如何创建一个组合嵌套在元组中的列表值的索引？

Question

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:-)

我看过示例 11.4。 （列出成员资格）在http://openbookproject.net/thinkcs/python/english3e/lists.html 上，这在某些方面非常接近我的目标。

项目是：

• 从引用 (key, [list of values]) 的元组列表开始

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

• 我想扫描“my_list”以附加嵌套列表，将 值列表组合为仅一个 键，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，但我找不到如何去做！ ^^

• 现在，这是我所拥有的：

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

• 很抱歉这行太长了，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，我会提供更多详细信息。

感谢您的时间和解释 :-)

Answer 1

使用OrderedDict（首先是import collections）看起来很简单：

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，如果您想恢复元组，只需遍历字典即可：

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 ('d', [4, 5]) 项扩展了您的原始列表，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

你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