NumPy结构化数组的真正递归`tolist()` [英] Truly recursive `tolist()` for NumPy structured arrays

问题描述

据我了解，将NumPy数组转换为本地Python列表的推荐方法是使用

From what I understand, the recommended way to convert a NumPy array into a native Python list is to use ndarray.tolist.

A，使用结构化数组时，这似乎无法递归工作.确实，某些ndarray对象在未转换的结果列表中被引用:

Alas, this doesn't seem to work recursively when using structured arrays. Indeed, some ndarray objects are being referenced in the resulting list, unconverted:

>>> dtype = numpy.dtype([('position', numpy.int32, 3)])
>>> values = [([1, 2, 3],)]
>>> a = numpy.array(values, dtype=dtype)
>>> a.tolist()
[(array([1, 2, 3], dtype=int32),)]

我确实编写了一个简单的函数来解决此问题:

I did write a simple function to workaround this issue:

def array_to_list(array):
    if isinstance(array, numpy.ndarray):
        return array_to_list(array.tolist())
    elif isinstance(array, list):
        return [array_to_list(item) for item in array]
    elif isinstance(array, tuple):
        return tuple(array_to_list(item) for item in array)
    else:
        return array

使用时可以提供预期的结果:

Which, when used, provides the expected result:

>>> array_to_list(a) == values
True

此功能的问题在于，它通过重新创建输出的每个列表/元组来复制ndarray.tolist的作业.并非最佳.

The problem with this function is that it duplicates the job of ndarray.tolist by recreating each list/tuple that it outputs. Not optimal.

所以问题是:

• ndarray.tolist的这种行为是预期的吗?
• 有没有更好的方法来实现这一目标?

• is this behaviour of ndarray.tolist to be expected?
• is there a better way to make this happen?

推荐答案

只是概括一下，我将在您的dtype中添加另一个字段

Just to generalize this a bit, I'll add an another field to your dtype

In [234]: dt = numpy.dtype([('position', numpy.int32, 3),('id','U3')])

In [235]: a=np.ones((3,),dtype=dt)

repr显示确实使用列表和元组:

The repr display does use lists and tuples:

In [236]: a
Out[236]: 
array([([1, 1, 1], '1'), ([1, 1, 1], '1'), ([1, 1, 1], '1')], 
  dtype=[('position', '<i4', (3,)), ('id', '<U3')])

但是请注意，tolist不会展开元素.

but as you note, tolist does not expand the elements.

In [237]: a.tolist()
Out[237]: [(array([1, 1, 1]), '1'), (array([1, 1, 1]), '1'), 
   (array([1, 1, 1]), '1')]

类似地，可以从完全嵌套的列表和元组创建这样的数组.

Similarly, such an array can be created from the fully nested lists and tuples.

In [238]: a=np.array([([1,2,3],'str')],dtype=dt)
In [239]: a
Out[239]: 
array([([1, 2, 3], 'str')], 
  dtype=[('position', '<i4', (3,)), ('id', '<U3')])
In [240]: a.tolist()
Out[240]: [(array([1, 2, 3]), 'str')]

通过这种不完全的递归重新创建数组没有问题:

There's no problem recreating the array from this incomplete recursion:

In [250]: np.array(a.tolist(),dtype=dt)
Out[250]: 
array([([1, 2, 3], 'str')], 
      dtype=[('position', '<i4', (3,)), ('id', '<U3')])

这是我第一次看到有人将tolist与这样的结构化数组一起使用，但是我并不感到惊讶.我不知道开发人员是否会认为这是一个错误.

This is the first that I've seen anyone use tolist with a structured array like this, but I'm not too surprised. I don't know if developers would consider this a bug or not.

为什么需要此数组的纯列表/元组渲染?

Why do you need a pure list/tuple rendering of this array?

我想知道numpy/lib/recfunctions.py中是否有一个函数可以解决这个问题.

I wonder if there's a function in numpy/lib/recfunctions.py that addresses this.

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