datetime.datetime(2021, 12, 21, 10, 1, 50, 600095)},
{'id': '2', 'VehicleType': 'Bike', 'CreationDate': datetime.datetime(2021, 12, 15, 11, 8, 21, 612000)},
{'id': '3', 'VehicleType': 'Truck', 'CreationDate': datetime.datetime(2021, 9, 13, 10, 1, 50, 350095)}]
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