Python 如何将嵌套在元组中的列表的值组合成索引？

我很难实现我需要实现的目标，所以我想知道这里是否有人可以帮助我：-）

我已经看到了示例11.4。（列出会员名单）在某些方面与我的目标非常接近

该项目是：

• 从引用的元组列表开始（键，[值列表]）

• 我想扫描“我的列表”，以便附加嵌套列表，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，我找不到如何做到这一点^^

• 现在，我得到的是：

  # 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, [...], [...], [...], [...]], [...], [...], [...]], [...], [...]], [...]])]
  

• 很抱歉线路太长了，但我认为这会更符合整个错误

如果我不够清楚，请毫不犹豫地告诉我，我会提供更多细节

---

感谢您的时间和解释：-）

使用

OrderedDict

（

首先导入集合

）这似乎非常简单：

现在，如果要恢复元组，只需在字典上迭代：

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

---

您可以使用字典来构造列表

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

for item in my_list:
    if item[0] in my_dict.keys():
        ### if key exists append to list
        my_dict[item[0]].extend(item[1][0:])
    else:
        ### if key does not exist create new key-value pair
        my_dict[item[0]] = [item[1][0]]

my_list = my_dict.items()
print my_list

您可以使用容器创建列表字典，然后只存储每个元组对的值。请注意，我使用

（'d'，[4,5]）

项扩展了原始列表，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])]

---

是的，使用字典是解决这个问题的最好方法。这里的顺序受到影响。事实上，在面对

setdefault

.Hmm时，defaultdict是过分的，不是很明确，但使用列表和元组则表明情况并非如此。。。无论如何，这是在OP上。看起来不错，但出于某种原因，它对我不起作用：python在第dd[k]行之后引发以下错误。extend（v[0]）>>>>>：Traceback（最近一次调用）：File“”，第3行，在TypeError中：“int”对象不是iterable@EtienneS它应该是

dd[k].extend（v）

而不是

dd[k].extend（v[0]）

。请注意，

dd[k].append（v[0]）

在每个元组对的列表中只有一个数字的情况下有效。如果列表中有多个元素，则会丢失数据。我建议改为

extend

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

for item in my_list:
    if item[0] in my_dict.keys():
        ### if key exists append to list
        my_dict[item[0]].extend(item[1][0:])
    else:
        ### if key does not exist create new key-value pair
        my_dict[item[0]] = [item[1][0]]

my_list = my_dict.items()
print my_list

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])]