繁体   English   中英

Python3.6:根据列表的给定元素将列表分离为子列表

[英]Python3.6: Separation of a list into sublists depending on given elements of the list

我有一些数据,如下所示。 我现在尝试了很长一段时间是将这个列表分成子列表,这样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为5个相应的列表。

这个问题似乎并不太复杂,但我现在尝试了一段时间,由于某些原因,我不能把它包裹起来。

我很感激你们的任何解决方案。 当然原始数据更大。

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']]}

要么:

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]

解释: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dict键不能重复的事实。 实际上,您可以从输入中获得一组唯一日期。

然后,对于每个输入字符串,将“有效负载”添加到给定密钥下的列表中。 同样,键将定义附加字符串的列表。

完成后, d是您想要的数据结构。

这与上面的ilia的解决方案非常相似。 这是没有列表理解的,最后输出是列表而不是字典。

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']

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM