Question

如果我有一个形状如下的python列表：

T = [
['07,07,2012 22:10', ['people','drama','melody','bun']],
['08,07,2012 21:04', ['queen','group']],
['08,07,2012 21:23', ['printing','market','shopping']],
['08,07,2012 21:04', ['people','bun']],
['08,11,2012 11:14', ['kangaroo']]
]

我需要的是将此列表转换为以下格式：

T =[
['07,07,2012 22:10', 'people'],
['07,07,2012 22:10', 'drama'],
['07,07,2012 22:10', 'melody'],
['07,07,2012 22:10', 'bun'],
['08,07,2012 21:04', 'queen'],
['08,07,2012 21:04', 'group'],
['08,07,2012 21:23', 'printing'],
['08,07,2012 21:23', 'market'],
['08,07,2012 21:23', 'shopping'],
['08,07,2012 21:04', 'people'],
['08,07,2012 21:04', 'bun'],
[''08,11,2012 11:14'', 'kangaroo']
]

即。对于具有大于1的第一子元素的长度的每个元素（在原始列表T中），分割第一子元素（a[1] for a in T if len(a[1] > 1)）并将其作为具有相同时间戳的另一列表附加。也许我的话缺乏解释，但上面的例子肯定是在解释我需要做什么。任何帮助将不胜感激。

Answer 1

使用列表理解：

[(timestamp, item) for timestamp, items in T for item in items]

这将为您提供元组列表，这可能适用于此处。但您可以修改它以获取列表列表：

[[timestamp, item] for timestamp, items in T for item in items]

Answer 2

在T：

上试试



def process(t):
    new = []
    for i in t:
        for j in i[1]:
            new.append([i[0], j])
    return new

在保持订单的同时重塑python列表

2 个答案: