该项目是：

从引用（键，[值列表]）的元组列表开始

my_list = [('a',[0]), ('b',[1]), ('c',[2]), ('a',[3])]

我想扫描一下my_list＆＃39;为了附加嵌套列表，只为一个键组合值列表，如下所示：
```
my_list = [('a',[0, 3]), ('b',[1]), ('c',[2])]
```
我成功地手动组合了值，但我想自动化它，我无法找到如何做到这一点！ ^^

目前，这就是我所拥有的：

# my_input == 'a b c a'

#splitting input to list
>>> raw_list = my_input.split()
>>> raw_list
['a', 'b', 'c', 'a']

#getting an enumeration for each entry
#### (in order of appearance, this is important!) ####
>>> enum_list = [(b,[a]) for a, b in enumerate(raw_list)]
>>> enum_list
[('a', [0]), ('b', [1]), ('c', [2]), ('a', [3])]

#trying to append the enum value of the second 'a' to the first tuple of 'a'
>>> for (x, y) in enum_list :
...     for (x, z) in enum_list :
...             enum_list[enum_list.index((x, z))][1].append(y)
... 
>>> enum_list
[('a', [0, [...], [1, [...], [...], [2, [...], [...], [...], [3, [...], [...], [...], [...]]], [3, [...], [...], [2, [...], [...], [...], [...]], [...]]], [2, [...], [1, [...], [...], [...], [3, [...], [...], [...], [...]]], [...], [3, [...], [1, [...], [...], [...], [...]], [...], [...]]], [3, [...], [1, [...], [...], [2, [...], [...], [...], [...]], [...]], [2, [...], [1, [...], [...], [...], [...]], [...], [...]], [...]]]), ('b', [1, [0, [...], [...], [2, [...], [...], [...], [3, [...], [...], [...], [...]]], [3, [...], [...], [2, [...], [...], [...], [...]], [...]]], [...], [2, [0, [...], [...], [...], [3, [...], [...], [...], [...]]], [...], [...], [3, [0, [...], [...], [...], [...]], [...], [...], [...]]], [3, [0, [...], [...], [2, [...], [...], [...], [...]], [...]], [...], [2, [0, [...], [...], [...], [...]], [...], [...], [...]], [...]]]), ('c', [2, [0, [...], [1, [...], [...], [...], [3, [...], [...], [...], [...]]], [...], [3, [...], [1, [...], [...], [...], [...]], [...], [...]]], [1, [0, [...], [...], [...], [3, [...], [...], [...], [...]]], [...], [...], [3, [0, [...], [...], [...], [...]], [...], [...], [...]]], [...], [3, [0, [...], [1, [...], [...], [...], [...]], [...], [...]], [1, [0, [...], [...], [...], [...]], [...], [...], [...]], [...], [...]]]), ('a', [3, [0, [...], [1, [...], [...], [2, [...], [...], [...], [...]], [...]], [2, [...], [1, [...], [...], [...], [...]], [...], [...]], [...]], [1, [0, [...], [...], [2, [...], [...], [...], [...]], [...]], [...], [2, [0, [...], [...], [...], [...]], [...], [...], [...]], [...]], [2, [0, [...], [1, [...], [...], [...], [...]], [...], [...]], [1, [0, [...], [...], [...], [...]], [...], [...], [...]], [...], [...]], [...]])]

很抱歉超长线，但我认为它会更符合整个错误......

如果我不够清楚，请不要犹豫告诉我，我会提供更多详情。

感谢您的时间和解释:-)

Answer 1

使用OrderedDict（import collections优先）时，这似乎非常简单：

In [438]: dict_ = collections.OrderedDict()

In [439]: for l in my_list:
     ...:     dict_.setdefault(l[0], []).extend(l[1])
     ...:     

In [440]: dict_
Out[440]: OrderedDict([('a', [0, 3]), ('b', [1]), ('c', [2])])

现在，如果你想让你的元组重新遍历字典：

In [441]: [(k, v) for k, v in dict_.items()]
Out[441]: [('a', [0, 3]), ('b', [1]), ('c', [2])]

Answer 2

您可以使用defaultdict容器创建列表字典，然后只存储每个元组对的值。请注意，我使用项('d', [4, 5])扩展原始列表，以说明此方法也适用于任意长度的列表。

from collections import defaultdict

my_list = [('a', [0]), ('b', [1]), ('c', [2]), ('a', [3]), ('d', [4, 5])]

dd = defaultdict(list)
for pair in my_list:
    k, v = pair
    dd[k].extend(v)  
>>> dd.items()
[('a', [0, 3]), ('c', [2]), ('b', [1]), ('d', [4, 5])]

Answer 3

您可以使用字典构建列表

@Transactional(propagation = Propagation.REQUIRED, rollbackFor = Exception.class, readOnly = false)
public class UserRolesService{
..........
}

如何组合嵌套在元组中的列表值的索引？

该项目是：

3 个答案: