Question

每当出现特定字符或一组字符时，我都会尝试将自己拥有的列表拆分为单个列表。

例如

Main_list = [ 'abcd 1233','cdgfh3738','hryg21','**L**','gdyrhr657','abc31637','**R**','7473hrtfgf'...]

我想打破此列表并将其保存到子列表中，只要遇到“ L”或“ R”

所需结果：

sublist_1 = ['abcd 1233','cdgfh3738','hryg21']
sublist_2 = ['gdyrhr657','abc31637']
sublist 3 = ['7473hrtfgf'...]

有内置的功能或快速的方法吗？

编辑：我不希望分隔符出现在列表中

Answer 1

使用字典存储可变数量的变量。

在这种情况下，您可以使用itertools.groupby来有效地分隔列表：

L = ['abcd 1233','cdgfh3738','hryg21','**L**',
     'gdyrhr657','abc31637','**R**','7473hrtfgf']

from itertools import groupby

# define separator keys
def split_condition(x):
    return x in {'**L**', '**R**'}

# define groupby object
grouper = groupby(L, key=split_condition)

# convert to dictionary via enumerate
res = dict(enumerate((list(j) for i, j in grouper if not i), 1))

print(res)

{1: ['abcd 1233', 'cdgfh3738', 'hryg21'],
 2: ['gdyrhr657', 'abc31637'],
 3: ['7473hrtfgf']}

Answer 2

考虑使用库中许多有用的工具之一，即more_itertools.split_at：

给出

window.addEventListener("beforeprint", beforePrint);
window.addEventListener("afterprint", afterPrint);
if (window.matchMedia) {
    var mediaQueryList = window.matchMedia('print');
    mediaQueryList.addListener(function(mql) {
        if (mql.matches) {
            beforePrint();
        } else {
            afterPrint();
        }
    });
}

function beforePrint() {
    //Resize for printing
}

function afterPrint() {
    //Resize again for after printing
}

代码

import more_itertools as mit


lst = [
    "abcd 1233", "cdgfh3738", "hryg21", "**L**",
    "gdyrhr657", "abc31637", "**R**", 
    "7473hrtfgf"
]

演示

result = list(mit.split_at(lst, pred=lambda x: set(x) & {"L", "R"}))

详细信息

sublist_1, sublist_2, sublist_3 = result sublist_1 # ['abcd 1233', 'cdgfh3738', 'hryg21'] sublist_2 # ['gdyrhr657', 'abc31637'] sublist_3 # ['7473hrtfgf']函数在满足特殊条件的位置拆分可迭代对象。条件函数（谓词）恰好是more_itertools.split_at函数，它等效于以下常规函数并可以替换为以下常规函数：

lambda

只要字符串def pred(x): a = set(x) b = {"L", "R"} return a.intersection(b)的字符与x或L相交，谓词就返回R，并在该位置进行拆分。

通过> pip install more_itertools在命令行中安装此软件包。

Answer 3

@Polyhedronic ，您也可以尝试一下。

>>> import re
>>> Main_list = [ 'abcd 1233','cdgfh3738','hryg21','**L**','gdyrhr657','abc31637','**R**','7473hrtfgf']
>>>
>>> s = ','.join(Main_list)
>>> s
'abcd 1233,cdgfh3738,hryg21,**L**,gdyrhr657,abc31637,**R**,7473hrtfgf'
>>>
>>> items = re.split('\*\*R\*\*|\*\*L\*\*', s)
>>> items
['abcd 1233,cdgfh3738,hryg21,', ',gdyrhr657,abc31637,', ',7473hrtfgf']
>>>
>>> output = [[a for a in item.split(',') if a] for item in items]
>>> output
[['abcd 1233', 'cdgfh3738', 'hryg21'], ['gdyrhr657', 'abc31637'], ['7473hrtfgf']]
>>>
>>> sublist_1 = output[0]
>>> sublist_2 = output[1]
>>> sublist_3 = output[2]
>>>
>>> sublist_1
['abcd 1233', 'cdgfh3738', 'hryg21']
>>>
>>> sublist_2
['gdyrhr657', 'abc31637']
>>>
>>> sublist_3
['7473hrtfgf']
>>>

Python：根据元素的子集将列表分为多个列表

3 个答案: