Question

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

我有像original_list这样的列表。如何根据这样的字符对它们进行分组？

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

Answer 1

根据时间结果（在答案中更新如下），最快的方法可能是使用collections.defaultdict。示例 -

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

演示 -

>>> 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]})

另一种方法可以是使用itertools.groupby。示例 -

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])}

演示 -

>>> 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]}

计时结果 -

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 2

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 3

在O(n*n)复杂度的情况下，仅使用一个list comprehension作为简短列表。

>>> 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]}

Answer 4

普通列表迭代和字典

>>> 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]}
>>>

通过 collections.defaultdict 方法：

这将删除内部if循环语句。

>>> 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]})
>>>

不要将保留关键字用作变量名称

dict是内置字典类型变量。

演示：

>>> 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 5

试试这个：pythonic方式

scaleShowHorizontalLines

从嵌套列表创建dict的有效方法

5 个答案: