Question

我有1000个示例的样本。每个样本包含一个18个长度可变的列表的列表，有些列表为空。

以下是示例：

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保持前5个元组

2-如果列表为空，则创建一个五个元组的列表，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-如果列表不为空但不包含5个元素，则将其完整以获取5个元素。由于My_list[12]仅包含一个元素list([(67, 156, 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)])

预期输出：

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)

我尝试了什么？

My_list=np.asarray(My_list)

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

但是，它没有完成任务。它只填充（0,0,0,0）的空列表，而且具有一个或多个元素的列表会跳过它们。并且期望用（0,0,0,0）填充所有空白列表或少于五个元素的列表，以使每个列表具有五个元素。

有提示吗？

Answer 1

这是一种方法：将5个元组粘到所有内容上，然后再修剪：

>>> 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dtype的数组委派了操作（例如，添加到其元素中）。因此，ml + z将每个原始列表与5x4零的副本结合在一起。

接下来，我们只需要将每个列表切成5个元素即可。操作somelist[:5]可以写为somelist.__getitem__(slice(5))，甚至可以写为list.__getitem__(somelist, slice(5))。最后一种形式是我们使用np.frompyfunc“向量化”的形式。

Answer 2

这是@PaulP答案（和@Eir的评论）的变体。它足够接近以至于我不会将其发布，除了它更快（并且可能更清晰）之外。

定义一次可在一个列表上操作的函数-使用添加填充板并剥离不需要的元素的想法：

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，如果您想要对象数组，@ Paul使用frompyfunc的方法效果很好：

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)

用零填充列表的空列表以获得5个元组的固定大小列表

2 个答案: