什么是最有效的pythonic方法：

Question

I've a list L of size N where each element of the list is between 0 to K-1 .
I want to create a 2D list S with K rows such that r ^th row contains all those indices i , such that L[i] == r .

For example if L was [0, 0, 1, 3, 0, 3]
then the new list S is [[0, 1, 4], [2], [], [3, 5]]

Solution should of course be O(N), it should also be as efficient as possible (Read : no useless append operations on list)

Answer 1

>>> L = [0, 0, 1, 3, 0, 3]
>>> import collections
>>> d = collections.defaultdict(list)
>>> for index, item in enumerate(L):
...   d[item].append(index)
... 
>>> d
defaultdict(<type 'list'>, {0: [0, 1, 4], 1: [2], 3: [3, 5]})
>>> [d[i] for i in xrange(1 + max(d))]
[[0, 1, 4], [2], [], [3, 5]]

Answer 2

>>> L = [0, 0, 1, 3, 0, 3]
>>> S = map(lambda x: [], L)
>>> S
[[], [], [], [], [], []]
>>> for index, item in enumerate(L):
    S[item].append(index)


>>> S
[[0, 1, 4], [2], [], [3, 5], [], []]
>>> 

With this solution, all indexes of S from 0 to K - 1 are filled in with an empty list.

EDIT: indeed wim is right, S[4] and S[5] are not desired so I reused wim's xrange to make it work as desired:

>>> L = [0, 0, 1, 3, 0, 3]
>>> S = map(lambda x: [], xrange(1 + max(L)))
...
>>> S
[[0, 1, 4], [2], [], [3, 5]]

Answer 3

Here's a simple but efficient way to do it:

K = 4
S = [ [] for _ in range(K) ]
for n, val in enumerate(L):
    S[val].append(n)