如何按元素长度拆分排序列表

Question

我有一个列表列表，它按子列表的长度排序。例如

[[str], [str1, str2], [str1, str2], [str1, str2, str3], [str1, str2, str3],...]

我想将此列表拆分为仅包含长度相同的子列表的子列表。例如

[[[str], [str], [str]],  [[str1, str2], [str1, str2], [str1, str2]], ...]

我想知道是否有比下面更有效的方法，希望少一些代码。

child_list = []
new_list = []
old_list = [['e3510000'], ['e2512001'], ['e3510000'], ['e92d4010'],
            ['e3a0b000', 'e3a0e000'], ['e92d4030', 'e59f5054'],
            ['e59f3038', 'e3530000'], ['e1a0c00d', 'e92dd800']]

# length of child
length = 1
for idx, i in enumerate(old_list):
    if idx == len(old_list)-1:
        child_list.append(i)
        new_list.append(child_list.copy())
    elif length == len(i):
        child_list.append(i)
    elif length < len(i):
        new_list.append(child_list.copy())
        del child_list[:]
        child_list.append(i)
        length = len(i)

输出：

[[['e3510000'], ['e2512001'], ['e3510000'], ['e92d4010']],
 [['e3a0b000', 'e3a0e000'], ['e92d4030', 'e59f5054'], 
  ['e59f3038', 'e3530000'], ['e1a0c00d', 'e92dd800']]]

Answer 1

您可以使用itertools.groupby按长度对old中的列表进行分组。请注意，如果原始列表已按照示例中的长度排序，则无需在此处进行排序。

from itertools import groupby
[list(v) for k,v in groupby(sorted(old_list, key=len), key=len)]

产量

[[['e3510000'], ['e2512001'], ['e3510000'], ['e92d4010']],
 [['e3a0b000', 'e3a0e000'],
  ['e92d4030', 'e59f5054'],
  ['e59f3038', 'e3530000'],
  ['e1a0c00d', 'e92dd800']]]

Answer 2

在这种情况下，groupby可能是最直观的，但是，您可以使用字典数据结构以不同方式解决问题：

from collections import defaultdict
data = [["str"], ["str1", "str2"], ["str1", "str2"], ["str1", "str2", "str3"], ["str1", "str2", "str3"]]
dct = defaultdict(list)
for el in data:
    dct[len(el)].append(el)

print(dct.values())

日期：

[[['str']],
 [['str1', 'str2'], ['str1', 'str2']],
 [['str1', 'str2', 'str3'], ['str1', 'str2', 'str3']]]

---

基准测试结果：使用基于字典的解决方案更快：

from itertools import groupby
from collections import defaultdict

data = [["str"], ["str1", "str2"], ["str1", "str2"], ["str1", "str2", "str3"], ["str1", "str2", "str3"]]

def solve_with_groupby(data):
     return [list(v) for k,v in groupby(sorted(data, key=len), key=len)]

def solve_with_dict(data):
    dct = defaultdict(list)
    for el in data:
        dct[len(el)].append(el)
    return dct.values() 

结果：

In [10]: timeit solve_with_groupby(data)
100000 loops, best of 3: 5.75 µs per loop

In [11]: timeit solve_with_dict(data)
100000 loops, best of 3: 2.56 µs per loop

Answer 3

只需使用groupby：

>>> l = [[1]*i for i in range(1, 5) for _ in range(3)]
>>> l
[[1], [1], [1], [1, 1], [1, 1], [1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]
>>> result = [list(g) for _, g in groupby(l, key=len)]
>>> result
[[[1], [1], [1]], [[1, 1], [1, 1], [1, 1]], [[1, 1, 1], [1, 1, 1], [1, 1, 1]], [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]]

或者，如果没有订购，您可以使用defaultdict：

>>> import random
>>> random.shuffle(l)
>>> l
[[1, 1, 1], [1, 1, 1, 1], [1], [1], [1, 1], [1, 1], [1, 1, 1], [1, 1, 1], [1, 1], [1], [1, 1, 1, 1], [1, 1, 1, 1]]
>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> for e in l:
...     d[len(e)].append(e)
... 
>>> result = list(d.values())
>>> result
[[[1, 1, 1], [1, 1, 1], [1, 1, 1]], [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]], [[1], [1], [1]], [[1, 1], [1, 1], [1, 1]]]