如何存储在不同的列表中？

Question

我有两个清单，如下

 l1 = [2,4,3]
l2 = [4,5,6,7,8,9,10,11,12]

我想从[4,5]中列出三个列表，例如[6,7,8,9]，[10,11,12]和l2。我可以用不同的方式来做，但是我需要使用循环来做。

我已经尝试过以下操作：

a = []
b = []
c = []
a.append(l2[0:2])
b.append(l2[2:4+2])
c.append(l2[4+2:])

我希望使用循环，其中l1的元素将是每个新列表的大小。

Answer 1

l1 = [2,4,3]
l2 = [4,5,6,7,8,9,10,11,12]

result = []
counter = 0
for x in l1:
   result += [l2[counter:counter+x]]
   counter += x

print(result)
[[4, 5], [6, 7, 8, 9], [10, 11, 12]]

Answer 2

您可以通过这种方式做到：

l1 = [2,4,3]
l2 = [4,5,6,7,8,9,10,11,12]
l2_ = l2.copy()
result = []

for item in l1:
    result.append(l2_[:item])
    del l2_[:item]

a,b,c = result
print (a)
print (b)
print (c)

输出：

[4, 5]
[6, 7, 8, 9]
[10, 11, 12]

Answer 3

使用列表推导和list.pop

Ex：

l1 = [2,4,3]
l2 = [4,5,6,7,8,9,10,11,12]
print([[l2.pop(0) for _ in range(i)] for i in l1])

输出：

[[4, 5], [6, 7, 8, 9], [10, 11, 12]]

Answer 4

您可以将l2转换为collections.deque，以允许使用popleft()从左侧弹出O（1）。

from collections import deque

l1 = [2,4,3]
l2 = deque([4,5,6,7,8,9,10,11,12])

result = [[l2.popleft() for _ in range(x)] for x in l1]

print(result)
# [[4, 5], [6, 7, 8, 9], [10, 11, 12]]

Answer 5

具有islice的列表理解

from itertools import islice
l1 = [2,4,3]
l2 = [4,5,6,7,8,9,10,11,12]
il2 = iter(l2)
l = [list(islice(il2, x)) for x in l1]
print(l)
# [[4, 5], [6, 7, 8, 9], [10, 11, 12]]

Performance

使用Jupyter笔记本中的timeit测试了这里发布的五种方法，并预加载了库。薄凤灿的帖子有最快的方法。

Method 1. pok fung chan -- 100000 loops, best of 3: 2.17 µs per loop
Method 2: ncica         -- 100000 loops, best of 3: 2.93 µs per loop
Method 3: darrylg       -- 100000 loops, best of 3: 4.02 µs per loop
Method 4: RoadRunner    -- 100000 loops, best of 3: 6.38 µs per loop
Method 5: Rakeesh       -- 100000 loops, best of 3: 8.76 µs per loop