在python中将列表分解为列表？

我有以下

字符串列表

：

data = ["1","0","1","<>","0","1","0","<>","1","0","1"]

---

例如，一个列表列表，其中每个列表由

分隔，结果字符串转换为整数。

您可以使用

itertools.groupby

在每个

处进行有效分区，然后在键不为

True

的结果组（例如：它是

）处转换为

int

，例如：

from itertools import groupby

data = ["1","0","1","<>","0","1","0","<>","1","0","1"]
new_date = [[int(i) for i in g] for k, g in groupby(data, '<>'.__ne__) if k]
# [[1, 0, 1], [0, 1, 0], [1, 0, 1]]

从itertools导入groupby
数据=[“1”、“0”、“1”、“0”、“1”、“0”、“0”、“1”、“0”、“1”、“0”、“1”]
新日期=[[int（i）for i in g]for k，g in groupby（数据，”。[u_ne___）如果k]
# [[1, 0, 1], [0, 1, 0], [1, 0, 1]]

您可以尝试将列表理解与连接和拆分结合使用：

your_list = [[int(j) for j in i.split()] for i in ' '.join(data).split('<>')]

your_list=[[int（j）for j in i.split（）]for i in'。加入（数据）.split（“”）]

我建议您使用更具功能性的解决方案：

data = ["1","0","1","<>","0","1","0","<>","1","0","1"]
reduce(lambda a, b: a[:-1] + [a[-1] + [int(b)]] if b != '<>' else a + [[]], data, [[]])

data=[“1”、“0”、“1”、“0”、“1”、“0”、“0”、“1”、“0”、“1”、“0”、“1”]
减少（λa，b:a[：-1]+[a[-1]+[int（b）]，如果b！=''，否则a+[[]]，数据，[[]]）

您也可以试试这个

l = ["1","0","1","<>","0","1","0","<>","1","0","1"]
m = []
x = []
j = []
for i in l:
    if  i == '<>':
        m.append(x)
        x = []
        j = []
    else:
        x.append(int(i))
        j.append(int(i))
m.append(j)
print(m)

---

你能分享一下你为解决这个问题所做的努力吗？我不认为这是一个重复，因为这些块的大小不一定是均匀的。@Andrewmage:的确，我的错误，重新打开了。糟糕的格式让我很失望。我测试了你的代码，但被以下错误阻止了：int（）的文本无效，以10为基数：'@mahya.khazae:你的数据中有分数？@mahya.khazaee，所以你有一个

，当列表中有一个数字超过一位数时，它会失败。它如何区分
[“1”，“0”，“，”，…]
来自
[“10”，“，…]
？我现在修复了它，但问题似乎只与1和0有关。这可能有效，但男孩，这是一种丑陋的方法，而且很脆弱。@MartijnPieters它怎么脆弱？它工作起来没有任何失败的可能性。
l = ["1","0","1","<>","0","1","0","<>","1","0","1"]
m = []
x = []
j = []
for i in l:
    if  i == '<>':
        m.append(x)
        x = []
        j = []
    else:
        x.append(int(i))
        j.append(int(i))
m.append(j)
print(m)

[[1, 0, 1], [0, 1, 0], [1, 0, 1]]