эффективный способ создания dict из вложенного списка

Question

original_list = [[a,1],[a,2],[a,3],[b,5],[a,4],[b,6],[c,7],[c,8],[c,9],[c,0]] 

У меня есть список как original_list. Как я могу группировать их на символы так?

dict = {'a': [1,2,3,4], 'b':[5,6], 'c':[7,8,9,0]} 

Answer 1

In [48]: result = {} 

In [49]: for x in original_list: result.setdefault(x[0], []).append(x[1]) 

In [50]: result 
Out[50]: {'a': [1, 2, 3, 4], 'b': [5, 6], 'c': [7, 8, 9, 0]} 

Answer 2

Нормальный Список итерация и словарь

>>> original_list = [["a",1],["a",2],["a",3],["b",5],["a",4],["b",6],["c",7],["c",8],["c",9],["c",0]] 
>>> original_list 
[['a', 1], ['a', 2], ['a', 3], ['b', 5], ['a', 4], ['b', 6], ['c', 7], ['c', 8], ['c', 9], ['c', 0]] 
>>> resut = {} 
>>> for i in original_list: 
...  if i[0] in resut: 
...   resut[i[0]].append(i[1]) 
...  else: 
...   resut[i[0]] = [i[1]] 
... 
>>> resut 
{'a': [1, 2, 3, 4], 'c': [7, 8, 9, 0], 'b': [5, 6]} 
>>> 

---

По collections.defaultdict метод:

Это приведет к удалению внутренней, если операторов цикла.

>>> import collections 
>>> result = collections.defaultdict(list) 
>>> result 
defaultdict(<type 'list'>, {}) 
>>> for i in original_list: 
...  result[i[0]].append(i[1]) 
... 
>>> result 
defaultdict(<type 'list'>, {'a': [1, 2, 3, 4], 'c': [7, 8, 9, 0], 'b': [5, 6]}) 
>>> 

---

Не используйте зарезервированное ключевое слово в качестве имен переменных

dict встроен словарь переменного типа.

Демо:

>>> dict 
<type 'dict'> 
>>> a = dict() 
>>> a 
{} 
>>> dict = {"a":1} 
>>> b = dict() 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
TypeError: 'dict' object is not callable 

Answer 3

В соответствии с результатами синхронизации (обновление ниже в ответе) быстрый метод может быть с помощью collections.defaultdict. Пример -

from collections import defaultdict 
result_dic = defaultdict(list) 
for a,b in original_list: 
    result_dic[a].append(b) 

Демо -

>>> original_list = [['a',1],['a',2],['a',3],['b',5],['a',4],['b',6],['c',7],['c',8],['c',9],['c',0]] 
>>> 
>>> from collections import defaultdict 
>>> result_dic = defaultdict(list) 
>>> for a,b in original_list: 
...  result_dic[a].append(b) 
... 
>>> result_dic 
defaultdict(<class 'list'>, {'b': [5, 6], 'c': [7, 8, 9, 0], 'a': [1, 2, 3, 4]}) 

---

Другой метод может быть использование itertools.groupby для этого. Пример -

from itertools import groupby 
result_dict = {key:list(b for a,b in group) for key,group in groupby(sorted(original_list),key=lambda x:x[0])} 

Демо -

>>> original_list = [['a',1],['a',2],['a',3],['b',5],['a',4],['b',6],['c',7],['c',8],['c',9],['c',0]] 
>>> from itertools import groupby 
>>> result_dict = {key:list(b for a,b in group) for key,group in groupby(sorted(original_list),key=lambda x:x[0])} 
>>> result_dict 
{'b': [5, 6], 'c': [0, 7, 8, 9], 'a': [1, 2, 3, 4]} 

---

результаты Timing -

In [21]: %paste 
def func1(): 
     original_list = [['a',1],['a',2],['a',3],['b',5],['a',4],['b',6],['c',7],['c',8],['c',9],['c',0]] 
     return {key:list(b for b in group) for key,group in groupby(sorted(original_list),key=lambda x:x[0])} 

def func2(): 
     original_list = [['a',1],['a',2],['a',3],['b',5],['a',4],['b',6],['c',7],['c',8],['c',9],['c',0]] 
     result_dic = defaultdict(list) 
     for a,b in original_list: 
       result_dic[a].append(b) 
     return result_dic 

def func3(): 
     original_list = [['a',1],['a',2],['a',3],['b',5],['a',4],['b',6],['c',7],['c',8],['c',9],['c',0]] 
     result = {} 
     for x in original_list: result.setdefault(x[0], []).append(x[1]) 
     return result 

## -- End pasted text -- 

In [22]: %timeit func1() 
The slowest run took 4.56 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 12.1 µs per loop 

In [23]: %timeit func2() 
100000 loops, best of 3: 3.82 µs per loop 

In [24]: %timeit func3() 
The slowest run took 4.77 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 4.31 µs per loop 

Answer 4

Попробуйте это: вещий способ

>>> import itertools 
>>> { x: map(lambda x:x[1], y) for x,y in itertools.groupby(sorted(original_list, key=lambda x:x[0]), key=lambda x:x[0])} 
{'a': [1, 2, 3, 4], 'c': [7, 8, 9, 0], 'b': [5, 6]} 

Answer 5

Используйте один list comprehension, только для краткого списка, учитывая сложность O(n*n).

>>> a = [['a',1],['a',2],['a',3],['b',5],['a',4],['b',6],['c',7],['c',8],['c',9],['c',0]] 
>>> {k:[v for k1, v in a if k == k1] for k, v in a} 
{'a': [1, 2, 3, 4], 'c': [7, 8, 9, 0], 'b': [5, 6]}