将子列表合并到基于python子列表项的列表中

Question

输入：

我有这个有序列表。

[[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 

希望的输出

[[1,['A','B','D']],[2, ['A','D']], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 

由于这两个子列表的第一项是一样的。

我也可以转换成一个关键字和这些值的字典对。像

{1:['A','B','D'],2:['A','D'],3:['C']} 

什么是最简单最简单的方法呢？

Answer 1

您可以使用groupby从itertools模块这样的例子：

a = [[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 

final = [] 
for k, v in groupby(sorted(a, key=lambda x: x[0]), lambda x: x[0]): 
     bb = list(v) 
     if len(bb) > 1: 
      final.append([k, [j for _, j in bb]]) 
     else: 
      final.append([k, bb[0][1]]) 

# OR: 
# Within a list comprehension 
# final = [[k, [j[1] for j in list(v)]] for k, v in groupby(sorted(a, key=lambda x: x[0]), lambda x: x[0])] 

print(final) 

输出：

[[1, ['A', 'B', 'D']], 
[2, ['A', 'D']], 
[3, 'C'], 
[4, 'D'], 
[5, 'B'], 
[6, 'D']] 

再到最后列表转换成一个字典，你可以这样做：

final_dict = {k:v if isinstance(v, list) else [v] for k, v in final} 
print(final_dict) 

输出：

{1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']}} 

Answer 2

in_list 
out_list = [] 
sublist = [] 
i = 0 
for l in in_list: 
    if l[0] != i: 
     i = l[0] 
     sublist = [] 
     out_list.append([i, sublist]) 
    sublist.append(l[1]) 
dico = dict(out_list) 

Answer 3

您可以直接从输入创建字典。

from collections import defaultdict 

input = [[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 

d = defaultdict(list) 
for el in input: d[el[0]].append(el[1]) 

的d输出将是：

{1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']} 

Answer 4

如果顺序并不重要，你想反正词典：

import collections 

your_list = [[1,'A'], [1,'B'], [1,'D'], [2,'A'], [2,'D'], [3,'C'], [4,'D'], [5,'B'], [6,'D']] 

result = collections.defaultdict(list) 
for k, v in your_list: 
    result[k].append(v) 

# {1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']} 

你也可以不用collections.defaultdict（可能在某些性能损失下，取决于关键频率）：

your_list = [[1,'A'], [1,'B'], [1,'D'], [2,'A'], [2,'D'], [3,'C'], [4,'D'], [5,'B'], [6,'D']] 

result = {} 
for k, v in your_list: 
    result[k] = result.get(k, []) + [v] 

# {1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']} 

Answer 5

如果数据是有序的，那么itertools.groupby是一个好办法：

>>> from itertools import groupby 
>>> from operator import itemgetter 
>>> data = [[1, 'A'], [1, 'B'], [2, 'A'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 
>>> final_data = [] 
>>> final_data = [] 
>>> for k, g in groupby(data, itemgetter(0)): 
...  group = list(g) 
...  if len(group) == 1: 
...   final_data.append(group[0]) 
...  else: 
...   final_data.append([k, [sub[1] for sub in group]]) 
... 
>>> final_data 
[[1, ['A', 'B']], [2, 'A'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 
>>> 

如果你想在一本字典的结果，那就是更容易：

>>> grouped_dict = {} 
>>> for num, letter in data: 
...  grouped_dict.setdefault(num, []).append(letter) 
... 
>>> grouped_dict 
{1: ['A', 'B'], 2: ['A'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']} 
>>> 

Answer 6

我找到了更好的做相反，而不是做一个列表，然后是一个字典，我做了字典，然后是一个列表。

输入：

in_list = [[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 

代码：

mydict = {} 
for sublist in in_list: 
    if sublist[0] in mydict.keys(): 
     mydict[sublist[0]] = [*mydict[sublist[0]],sublist[1]] 
    else: 
     mydict[sublist[0]] = sublist[1] 

输出：

>>> mydict 
{1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: 'C', 4: 'D', 5: 'B', 6: 'D'} 

做一个简单的列表，从词典：

mylist = list(mydict.items()) 

输出：

>>> mylist 
[(1, ['A', 'B', 'D']), (2, ['A', 'D']), (3, 'C'), (4, 'D'), (5, 'B'), (6, 'D')] 

让另一个列表，从词典：

mylist = = [[k,v] for k,v in mydict.items()] 

同：

mylist = [] 
for key, value in mydict.items(): 

输出：

>>> mylist 
[[1, ['A', 'B', 'D']], [2, ['A', 'D']], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']] 

Answer 7

在python文档示例中，https://docs.python.org/2/library/collections.html#defaultdict-examples 他们在您的帖子中使用了相同的问题。

>>> s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)] 
>>> d = defaultdict(list) 
>>> for k, v in s: 
...  d[k].append(v) 
... 
>>> d.items() 
[('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])] 

然后我投了一个答案与defaultdict。