Python 在每['；-1'；]处拆分嵌套列表

如果我有这样一个嵌套列表：

[['01'], ['02'], ['-1'], ['03'], ['04']]

[[['01'], ['02']], [['03'], ['04']]]

是否有办法在每个

['-1']

处拆分此嵌套列表

所以它看起来像这样：

[['01'], ['02'], ['-1'], ['03'], ['04']]

[[['01'], ['02']], [['03'], ['04']]]

任何形式的帮助都将不胜感激：）

试试这个

lists = [['01'], ['02'], ['-1'], ['03'], ['04'], ['-1'], ['05'], ['-1']]

results = list()
prev_idx = 0
for idx, l in enumerate(lists):
    if l == ['-1']:
        results.append(lists[prev_idx:idx])
        prev_idx = idx+1

if prev_idx <= idx:     # the last group might be [] as shown in this case
    results.append(lists[prev_idx:])

print(results)
# Output
[[['01'], ['02']], [['03'], ['04']], [['05']]]

list=['01']、['02']、['-1']、['03']、['04']、['-1']、['05']、['-1']]
结果=列表（）
上一个idx=0
对于idx，枚举中的l（列表）：
如果l==['-1']：
results.append（列出[prev_idx:idx]）
上一个idx=idx+1
如果上一个idx试试这个

lists = [['01'], ['02'], ['-1'], ['03'], ['04'], ['-1'], ['05'], ['-1']]

results = list()
prev_idx = 0
for idx, l in enumerate(lists):
    if l == ['-1']:
        results.append(lists[prev_idx:idx])
        prev_idx = idx+1

if prev_idx <= idx:     # the last group might be [] as shown in this case
    results.append(lists[prev_idx:])

print(results)
# Output
[[['01'], ['02']], [['03'], ['04']], [['05']]]

list=['01']、['02']、['-1']、['03']、['04']、['-1']、['05']、['-1']]
结果=列表（）
上一个idx=0
对于idx，枚举中的l（列表）：
如果l==['-1']：
results.append（列出[prev_idx:idx]）
上一个idx=idx+1
如果prev_idx一个好的老式循环应该做到这一点：

l = [['01'], ['02'], ['-1'], ['03'], ['04']]

new = []

current = []  # Build a new list here
for i, item in enumerate(l):
    if item != ['-1']:
        current.append(item)
        if i == len(l) - 1:  # If the item is the last in the list
            new.append(current)
    else:
        new.append(current)
        current = []

>>>  [[['01'], ['02']], [['03'], ['04']]]

一个好的老式循环应该可以做到这一点：

l = [['01'], ['02'], ['-1'], ['03'], ['04']]

new = []

current = []  # Build a new list here
for i, item in enumerate(l):
    if item != ['-1']:
        current.append(item)
        if i == len(l) - 1:  # If the item is the last in the list
            new.append(current)
    else:
        new.append(current)
        current = []

>>>  [[['01'], ['02']], [['03'], ['04']]]

似乎是groupby的一个用例

似乎是groupby的一个用例

您可以使用在每次出现分割值时进行分组（此处
['-1']
）
如果不是k
则确保忽略分割值本身

orig = [['01'], ['02'], ['-1'], ['03'], ['04']]
from itertools import groupby
n = [list(g) for k, g in groupby(orig, lambda x: x == ['-1']) if not k]

您可以使用在每次出现拆分值时进行分组（此处
['-1']
）
如果不是k
则确保忽略分割值本身

orig = [['01'], ['02'], ['-1'], ['03'], ['04']]
from itertools import groupby
n = [list(g) for k, g in groupby(orig, lambda x: x == ['-1']) if not k]