Python - 从嵌套列表中提取值

Question

我有一个列表，如下所示：

[{'id': 'id_123',

  'type': 'type_1',

  'created_at': '2020-02-12T17:45:00Z'},

 {'id': 'id_124',

  'type': 'type_2',

  'created_at': '2020-02-12T18:15:00Z'},

 {'id': 'id_125',

  'type': 'type_1',

  'created_at': '2020-02-13T19:43:00Z'},

 {'id': 'id_126',

  'type': 'type_3',

  'created_at': '2020-02-13T07:00:00Z'}]

我试图找出发生多少次，以及该列表中最早的时间戳是什么type : type_1created_attype_1

Answer 1

我们可以通过几个步骤来实现这一目标。要查找发生的次数，我们可以将内置过滤器与 itemgetter 结合使用。type_1from operator import itemgetterdef my_filter(item):    return item['type'] == 'type_1'key = itemgetter('created_at')items = sorted(filter(my_filter, data), key=key)print(f"Num records is {len(items)}")print(f"Earliest record is {key(items[0])}")Num records is 2Earliest record is 2020-02-12T17:45:00Z相反，您可以使用生成器理解，然后对生成器进行排序。gen = (item for item in data if item['type'] == 'type_1')items = sorted(gen, key=key)# rest of the steps are the same...

Answer 2

您可以使用列表理解来获取您感兴趣的所有子列表，然后按“created_at”进行排序。l = [{'id': 'id_123',  'type': 'type_1',  'created_at': '2020-02-12T17:45:00Z'}, {'id': 'id_124',  'type': 'type_2',  'created_at': '2020-02-12T18:15:00Z'}, {'id': 'id_125',  'type': 'type_1',  'created_at': '2020-02-13T19:43:00Z'}, {'id': 'id_126',  'type': 'type_3',  'created_at': '2020-02-13T07:00:00Z'}]ll = [x for x in l if x['type'] == 'type_1']ll.sort(key=lambda k: k['created_at'])print(len(ll))print(ll[0]['created_at'])输出：202/12/2020 17:45:00

Answer 3

您可以使用 生成所有type_1s的列表，然后它们使用 with 对值进行相应的排序list_comprehensionsortdatetime.strptimefrom datetime import datetime# Generate a list with only the type_1s' created_at valuestype1s = [val['created_at'] for val in vals if val['type']=="type_1"]# Sort them based on the timestampstype1s.sort(key=lambda date: datetime.strptime(date, "%Y-%m-%dT%H:%M:%SZ"))# Print the lowest valueprint(type1s[0])#'2020-02-12T17:45:00Z'

Answer 4

您可以使用以下函数获取所需的输出：from datetime import datetimedef sol(l):    sum_=0    dict_={}    for x in l:        if x['type']=='type_1':            sum_+=1            dict_[x['id']]=datetime.strptime(x['created_at'], "%Y-%m-%dT%H:%M:%SZ")    date =sorted(dict_.values())[0]    for key,value in dict_.items():        if value== date: id_=key    return sum_,date,id_sol(l)此函数给出类型 ='type_1'的次数，分别给出相应的最小日期及其 ID。希望这有帮助！

Answer 5

这是使用 和 的一种方法。filtermin前任：data = [{'id': 'id_123',  'type': 'type_1',  'created_at': '2020-02-12T17:45:00Z'}, {'id': 'id_124',  'type': 'type_2',  'created_at': '2020-02-12T18:15:00Z'}, {'id': 'id_125',  'type': 'type_1',  'created_at': '2020-02-13T19:43:00Z'}, {'id': 'id_126',  'type': 'type_3',  'created_at': '2020-02-13T07:00:00Z'}]onlytype_1 = list(filter(lambda x: x['type'] == 'type_1', data))print(len(onlytype_1))print(min(onlytype_1, key=lambda x: x['created_at']))艺术temp = {}for i in data:    temp.setdefault(i['type'], []).append(i)print(len(temp['type_1']))print(min(temp['type_1'], key=lambda x: x['created_at']))输出：2{'id': 'id_123', 'type': 'type_1', 'created_at': '2020-02-12T17:45:00Z'}