基于子字符串拆分列表

Question

我有以下列表:

['1(Reg)', '100', '103', '102', '100', '2(Reg)', '98', '101', '100', '3(Reg)', '96', '99', '98', '4(Reg)', '100', '100', '100', '100', '5(Reg)', '98', '99', '99', '100', '6(Reg)', '99.47', '99.86', '99.67', '100']

我想将此列表拆分为多个列表,以便每个子列表都有子串"(Reg)"出现一次:

[['1(Reg)', '100', '103', '102', '100'],
['2(Reg)', '98', '101', '100'],
['3(Reg)', '96', '99', '98'],
['4(Reg)', '100', '100', '100', '100'],
['5(Reg)', '98', '99', '99', '100'],
['6(Reg)', '99.47', '99.86', '99.67', '100']]

我已尝试使用分隔符加入列表并将其拆分为(Reg),但这不起作用.如何将列表拆分为上面的嵌套列表？

Answer 1

WVO答案略有不同(优化)版本:

splitted = []

for item in l:
    if '(Reg)' in item:
        splitted.append([])
    splitted[-1].append(item)

#[['1(Reg)', '100', '103', '102', '100'], ['2(Reg)', '98', '101', '100'], 
# ['3(Reg)', '96', '99', '98'], ['4(Reg)', '100', '100', '100', '100'], 
# ['5(Reg)', '98', '99', '99', '100'], 
# ['6(Reg)', '99.47', '99.86', '99.67', '100']]

Answer 2

您可以使用itertools.groupby正则表达式:

import itertools
import re
s = ['1(Reg)', '100', '103', '102', '100', '2(Reg)', '98', '101', '100', '3(Reg)', '96', '99', '98', '4(Reg)', '100', '100', '100', '100', '5(Reg)', '98', '99', '99', '100', '6(Reg)', '99.47', '99.86', '99.67', '100']
new_data = [list(b) for _, b in itertools.groupby(s, key=lambda x:bool(re.findall('\d+\(', x)))]
final_data = [new_data[i]+new_data[i+1] for i in range(0, len(new_data), 2)]

输出:

[['1(Reg)', '100', '103', '102', '100'], 
 ['2(Reg)', '98', '101', '100'], 
 ['3(Reg)', '96', '99', '98'], 
 ['4(Reg)', '100', '100', '100', '100'], 
 ['5(Reg)', '98', '99', '99', '100'], 
 ['6(Reg)', '99.47', '99.86', '99.67', '100']]

Answer 3

这是一种方式,但不一定是最佳方式:

from itertools import zip_longest

lst = ['1(Reg)', '100', '103', '102', '100', '2(Reg)', '98', '101', '100',
       '3(Reg)', '96', '99', '98', '4(Reg)', '100', '100', '100', '100',
       '5(Reg)', '98', '99', '99', '100', '6(Reg)', '99.47', '99.86', '99.67', '100']

indices = [i for i, j in enumerate(lst) if '(Reg)' in j]
lst_new = [lst[i:j] for i, j in zip_longest(indices, indices[1:])]

# [['1(Reg)', '100', '103', '102', '100'],
#  ['2(Reg)', '98', '101', '100'],
#  ['3(Reg)', '96', '99', '98'],
#  ['4(Reg)', '100', '100', '100', '100'],
#  ['5(Reg)', '98', '99', '99', '100'],
#  ['6(Reg)', '99.47', '99.86', '99.67', '100']]