如何将列表与嵌套列表python的长度分开？答案

【问题标题】：how to separate list with the length of nested list python?如何将列表与嵌套列表python的长度分开？
【发布时间】：2020-03-16 07:03:01
【问题描述】：

我有一个这样的嵌套列表：

 a = [[0.00069053395], [0.7278625], [0.8591849, 0.86290157, 0.8703022], [0.9041279, 0.9102304, 0.91197634], [0.93096334, 0.93327045], [0.9456768, 0.95339334], [0.98164046, 0.9836741]]

还有一个如下列表：

b = ['/home/shape/13.jpg', '/home/shape/5.jpg', '/home/shape/6.jpg', '/home/shape/0.jpg', '/home/shape/2.jpg', '/home/shape/1.jpg', '/home/shape/7.jpg', '/home/shape/11.jpg', '/home/cuts/shape/10.jpg', '/home/shape/4.jpg', '/home/shape/14.jpg', '/home/shape/12.jpg', '/home/shape/16.jpg', '/home/shape/8.jpg']

我想让列表 b 与嵌套列表 a 相似，并具有其元素的长度。

预期输出：

c = [['/home/shape/13.jpg'],['/home/shape/5.jpg'],['/home/shape/6.jpg', '/home/shape/0.jpg', '/home/shape/2.jpg'],['/home/shape/1.jpg', '/home/shape/7.jpg', '/home/shape/11.jpg'],['/home/cuts/shape/10.jpg', '/home/shape/4.jpg'],['/home/shape/14.jpg', '/home/shape/12.jpg'],['/home/shape/16.jpg', '/home/shape/8.jpg']]

任何建议都会有所帮助！

【问题讨论】：

你已经尝试了什么？

标签： python python-3.x list nested-lists

【解决方案1】：

你需要：

c = []
count = 0
for i in a:
     c.append(b[count:count+len(i)])
     count = count+len(i)

print(c)

输出：

[['/home/shape/13.jpg'], ['/home/shape/5.jpg'], ['/home/shape/6.jpg', 
 '/home/shape/0.jpg', '/home/shape/2.jpg'], ['/home/shape/1.jpg', 
 '/home/shape/7.jpg', '/home/shape/11.jpg'], ['/home/cuts/shape/10.jpg', 
 '/home/shape/4.jpg'], ['/home/shape/14.jpg', '/home/shape/12.jpg'], 
 ['/home/shape/16.jpg', '/home/shape/8.jpg']]

【讨论】：

【解决方案2】：

您可以将iter 内置函数与列表理解一起使用：

it = iter(b)

[[next(it) for _ in l] for l in a]

输出：

[['/home/shape/13.jpg'],
 ['/home/shape/5.jpg'],
 ['/home/shape/6.jpg', '/home/shape/0.jpg', '/home/shape/2.jpg'],
 ['/home/shape/1.jpg', '/home/shape/7.jpg', '/home/shape/11.jpg'],
 ['/home/cuts/shape/10.jpg', '/home/shape/4.jpg'],
 ['/home/shape/14.jpg', '/home/shape/12.jpg'],
 ['/home/shape/16.jpg', '/home/shape/8.jpg']]

【讨论】：