根据特定元素拆分具有多个列表的列表答案

【问题标题】：Split list with multiple list according with a specific element根据特定元素拆分具有多个列表的列表
【发布时间】：2022-01-04 18:38:27
【问题描述】：

我有一个由安卓应用生成的坐标列表。我想要的是每次找到元素（0.0,0.0）时自动将列表拆分到另一个列表中，当它找到下一个（0.0,0.0）时，它会将其拆分到另一个列表中，依此类推。元素的数量（0.0,0.0）取决于我运行应用程序的次数，因此在示例中，我运行了应用程序 3 次，但如果我运行了 4 次，我需要将 big_list 拆分为 4 个列表。这样，决定拆分次数的 big_list 的大小和 (0.0,0.0) 元素的数量取决于我在从数据库中导出文件并将其加载到 python 上进行分析之前运行应用程序的次数.

示例如下：

big_list = [**(0.0, 0.0)**, (0.7061503529548645, -0.5579889416694641), (1.412300705909729, -1.1159778833389282), (2.1184511184692383, -1.673966884613037), (2.824601411819458, -2.2319557666778564), **(0.0, 0.0)**, (0.6936703324317932, -0.573429524898529), (1.3873406648635864, -1.146859049797058), (2.0810110569000244, -1.7202885150909424), (2.7677958011627197, -2.3019471168518066), **(0.0, 0.0)**, (0.6973708868026733, -0.5689234137535095), (1.3947417736053467, -1.137846827507019), (2.0921125411987305, -1.7067701816558838), (2.7894835472106934, -2.275693655014038), (3.4868545532226562, -2.8446171283721924), (4.184225559234619, -3.4135406017303467)]

我想要的是从元素 (0.0,0.0) 和以下元素开始的 3 个列表，直到找到下一个 (0.0,0.0) 再次拆分：

list1 = [(0.0, 0.0), (0.7061503529548645, -0.5579889416694641), (1.412300705909729, -1.1159778833389282), (2.1184511184692383, -1.673966884613037), (2.824601411819458, -2.2319557666778564)]

list2= [(0.0, 0.0), (0.6936703324317932, -0.573429524898529), (1.3873406648635864, -1.146859049797058), (2.0810110569000244, -1.7202885150909424), (2.7677958011627197, -2.3019471168518066)]

list3 = [(0.0, 0.0), (0.6973708868026733, -0.5689234137535095), (1.3947417736053467, -1.137846827507019), (2.0921125411987305, -1.7067701816558838), (2.7894835472106934, -2.275693655014038), (3.4868545532226562, -2.8446171283721924), (4.184225559234619, -3.4135406017303467)]

【问题讨论】：

标签： python arrays list

【解决方案1】：

这个方法会给你一个你需要的答案

def split_list(input_list:list) -> list:
    start_ind=0
    result = []
    for ind, x in enumerate(input_list):
        if ( x == (0.0, 0.0) or ind == len(input_list)-1 ) and ind != 0:
            result.append(input_list[start_ind:ind])
            start_ind=ind
    return result

【讨论】：

【解决方案2】：

我同意@conol 的回答，但您可以添加这个额外的步骤并获得您想要的结果。

基本上，我们在您编写的分隔符上拆分列表，然后我们使用代码更新语言环境，以便您在运行时获得动态创建的变量

def split_list(input_list:list) -> list:
    start_ind=0
    dict = {}
    varname= 'list'
    counter = 0
    for ind, x in enumerate(input_list):
        if ( x == (0.0, 0.0) or ind == len(input_list)-1 ) and ind != 0:
            dict[varname+str(counter)] = input_list[start_ind:ind]
            start_ind=ind
            counter+=1
    return dict

locals().update(split_list(a))
print(list0,list1,list2)

【讨论】：

【解决方案3】：

例如，可以定义一个函数，循环遍历大列表的条目并创建子列表（带有类型注释）：

from pprint import pprint
from typing import List, Optional, Tuple


def split_big_list(
        big_list: List[Tuple[float, float]],
) -> List[List[Tuple[float, float]]]:
    result: List[List[Tuple[float, float]]] = []
    current_list: Optional[List[Tuple[float, float]]] = None
    for entry in big_list:
        if entry == (0.0, 0.0):
            current_list = [entry]
            result.append(current_list)
        elif current_list is None:
            # ignore entries before first tuple with zeros
            continue
        else:
            current_list.append(entry)

    return result


BIG_LIST = [(0.0, 0.0), (0.7061503529548645, -0.5579889416694641), (1.412300705909729, -1.1159778833389282), (2.1184511184692383, -1.673966884613037), (2.824601411819458, -2.2319557666778564), (0.0, 0.0), (0.6936703324317932, -0.573429524898529), (1.3873406648635864, -1.146859049797058), (2.0810110569000244, -1.7202885150909424), (2.7677958011627197, -2.3019471168518066), (0.0, 0.0), (0.6973708868026733, -0.5689234137535095), (1.3947417736053467, -1.137846827507019), (2.0921125411987305, -1.7067701816558838), (2.7894835472106934, -2.275693655014038), (3.4868545532226562, -2.8446171283721924), (4.184225559234619, -3.4135406017303467)]
lists = split_big_list(BIG_LIST)
pprint(lists)

结果：

[[(0.0, 0.0),
  (0.7061503529548645, -0.5579889416694641),
  (1.412300705909729, -1.1159778833389282),
  (2.1184511184692383, -1.673966884613037),
  (2.824601411819458, -2.2319557666778564)],
 [(0.0, 0.0),
  (0.6936703324317932, -0.573429524898529),
  (1.3873406648635864, -1.146859049797058),
  (2.0810110569000244, -1.7202885150909424),
  (2.7677958011627197, -2.3019471168518066)],
 [(0.0, 0.0),
  (0.6973708868026733, -0.5689234137535095),
  (1.3947417736053467, -1.137846827507019),
  (2.0921125411987305, -1.7067701816558838),
  (2.7894835472106934, -2.275693655014038),
  (3.4868545532226562, -2.8446171283721924),
  (4.184225559234619, -3.4135406017303467)]]

【讨论】：

【解决方案4】：

使用 enumerate 查找 (0,0) 项的索引。然后 zip() 将它们与 teir 后继者配对以形成子范围：

starts   = [i for i,v in enumerate(big_list) if v == (0,0)]
sublists = [big_list[s:e] for s,e in zip(starts,starts[1:]+[None])]

print(sublists)
[[(0.0, 0.0), (0.7061503529548645, -0.5579889416694641), (1.412300705909729, -1.1159778833389282), (2.1184511184692383, -1.673966884613037), (2.824601411819458, -2.2319557666778564)], 
 [(0.0, 0.0), (0.6936703324317932, -0.573429524898529), (1.3873406648635864, -1.146859049797058), (2.0810110569000244, -1.7202885150909424), (2.7677958011627197, -2.3019471168518066)], 
 [(0.0, 0.0), (0.6973708868026733, -0.5689234137535095), (1.3947417736053467, -1.137846827507019), (2.0921125411987305, -1.7067701816558838), (2.7894835472106934, -2.275693655014038), (3.4868545532226562, -2.8446171283721924), (4.184225559234619, -3.4135406017303467)]]


# if you want them in variables 
# (but then you need to know how many subslits there are ahead of time):

lst1,lst2,lst3 = sublists

【讨论】：