什麼是有效的方式來分割一個列表到Python中的幾個列表

Question

我想根據一些表達式將列表拆分爲多個列表。

例如：

for elem in elemList: 
    if elem>0 and elem<=3 add element to Partition1 
    if elem>3 and elem<=6 add element to Partition2 
    else     add element to Partition3 

如何有效地做到這一點？ 在複雜度爲O（1）的C++列表中是否存在類似erase和push_back的東西？

還有其他一些可以做到的集合嗎？

Answer 1

首先，here是Python中各種操作的時間複雜性的好備忘單。

鑑於您對我的聲明的最後一次回覆，我假設您想要使用並將給定列表拆分爲三個單獨的列表，而不通過理解創建任何中間列表。完成此操作的一種方法是將數據結構更改爲collections.deque，因爲list.remove()需要O(n)（因爲它需要移位元素）而deque.popleft()需要O(1)。

>>> def partition(queue): 
...  partition1, partition2, partition3 = [], [], [] 
...  while queue: 
...   ele = queue.popleft() 
...   if 0 < ele <= 3: 
...    partition1.append(ele) 
...   elif 3 < ele <= 6: 
...    partition2.append(ele) 
...   else: 
...    partition3.append(ele) 
...  return partition1, partition2, partition3 

>>> import collections 
>>> p1, p2, p3 = partition(collections.deque(range(10))) 

>>> print p1 
[1, 2, 3] 

>>> print p2 
[4, 5, 6] 

>>> print p3 
[0, 7, 8, 9] 

Answer 2

如果我理解正確的問題，你可以使用list comprehensions：

Partition1 = [elem for elem in elemList if elem <= 3] 
Partition2 = [elem for elem in elemList if elem > 3 and elem <= 6] 
Partition3 = [elem for elem in elemList if elem > 6] 

這是做（對分區1）的Python的方式：

Partition1 = [] # creates an empty list 
for elem in elemList: # for each element in list 
    if elem > 0 and elem <= 3: # if they satisfy the condition 
     Partition1.append(elem) # append them to the list 

編輯：我假設上面所有元素都大於零...如果其中一些元素不是，則Partition1和Partition3的列表解析必須是：

Partition1 = [elem for elem in elemList if elem > 0 and elem <= 3] 
Partition3 = [elem for elem in elemList if elem > 6 or elem <= 0]