我有一些数据，如下所示。 我尝试了很长一段时间，就是将此列表分成子列表，以便每个子列表代表第一行中给出的一个日期。我们在示例中有 5 个不同的日子，我希望将原始列表分成 5 个单独的列表。
这个问题似乎不太复杂，但是我已经尝试了一段时间，由于某种原因，我无法解决它。
我将不胜感激你们的任何解决方案。当然，原始数据要大得多。
listofstrings=[
"17.02.2018 14:30:24 00000000 23,7 23,9 -2,0 1,1",
"17.02.2018 15:00:21 00000000 23,7 23,8 -4,0 1,1",
"19.02.2018 18:30:24 00000000 23,6 23,7 -3,0 1,1",
"19.02.2018 19:00:21 00000000 23,6 23,6 -7,0 1,1",
"19.02.2018 19:30:22 00000000 23,5 23,5 -5,0 1,1",
"20.02.2018 05:30:21 00000000 23,5 23,8 -3,0 1,1",
"20.02.2018 06:00:21 00000000 23,5 23,8 1,0 1,1",
"20.02.2018 16:00:22 00000000 23,6 23,8 -4,0 1,1",
"21.02.2018 05:00:22 00000000 23,6 23,7 0,0 1,1",
"21.02.2018 05:30:23 00000000 23,6 23,8 -6,0 1,1",
"22.02.2018 07:30:23 00000000 23,6 23,8 -6,0 1,1",
"22.02.2018 08:00:21 00000000 23,6 23,9 -3,0 1,1",
"22.02.2018 13:30:25 00000000 23,6 23,8 -3,0 1,1"]
listoflists=[]
locallist=[]
for i in range(0, len(listofstrings)):
current_string=listofstrings[i]
current_date=current_string.split()[0]
if not i==0:
recent_string=listofstrings[i-1]
recent_date=recent_string.split()[0]
if current_date==recent_date:
locallist.append(current_string)
locallist.append(recent_string)
listoflists.append(locallist)
locallist.clear()
预期的输出将是这样的：
list1=["17.02.2018 14:30:24 00000000 23,7 23,9 -2,0 1,1",
"17.02.2018 15:00:21 00000000 23,7 23,8 -4,0 1,1"]
list2=["19.02.2018 18:30:24 00000000 23,6 23,7 -3,0 1,1",
"19.02.2018 19:00:21 00000000 23,6 23,6 -7,0 1,1",
"19.02.2018 19:30:22 00000000 23,5 23,5 -5,0 1,1",]
....
看起来你需要itertools.groupby
演示：
from itertools import groupby
listofstrings=[
"17.02.2018 14:30:24 00000000 23,7 23,9 -2,0 1,1",
"17.02.2018 15:00:21 00000000 23,7 23,8 -4,0 1,1",
"19.02.2018 18:30:24 00000000 23,6 23,7 -3,0 1,1",
"19.02.2018 19:00:21 00000000 23,6 23,6 -7,0 1,1",
"19.02.2018 19:30:22 00000000 23,5 23,5 -5,0 1,1",
"20.02.2018 05:30:21 00000000 23,5 23,8 -3,0 1,1",
"20.02.2018 06:00:21 00000000 23,5 23,8 1,0 1,1",
"20.02.2018 16:00:22 00000000 23,6 23,8 -4,0 1,1",
"21.02.2018 05:00:22 00000000 23,6 23,7 0,0 1,1",
"21.02.2018 05:30:23 00000000 23,6 23,8 -6,0 1,1",
"22.02.2018 07:30:23 00000000 23,6 23,8 -6,0 1,1",
"22.02.2018 08:00:21 00000000 23,6 23,9 -3,0 1,1",
"22.02.2018 13:30:25 00000000 23,6 23,8 -3,0 1,1"]
listofstrings = [i.split() for i in listofstrings]
result = dict((k, list(v)) for k, v in groupby(listofstrings, lambda x: x[0]))
print(result)
输出：
{'17.02.2018': [['17.02.2018', '14:30:24', '00000000', '23,7', '23,9', '-2,0', '1,1'], ['17.02.2018', '15:00:21', '00000000', '23,7', '23,8', '-4,0', '1,1']],
'19.02.2018': [['19.02.2018', '18:30:24', '00000000', '23,6', '23,7', '-3,0', '1,1'], ['19.02.2018', '19:00:21', '00000000', '23,6', '23,6', '-7,0', '1,1'], ['19.02.2018', '19:30:22', '00000000', '23,5', '23,5', '-5,0', '1,1']],
'22.02.2018': [['22.02.2018', '07:30:23', '00000000', '23,6', '23,8', '-6,0', '1,1'], ['22.02.2018', '08:00:21', '00000000', '23,6', '23,9', '-3,0', '1,1'], ['22.02.2018', '13:30:25', '00000000', '23,6', '23,8', '-3,0', '1,1']],
