问题描述

下面的x[...]是什么意思?

a = np.arange(6).reshape(2,3)
for x in np.nditer(a, op_flags=['readwrite']):
    x[...] = 2 * x

推荐答案

虽然建议重复Python Ellipsis 对象做什么? 在一般 python 上下文中回答了这个问题，我认为它在 nditer 循环中的使用需要添加信息.

While the proposed duplicate What does the Python Ellipsis object do? answers the question in a general python context, its use in an nditer loop requires, I think, added information.

https://docs.scipy.org/doc/numpy/reference/arrays.nditer.html#modifying-array-values

Python 中的正则赋值只是更改本地或全局变量字典中的引用，而不是修改现有变量.这意味着简单地分配给 x 不会将值放入数组的元素中，而是将 x 从数组元素引用切换为对您分配的值的引用.要实际修改数组的元素，x 应使用省略号进行索引.

Regular assignment in Python simply changes a reference in the local or global variable dictionary instead of modifying an existing variable in place. This means that simply assigning to x will not place the value into the element of the array, but rather switch x from being an array element reference to being a reference to the value you assigned. To actually modify the element of the array, x should be indexed with the ellipsis.

该部分包含您的代码示例.

That section includes your code example.

所以用我的话来说，x[...] = ... 就地修改了x；x = ... 会破坏到 nditer 变量的链接，而不是更改它.它类似于 x[:] = ... 但适用于任何维度的数组(包括 0d).在这种情况下，x 不仅仅是一个数字，它还是一个数组.

So in my words, the x[...] = ... modifies x in-place; x = ... would have broken the link to the nditer variable, and not changed it. It's like x[:] = ... but works with arrays of any dimension (including 0d). In this context x isn't just a number, it's an array.

也许最接近这个 nditer 迭代的东西，没有 nditer 是:

Perhaps the closest thing to this nditer iteration, without nditer is:

In [667]: for i, x in np.ndenumerate(a):
     ...:     print(i, x)
     ...:     a[i] = 2 * x
     ...:     
(0, 0) 0
(0, 1) 1
...
(1, 2) 5
In [668]: a
Out[668]: 
array([[ 0,  2,  4],
       [ 6,  8, 10]])

请注意，我必须直接索引和修改 a[i].我无法使用 x = 2*x.在这个迭代中，x 是一个标量，因此不可变

Notice that I had to index and modify a[i] directly. I could not have used, x = 2*x. In this iteration x is a scalar, and thus not mutable

In [669]: for i,x in np.ndenumerate(a):
     ...:     x[...] = 2 * x
  ...
TypeError: 'numpy.int32' object does not support item assignment

但在 nditer 的情况下，x 是一个 0d 数组，并且是可变的.

But in the nditer case x is a 0d array, and mutable.

In [671]: for x in np.nditer(a, op_flags=['readwrite']):
     ...:     print(x, type(x), x.shape)
     ...:     x[...] = 2 * x
     ...:     
0 <class 'numpy.ndarray'> ()
4 <class 'numpy.ndarray'> ()
...

而且因为是0d，所以不能用x[:]代替x[...]

And because it is 0d, x[:] cannot be used instead of x[...]

----> 3     x[:] = 2 * x
IndexError: too many indices for array

更简单的数组迭代也可能提供洞察力:

A simpler array iteration might also give insight:

In [675]: for x in a:
     ...:     print(x, x.shape)
     ...:     x[:] = 2 * x
     ...:     
[ 0  8 16] (3,)
[24 32 40] (3,)

这会在 a 的行(第一个暗淡)上进行迭代.x 是一维数组，可以使用 x[:]=... 或 x[...]=....

如果我从下一个 external_loop 标志-external-loop" rel="noreferrer">section，x 现在是一维数组，x[:] = 可以工作.但是 x[...] = 仍然有效并且更通用.x[...] 用于所有其他 nditer 示例.

And if I add the external_loop flag from the next section, x is now a 1d array, and x[:] = would work. But x[...] = still works and is more general. x[...] is used all the other nditer examples.

In [677]: for x in np.nditer(a, op_flags=['readwrite'], flags=['external_loop']):
     ...:     print(x, type(x), x.shape)
     ...:     x[...] = 2 * x
[ 0 16 32 48 64 80] <class 'numpy.ndarray'> (6,)

比较这个简单的行迭代(在二维数组上):

Compare this simple row iteration (on a 2d array):

In [675]: for x in a:
     ...:     print(x, x.shape)
     ...:     x[:] = 2 * x
     ...:     
[ 0  8 16] (3,)
[24 32 40] (3,)

这会在 a 的行(第一个暗淡)上进行迭代.x 是一维数组，可以使用 x[:] = ... 或 x[...] = ....