从列表中排序和删除Python

我有一个dict列表，列表中的每个dict都有一个字符串格式的时间戳和一个键。一个特定的键可以在列表中重复多次。我只想保留带有最新时间戳的密钥的dict，并从列表中删除所有其他dict。我实现soluion的一种方法是使用另一个变量，在所有键上循环，并与现有的一个进行比较

有没有更好的方法来解决这个问题，使用列表理解或itertools或任何其他方式

下面是示例输入数据

data = [
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:21.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:22.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:23.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:19.762278'},
    {'key': 'key3', 'timestamp': '2017-08-03T10:24:25.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:11.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:45.762278'},
    {'key': 'key4', 'timestamp': '2017-08-03T10:24:39.762278'}
]

以下是预期的输出

data = [
    {'key': 'key3', 'timestamp': '2017-08-03T10:24:25.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:22.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:45.762278'},
    {'key': 'key4', 'timestamp': '2017-08-03T10:24:39.762278'}
]

我在python中的实现如下

from dateutil.parser import parse
def sort_and_eliminate(data):
    processed_data = {}
    for cur_item in data:
        key = cur_item.get('key')
        if key not in processed_data:
            processed_data[key] = cur_item
        else:
            ex_item = processed_data.get(key)
            ex_ts = parse(ex_item.get("timestamp"))
            cur_ts = parse(cur_item.get("timestamp"))
            if cur_ts > ex_ts:
                processed_data[key] = cur_item
    return processed_data.values()

有没有更好的方法来解决这个问题，使用列表理解或itertools或任何其他方法

这里有一种方法

根据键和时间戳对字典进行排序

x=sorted(data, key=lambda k: (k['key'],k['timestamp']), reverse=True)
print(x)

[{'key': 'key4', 'timestamp': '2017-08-03T10:24:39.762278'}, 
 {'key': 'key3', 'timestamp': '2017-08-03T10:24:25.762278'}, 
 {'key': 'key2', 'timestamp': '2017-08-03T10:24:22.762278'}, 
 {'key': 'key2', 'timestamp': '2017-08-03T10:24:19.762278'}, 
 {'key': 'key2', 'timestamp': '2017-08-03T10:24:11.762278'}, 
 {'key': 'key1', 'timestamp': '2017-08-03T10:24:45.762278'},  
 {'key': 'key1', 'timestamp': '2017-08-03T10:24:23.762278'}, 
 {'key': 'key1', 'timestamp': '2017-08-03T10:24:21.762278'}]

创建一个新列表并仅插入该键的第一个匹配项

new_list=[]
temp=None
for values in x:
  if values['key']!=temp:
    new_list.append(values)
    temp=values['key']
print(new_list)

[{'key': 'key4', 'timestamp': '2017-08-03T10:24:39.762278'}, 
 {'key': 'key3', 'timestamp': '2017-08-03T10:24:25.762278'}, 
 {'key': 'key2', 'timestamp': '2017-08-03T10:24:22.762278'}, 
 {'key': 'key1', 'timestamp': '2017-08-03T10:24:45.762278'}]

希望这有帮助

from datetime import datetime
from operator import itemgetter
from itertools import groupby
from dateutil.parser import parse

expected = [
    {'key': 'key3', 'timestamp': '2017-08-03T10:24:25.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:22.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:45.762278'},
    {'key': 'key4', 'timestamp': '2017-08-03T10:24:39.762278'}
]

data = [
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:21.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:22.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:23.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:19.762278'},
    {'key': 'key3', 'timestamp': '2017-08-03T10:24:25.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:11.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:45.762278'},
    {'key': 'key4', 'timestamp': '2017-08-03T10:24:39.762278'}
]


# alt way without dateutil
def dtconv(s):
    return datetime.strptime(s, "%Y-%m-%dT%H:%M:%S.%f")

ds = sorted(data, key=lambda x: (x['key'], parse(x['timestamp'])), reverse=True)

result = []
for grouper, group in groupby(ds, key=itemgetter('key')):
    result.append(next(group))

print("result:")
for r in result:
    print(r)

print("expected")
for e in expected:
    print(e)

# demonstrate it's equal to expected value
print(sorted(result, key=itemgetter('key')) == sorted(expected, key=itemgetter('key')))

---

尝试按键和日期戳对列表进行排序。然后，您可以执行一个

groupby

并获取第一个元素，这将是您想要保留的元素。

按时间戳字符串的相反顺序对数据进行排序，然后每个唯一键的第一个外观将是您想要保留的

from dateutil.parser import parse

data = [
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:21.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:22.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:23.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:19.762278'},
    {'key': 'key3', 'timestamp': '2017-08-03T10:24:25.762278'},
    {'key': 'key2', 'timestamp': '2017-08-03T10:24:11.762278'},
    {'key': 'key1', 'timestamp': '2017-08-03T10:24:45.762278'},
    {'key': 'key4', 'timestamp': '2017-08-03T10:24:39.762278'}]


all_keys = [k['key'] for k in data]

all_keys_unique = set(all_keys)

new_dict = {}

for k in all_keys_unique:

    #find all values for that key and parse them
    values_of_key = [j['timestamp'] for j in data if k == j['key']]

    parsed_values = [parse(k2) for k2 in values_of_key]

    #use max to find latest time step, works on datetimes
    #and add to dictionary
    new_dict[k] = max(parsed_values)

print(new_dict)

data = sorted(data, key=lambda x: x["timestamp"], reverse=True) 
used_keys, cleaned_data = [ ], [ ]
for item in data:
    if not item['key'] in used_keys:
        # if a key that we encounter in the list isn't used yet,
        # add its corresponding item to cleaned_data and add it to
        # used_keys so we know not to use it again.
        cleaned_data.append(item)
        used_keys.append(item['key'])

---

只是创建另一个dict，将key值作为key，比较时间戳并插入latest timestamp作为value。

刚刚注意到有人发布的基本上就是这个。哦，好的，修好了。这个问题没有提到任何关于保留剩余键的原始顺序的内容，所以我假设按时间戳排序是可以的。是的，按时间戳排序是可以的。不需要订购请解释代码解决问题的原因和方式，而不是只发布代码答案。与中提供的实现相比，这需要更多的时间question@akashdeep即使这是真的。它更清楚，更容易推理。OP要求更好的解决方案，但这并不一定意味着它必须更快。几乎没有理由投票，但这是你的特权。最后两种是为了演示目的。我希望你没有把这些包括在你的时间里？