如何过滤python中的字典列表？

Question

我有一个字典列表，如下-

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)
        }
    ]

如何根据“CreationDate”获取每个“VehicleType”的最新车辆列表？

我期待这样的事情-

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)
        }
    ]

我尝试根据每个字典的“VehicleType”将它们分离到不同的列表中，然后根据它们的“CreationDate”对它们进行排序，然后选择最新的。

我相信可能有更好的方法来做到这一点。

Answer 1

Blckknght's answer 的一种变体，使用 defaultdict 来避免长 if 条件：

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())

最新车辆：

[{'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)}]

Answer 2

一个更易读的代码的小请求：

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)
]

Answer 3

这是使用max 和filter 的解决方案：

VehicleLatest = [
    max(
        filter(lambda _: _["VehicleType"] == t, VehicleList), 
        key=lambda _: _["CreationDate"]
    ) for t in {_["VehicleType"] for _ in VehicleList}
]

结果

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)}]

Answer 4

您可以使用运算符来实现该目标：

import operator
my_sorted_list_by_type_and_date = sorted(VehicleList, key=operator.itemgetter('VehicleType', 'CreationDate'))

Answer 5

这在pandas 中非常简单明了。首先将 dicts 列表加载为 pandas 数据框，然后按日期对值进行排序，取出前 n 个项目（下例中为 3 个），然后导出到 dict。

import pandas as pd

df = pd.DataFrame(VehicleList)
df.sort_values('CreationDate', ascending=False).head(3).to_dict(orient='records')

Answer 6

按'VehicleType' 和'CreationDate' 排序，然后根据'VehicleType' 和车辆创建字典，以获取每种类型的最新车辆：

VehicleList.sort(key=lambda x: (x.get('VehicleType'), x.get('CreationDate')))
out = list(dict(zip([item.get('VehicleType') for item in VehicleList], VehicleList)).values())

输出：

[{'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)}]

Answer 7

使用从VehicleType 值到最终列表中所需的字典的字典映射。将输入列表中每个项目的日期与您的字典中的日期进行比较，并保留后一个。

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())

Answer 8

我认为你可以使用 itertools 中的 groupby 函数来实现你想要的。

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])