Python中的拆分整数列表

我有一个整数列表。我想将主列表拆分为多个非零整数的子列表

例如：

main_lists = [0,0,0,0,0,0,123,432,5,54,0,654,645,34,23,12,0,0,0,0,0,0,123,1,312,312,132,3,123,0,0,0,0,0,0,0]

output = 

[123,432,5,54,0,654,645,34,23,12]
[123,1,312,312,132,3,123]

情况是,， 如果在两个非零值之间发现一个或两个（小于5）零，则应在输出列表中包含这些零。如果发现超过5个连续零，则应将列表从该位置拆分

main_lists = [0,0,0,0,0,0,123,432,5,54,0,654,645,34,23,12,0,0,0,0,0,0,123,1,312,312,132,3,123,0,0,0,0,0,0,0]
zero_count = 0
non_zero_temp = []
for i in main_lists:
    if i == 0:
        zero_count = zero_count + 1
    else:
        if zero_count < 5:
            non_zero_temp.append(i)
            zero_count = 0

main_list=[0,0,0,0,0123432,5,54,0654645,34,23,12,0,0,0,0,0,0123,1312312132,3123,0,0,0,0,0]
零计数=0
非零温度=[]
对于主列表中的i：
如果i==0：
零计数=零计数+1
其他：
如果零计数小于5：
非零温度附加（i）
零计数=0

---

提前感谢，

我将发布一个备选方案：

from itertools import groupby

main_lists = [
    0,0,0,0,0,0,123,432,5,54,0,654,645,34,23,12,0,
    0,0,0,0,0,123,1,312,312,132,3,123,0,0,0,0,0,0,0
]

# group numbers into contiguous lists (by 0 or not-0)
is_zero = lambda n: not n
groups = (list(nums) for zero,nums in groupby(main_lists, key=is_zero))

# group lists into contiguous chunks (by to-keep or to-discard)
is_keeper = lambda lst: bool(lst[0]) or len(lst) < 5
chunks = (chunk for keep,chunk in groupby(grouped, key=is_keeper) if keep)

# reassemble chunks
final = [[i for lst in chunk for i in lst] for chunk in chunks]

output = []
non_zero_temp = []
length = len(main_lists)
i = 0
while i < length:
    zeros = []
    while i < length and main_lists[i] == 0:
        i += 1
        zeros.append(0)
    if len(zeros) != 0 and len(zeros) < 5:
        non_zero_temp += zeros
    elif len(zeros) > 5:
        if len(non_zero_temp) > 0 and i < length:
            output.append(non_zero_temp)
            non_zero_temp = []
    else:
        non_zero_temp.append(main_lists[i])
        i += 1
if len(non_zero_temp) > 0:
    output.append(non_zero_temp)
print(output)

[[123, 432, 5, 54, 0, 654, 645, 34, 23, 12], [123, 1, 312, 312, 132, 3, 123]]

output=[]
非零温度=[]
长度=长度（主列表）
i=0
而我<长度：
零=[]
而i5：
如果len（非零温度）>0且i0：
输出.附加（非零温度）
打印（输出）
[[123, 432, 5, 54, 0, 654, 645, 34, 23, 12], [123, 1, 312, 312, 132, 3, 123]]

main_list=[0,0,0,0,0,0123432,5,54,0654645,34,23,12,0,0,0,0,0,0123,1312312132,3123,0,0,0,0,0,0,0,0]zero_count=0非zero_temp=[]对于主_list中的i:if i=0:zero_count=0+1其他：if zero_count<5:non_temp.append（i）zero_cout=0@Ffisegydd完成！这和？@Ffisegydd不完全一样，我在那个问题上出错了……我能做些什么……删除这个问题或将其标记为不相关？？对不起…“我在那个问题上错了。”你这是什么意思？如果可能的话，尝试通过编辑来更正问题。您可以使用

bool

作为

groupby

中的键，而不是lambda。@Blckknght:是的，我使用的是

运算符。不是

，但我认为这“更明显”。

output = []
non_zero_temp = []
length = len(main_lists)
i = 0
while i < length:
    zeros = []
    while i < length and main_lists[i] == 0:
        i += 1
        zeros.append(0)
    if len(zeros) != 0 and len(zeros) < 5:
        non_zero_temp += zeros
    elif len(zeros) > 5:
        if len(non_zero_temp) > 0 and i < length:
            output.append(non_zero_temp)
            non_zero_temp = []
    else:
        non_zero_temp.append(main_lists[i])
        i += 1
if len(non_zero_temp) > 0:
    output.append(non_zero_temp)
print(output)

[[123, 432, 5, 54, 0, 654, 645, 34, 23, 12], [123, 1, 312, 312, 132, 3, 123]]