列表Python中的排序和删除
列表Python中的排序和删除
提问于 2017-08-03 05:13:32
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dict，并从列表中删除/删除所有其他dict。我实现解决方案的一种方法是使用另一个变量，对所有键进行循环，并将其与现有的键进行比较。
是否有更好的方法来解决这个问题，使用列表理解或迭代工具或其他方法？
这是示例输入数据
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()是否有更好的方法来解决这个问题，使用列表理解或迭代工具或其他方法？
回答 5
Stack Overflow用户
回答已采纳
发布于 2017-08-03 05:38:06
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')))尝试按键和日期标记对列表进行排序。然后您可以执行一个groupby并接受第一个元素，这将是您想要保留的。
Stack Overflow用户
发布于 2017-08-03 05:35:29
这是一种方法。
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'}]希望这能有所帮助！
Stack Overflow用户
发布于 2017-08-03 05:48:49
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)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/45475343
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