Python 将列表拆分为大小在另一个列表中给定的子列表

我有以下清单：

l1 = [a,b,c,d,e,f,g,h,i]
l2 = [1,3,2,3]

如何以这种方式拆分

l1

[[a],[b,c,d],[e,f],[g,h,i]]

-The first element is a list of one element because of the 1 of l2
-The second element is a list of three elements because of the 3 of l2
-The third element is a list of two elements because of the 2 of l2
-The fourth element is a list of three elements because of the 3 of l2

我尝试过使用一个函数：

def divide(num, n):
        return [num[i*n : (i+1)*n] for i in range(len(num)//n)]  

x = divide(l2,1)
for i in range(len(l2)):
    z = x[i]
    divide(l1,z)

但它不起作用…

代码 输出 输出

[['a'], ['b', 'c', 'd'], ['e', 'f'], ['g', 'h', 'i']]

---

你可以这样做=

start = 0
sublists = []
for n in l2:
   to = start + n
   sublists.append(l1[start:to])
   start = to

---

结果是您想要的列表

使用显式迭代器和

itertools.islice

>>> l1 = list("abcdefghi")
>>> l2 = [1,3,2,3]
>>> itr = iter(l1)
>>> [list(islice(itr, None, x)) for x in l2]
[['a'], ['b', 'c', 'd'], ['e', 'f'], ['g', 'h', 'i']]

使用

itr

而不是

l1

意味着

islice

不会在每次调用时从

l1

开始读取

您可以使用

functools.partial

对其进行重构，使列表更具可读性：

>>> from functools import partial
>>> take = partial(islice, iter(l1), None)
>>> [list(take(x)) for x in l2]
[['a'], ['b', 'c', 'd'], ['e', 'f'], ['g', 'h', 'i']]

---

使用

枚举

和

切片

，可以这样做：

l1 = ['a','b','c','d','e','f','g','h','i']
l2 = [1,3,2,3]

for index, num in enumerate(l2):
  sum_up = sum(l2[:index])
  print(l1[sum_up:sum_up+num])

---

到目前为止，您尝试了什么来解决这个问题？这更像是为我做这件事，你能做到吗。

>>> l1 = list("abcdefghi")
>>> l2 = [1,3,2,3]
>>> itr = iter(l1)
>>> [list(islice(itr, None, x)) for x in l2]
[['a'], ['b', 'c', 'd'], ['e', 'f'], ['g', 'h', 'i']]

>>> from functools import partial
>>> take = partial(islice, iter(l1), None)
>>> [list(take(x)) for x in l2]
[['a'], ['b', 'c', 'd'], ['e', 'f'], ['g', 'h', 'i']]

l1 = ['a','b','c','d','e','f','g','h','i']
l2 = [1,3,2,3]

for index, num in enumerate(l2):
  sum_up = sum(l2[:index])
  print(l1[sum_up:sum_up+num])