'21.02.2018': [['21.02.2018', '05:00:22', '00000000', '23,6', '23,7', '0,0', '1,1'], ['21.02.2018', '05:30:23', '00000000', '23,6', '23,8', '-6,0', '1,1']],
'20.02.2018': [['20.02.2018', '05:30:21', '00000000', '23,5', '23,8', '-3,0', '1,1'], ['20.02.2018', '06:00:21', '00000000', '23,5', '23,8', '1,0', '1,1'], ['20.02.2018', '16:00:22', '00000000', '23,6', '23,8', '-4,0', '1,1']]}
或：
result = dict((k, list(v)) for k, v in groupby(listofstrings, lambda x: x[:10]))
输出：
{'17.02.2018': ['17.02.2018 14:30:24 00000000 23,7 23,9 -2,0 1,1', '17.02.2018 15:00:21 00000000 23,7 23,8 -4,0 1,1'],
'19.02.2018': ['19.02.2018 18:30:24 00000000 23,6 23,7 -3,0 1,1', '19.02.2018 19:00:21 00000000 23,6 23,6 -7,0 1,1', '19.02.2018 19:30:22 00000000 23,5 23,5 -5,0 1,1'],
'22.02.2018': ['22.02.2018 07:30:23 00000000 23,6 23,8 -6,0 1,1', '22.02.2018 08:00:21 00000000 23,6 23,9 -3,0 1,1', '22.02.2018 13:30:25 00000000 23,6 23,8 -3,0 1,1'],
'21.02.2018': ['21.02.2018 05:00:22 00000000 23,6 23,7 0,0 1,1', '21.02.2018 05:30:23 00000000 23,6 23,8 -6,0 1,1'],
'20.02.2018': ['20.02.2018 05:30:21 00000000 23,5 23,8 -3,0 1,1', '20.02.2018 06:00:21 00000000 23,5 23,8 1,0 1,1', '20.02.2018 16:00:22 00000000 23,6 23,8 -4,0 1,1']}
这是一个不需要导入模块的解决方案。
l = listofstrings # an alias for conciseness
d={st[:10]:[] for st in l}
for st in l:
d[st[:10]] += [st]
说明：首先在字典 D 中创建一个空列表，其中 key 是每个输入字符串的前 10 个字符，即日期。 这利用了字典键无法复制的事实。实际上，您可以从输入中获得唯一日期的集合。
然后，对于每个输入字符串，将"payload"添加到给定键下的列表中。同样，键将定义字符串附加到哪个列表。
完成后，d是您想要的数据结构。
这与上面的 ilia 解决方案非常相似。这是没有列表理解的，最后输出是列表列表而不是字典。
listofstrings = [
"17.02.2018 14:30:24 00000000 23,7 23,9 -2,0 1,1",
"17.02.2018 15:00:21 00000000 23,7 23,8 -4,0 1,1",
"19.02.2018 18:30:24 00000000 23,6 23,7 -3,0 1,1",
"19.02.2018 19:00:21 00000000 23,6 23,6 -7,0 1,1",
"19.02.2018 19:30:22 00000000 23,5 23,5 -5,0 1,1",
"20.02.2018 05:30:21 00000000 23,5 23,8 -3,0 1,1",
"20.02.2018 06:00:21 00000000 23,5 23,8 1,0 1,1",
"20.02.2018 16:00:22 00000000 23,6 23,8 -4,0 1,1",
"21.02.2018 05:00:22 00000000 23,6 23,7 0,0 1,1",
"21.02.2018 05:30:23 00000000 23,6 23,8 -6,0 1,1",
"22.02.2018 07:30:23 00000000 23,6 23,8 -6,0 1,1",
"22.02.2018 08:00:21 00000000 23,6 23,9 -3,0 1,1",
"22.02.2018 13:30:25 00000000 23,6 23,8 -3,0 1,1"]
_list = {}
for d in listofstrings:
if d[:10] not in _list:
_list[d[: 10]] = [d]
else:
_list[d[:10]].append(d)
_list_of_lists = []
for k, v in _list.items():
_list_of_lists.append(v)
print(*_list_of_lists, sep="n")
输出：
['17.02.2018 14:30:24 00000000 23,7 23,9 -2,0 1,1', '17.02.2018 15:00:21 00000000 23,7 23,8 -4,0 1,1']
['19.02.2018 18:30:24 00000000 23,6 23,7 -3,0 1,1', '19.02.2018 19:00:21 00000000 23,6 23,6 -7,0 1,1', '19.02.2018 19:30:22 00000000 23,5 23,5 -5,0 1,1']
['20.02.2018 05:30:21 00000000 23,5 23,8 -3,0 1,1', '20.02.2018 06:00:21 00000000 23,5 23,8 1,0 1,1', '20.02.2018 16:00:22 00000000 23,6 23,8 -4,0 1,1']
['21.02.2018 05:00:22 00000000 23,6 23,7 0,0 1,1', '21.02.2018 05:30:23 00000000 23,6 23,8 -6,0 1,1']
['22.02.2018 07:30:23 00000000 23,6 23,8 -6,0 1,1', '22.02.2018 08:00:21 00000000 23,6 23,9 -3,0 1,1', '22.02.2018 13:30:25 00000000 23,6 23,8 -3,0 1,1']