Arrays 用零填充空列表以获得5个元组的固定大小列表

我有1000个例子。每个示例包含一个包含18个列表的列表，这些列表长度可变，有些列表为空

以下是一个示例：

len(My_list)
18
print(My_list)
array([list([(17, 163, 0.11258018, 15),(78, 193, 0.99713018, 17),(478, 94, 0.7299528, 2), (63, 268, 0.77531445, 3), (169, 279, 0.7947326, 4),(456, 140, 0.65013665, 7), (61, 301, 0.7433308, 8)]), 
list([]),
list([]), 
list([]), 
list([]), 
list([]),
list([]),
list([]), 
list([(63, 176, 0.18713018, 0),(199, 185, 0.88743243, 79), (282, 75, 0.752135, 84)]),
list([(62, 185, 0.13743243, 1)]), 
list([]),
list([(67, 156, 0.14346971, 2)]), 
list([(2, 15, 0.00639179, 3)]),
list([]),
list([]), 
list([]), 
list([]), 
list([])], 
dtype=object)

我想做什么？

对于每个列表：

1-保留前5个元组

2-如果列表为空，则创建一个包含五个元组的列表，如下所示

list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]).

3-如果列表不是空的，但不包含5个元素，则完成它以获得5个元素。由于

My_list[12]

只包含一个元素

list（[（67156,0.14346971,2）]）

因此：

My_list[12]=list([(67, 156, 0.14346971, 2),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)])

预期产出：

array([list([(17, 163, 0.11258018, 15),(78, 193, 0.99713018, 17),(478, 94, 0.7299528, 2), (63, 268, 0.77531445, 3), (169, 279, 0.7947326, 4)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]),
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]),
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]),
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(63, 176, 0.18713018, 0),(199, 185, 0.88743243, 79), (282, 75, 0.752135, 84),(0,0,0,0),(0,0,0,0)]),
list([(62, 185, 0.13743243, 1),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]),
list([(67, 156, 0.14346971, 2),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(2, 15, 0.00639179, 3),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]),
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]),
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)]), 
list([(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0)])], 
dtype=object)

我试过什么？

My_list=np.asarray(My_list)

My_list = [joint if len(joint) != 0 else [(0, 0, 0,0)] for joint in My_list]

然而，这并不意味着这项工作。它仅用（0,0,0,0）填充空列表。此外，包含一个或多个元素的列表将跳过它们。并且它希望用（0,0,0,0）填充所有空列表或少于五个元素的列表，以获得每个列表的五个元素

---

有什么提示吗？

这里有一种方法：将5个元组粘贴到所有内容上，然后修剪：

>>> ml
array([list([(17, 163, 0.11258018, 15), (78, 193, 0.99713018, 17), (478, 94, 0.7299528, 2), (63, 268, 0.77531445, 3), (169, 279, 0.7947326, 4), (456, 140, 0.65013665, 7), (61, 301, 0.7433308, 8)]),
       list([]), list([]), list([]), list([]), list([]), list([]),
       list([]),
       list([(63, 176, 0.18713018, 0), (199, 185, 0.88743243, 79), (282, 75, 0.752135, 84)]),
       list([(62, 185, 0.13743243, 1)]), list([]),
       list([(67, 156, 0.14346971, 2)]), list([(2, 15, 0.00639179, 3)]),
       list([]), list([]), list([]), list([]), list([])], dtype=object)
>>> 
>>> z = np.array([None, 5*[4*(0,)]])[[1]]
>>> z
array([list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)])],
      dtype=object)
>>> 
>>> res = np.frompyfunc(list.__getitem__, 2, 1)(ml + z, slice(5))
>>> res
array([list([(17, 163, 0.11258018, 15), (78, 193, 0.99713018, 17), (478, 94, 0.7299528, 2), (63, 268, 0.77531445, 3), (169, 279, 0.7947326, 4)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(63, 176, 0.18713018, 0), (199, 185, 0.88743243, 79), (282, 75, 0.752135, 84), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(62, 185, 0.13743243, 1), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(67, 156, 0.14346971, 2), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(2, 15, 0.00639179, 3), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)])],
      dtype=object)

说明：对象数据类型的数组委托操作，如添加到其元素。因此，

ml+z

将每个原始列表与5x4个零的副本组合在一起

---

接下来，我们只需要将每个列表缩减为5个元素。操作

somelist[：5]

可以写成

somelist.\uu getitem\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。最后一种形式是我们使用
np“矢量化”的形式。frompyfunc
这是@PaulP answer（和@Eir的评论）的一个变体。它足够近了，我不会发布它，除非它更快（可能更清晰）

定义一个一次在一个列表上运行的函数-使用添加pad和剥离不需要的元素的思想：

In [209]: z = [4*(0,) for _ in range(5)]
In [210]: def foo(alist):
     ...:     return  (alist + z)[:5]

这可以通过列表理解应用于每个列表：

In [211]: [foo(row) for row in arr]
Out[211]: 
[[(17, 163, 0.11258018, 15),
  (78, 193, 0.99713018, 17),
  (478, 94, 0.7299528, 2),
  (63, 268, 0.77531445, 3),
  (169, 279, 0.7947326, 4)],
 [(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)],
 ....
 [(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]]

但是如果你想要一个对象数组，@Paul使用
frompyfunc
的方法非常有效：

In [212]: np.frompyfunc(foo,1,1)(arr)
Out[212]: 
array([list([(17, 163, 0.11258018, 15), (78, 193, 0.99713018, 17), (478, 94, 0.7299528, 2), (63, 268, 0.77531445, 3), (169, 279, 0.7947326, 4)]),
       list([(0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 0)]),
  ....          dtype=object)

时间：

In [176]: timeit np.frompyfunc(list.__getitem__, 2, 1)(arr + z, slice(5))
14.8 µs ± 18.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [184]: timeit [foo(row) for row in arr]
7.6 µs ± 26.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [213]: timeit np.frompyfunc(foo,1,1)(arr)
8.49 µs ± 27.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

您可能希望向每个列表中添加一个包含5个零元素的常量列表，并只取前5个元素。这可能会消耗一些内存，但这项工作是否小心？您的
z
会复制元组。虽然是元组可能并不重要，但速度确实更快。