命名数据类型数组：在[0]['name']和['name'][0]之间的区别？

Question

我在numpy中遇到了以下奇怪的东西，可能是也可能不是bug：

import numpy as np
dt = np.dtype([('tuple', (int, 2))])
a = np.zeros(3, dt)
type(a['tuple'][0])  # ndarray
type(a[0]['tuple'])  # ndarray

a['tuple'][0] = (1,2)  # ok
a[0]['tuple'] = (1,2)  # ValueError: shape-mismatch on array construction

我本以为下面的两个选项都有效。 意见？在

Answer 1

我是在numpy讨论列表上问的。特拉维斯·奥列芬特回答here。在

引用他的回答：

The short answer is that this is not really a "normal" bug, but it could be considered a "design" bug (although the issues may not be straightforward to resolve). What that means is that it may not be changed in the short term - and you should just use the first spelling.

Structured arrays can be a confusing area of NumPy for several of reasons. You've constructed an example that touches on several of them. You have a data-type that is a "structure" array with one member ("tuple"). That member contains a 2-vector of integers.

First of all, it is important to remember that with Python, doing

a['tuple'][0] = (1,2)

is equivalent to

b = a['tuple']; b[0] = (1,2)

In like manner,

a[0]['tuple'] = (1,2)

is equivalent to

b = a[0]; b['tuple'] = (1,2)

To understand the behavior, we need to dissect both code paths and what happens. You built a (3,) array of those elements in 'a'. When you write b = a['tuple'] you should probably be getting a (3,) array of (2,)-integers, but as there is currently no formal dtype support for (n,)-integers as a general dtype in NumPy, you get back a (3,2) array of integers which is the closest thing that NumPy can give you. Setting the [0] row of this object via

a['tuple'][0] = (1,2)

works just fine and does what you would expect.

On the other hand, when you type:

b = a[0]

you are getting back an array-scalar which is a particularly interesting kind of array scalar that can hold records. This new object is formally of type numpy.void and it holds a "scalar representation" of anything that fits under the "VOID" basic dtype.

For some reason:

b['tuple'] = [1,2]

is not working. On my system I'm getting a different error: TypeError: object of type 'int' has no len()

I think this should be filed as a bug on the issue tracker which is for the time being here: http://projects.scipy.org/numpy

The problem is ultimately the void->copyswap function being called in voidtype_setfields if someone wants to investigate. I think this behavior should work.

对此的解释见a numpy bug report。在

Answer 2

这是一个上游错误，从NumPy PR #5947开始修复，在1.9.3中进行了修复。在

Answer 3

我得到的错误与您不同（使用numpy 1.7.0.dev）：

ValueError: setting an array element with a sequence.

所以下面的解释可能对您的系统不正确（或者对我所看到的可能是错误的解释）。在

首先，注意索引structured array的一行会得到一个numpy.void对象（请参见data type docs）

^{pr2}$

据我所知，void有点像Python列表，因为它可以保存不同数据类型的对象，这是有意义的，因为结构化数组中的列可以是不同的数据类型。在

如果不是索引，而是切掉第一行，则会得到一个ndarray：

print type(a[:1]) # = numpy.ndarray

这类似于Python列表的工作方式：

b = [1, 2, 3]
print b[0] # 1
print b[:1] # [1]

切片返回原始序列的缩短版本，但索引返回一个元素（这里是int；上面是void类型）。在

因此，当您对结构化数组的行进行切片时，应该希望它的行为与原始数组相同（只是行数较少）。继续您的示例，现在可以为第一行的“tuple”列赋值：

a[:1]['tuple'] = (1, 2)

所以，。。。为什么a[0]['tuple'] = (1, 2)不起作用？在

好吧，回想一下a[0]返回一个void对象。所以，当你打电话

a[0]['tuple'] = (1, 2) # this line fails

您将一个tuple分配给void对象的“tuple”元素。注意：尽管您将此索引称为“元组”，但它存储为ndarray：

print type(a[0]['tuple']) # = numpy.ndarray

所以，这意味着元组需要转换成ndarray。但是，void对象不能强制转换赋值（这只是猜测），因为它可以包含任意的数据类型，所以它不知道要转换到什么类型。要解决这一问题，您可以自己投射输入：

a[0]['tuple'] = np.array((1, 2))

事实上我们得到了不同的错误，这表明上面的一行可能不适合您，因为casting处理的是我收到的错误，而不是您收到的错误。在

附录：

那么，为什么下面的工作呢？在

a[0]['tuple'][:] = (1, 2)

在这里，当您添加[:]时，您将索引到数组中，但是如果没有这个，您将索引到void对象。换句话说，a[0]['tuple'][:]表示“替换存储数组的元素”（由数组处理），a[0]['tuple']表示“替换存储数组”（由void处理）。在

结语：

奇怪的是，访问行（即用0索引）似乎会删除基数组，但它仍然允许您分配给基数组。在

print a['tuple'].base is a # = True
print a[0].base is a # = False
a[0] = ((1, 2),) # `a` is changed

也许void不是真正的数组，所以它没有基数组，。。。但是为什么它有一个base属性？在