问题

Here is the translation of the code portion you provided:

好的，我尽力在标题中描述我的问题。

我的问题如下：

我有一个numpy数组，它的形状/维度可能不总是一致的（范围从1到3）。以最简单的情况，当数组的形状为[100]时，我可以执行以下操作（并获得所需的结果）：

```python
for i, bounds in enumerate(values):
    low, high = bounds
    arr[i] *= high - low

当数组的形状为[100, 200]时，我可以执行以下操作：

for i, bounds in enumerate(values):
    low, high = bounds
    arr[i, :] *= high - low

或者如果数组的形状为[200, 100]，我可以执行以下操作：

for i, bounds in enumerate(values):
    low, high = bounds
    arr[:, i] *= high - low

而在3D情况下，如果数组的形状为[300, 100, 200]，我会这样做：

for i, bounds in enumerate(values):
    low, high = bounds
    arr[:, i, :] *= high - low

我的问题是，我不知道如何改变i在索引中的位置，或者如何在迭代与i对应的轴上索引所有元素（当arr的形状正在改变时）。在我的示例中，i的“位置”是基于数组形状中100的位置。这是NumPy可以做到的吗，还是我必须使用多个if语句？

<details>
<summary>英文:</summary>

Ok I did my best to describe my problem in the title.

My problem is as follows:

I have a numpy array which may not always have a consistent shape/dimension (ranges from 1 to 3). Taking the simplest case when the array is of shape [100], I can perform the following (and obtain the desired result):

    for i, bounds in enumerate(values):
        low, high = bounds
        arr[i] *= high - low

when the array is of shape [100, 200], I can do the following:

    for i, bounds in enumerate(values):
        low, high = bounds
        arr[i, :] *= high - low

or if the array is of shape [200, 100], I can instead do:

    for i, bounds in enumerate(values):
        low, high = bounds
        arr[:, i] *= high - low

and in the 3d case if the array is of shape [300, 100, 200], I would do:

    for i, bounds in enumerate(values):
        low, high = bounds
        arr[:, i, :] *= high - low

My problem is I don&#39;t know how to alter the position of `i` in the index or how to index all the elements when iterating over the axis that corresponds to `i` (when the shape of `arr` is changing). In my example, the &quot;location&quot; of `i` is based on where `100` falls in the shape of the array. Is this something numpy can do or am I stuck with a number of if statements? 

</details>


# 答案1
**得分**: 2

有一种可能性是将轴移动，以将您想要的轴（根据您的标准）放在新视图 `a2` 中的第一个位置。然后，`a2[i,:,:]` 就等同于 `arr[:,i,:]`（例如），如果正确的轴是1。此外，通过这种方式，您可以使用 `...` 表示法来替代所有的 `:`。所以 `a2[i,...]`。

```python
# 将大小为100的轴移动到第一个位置（索引0）在a2视图中
# （a2是arr的视图。因此，修改其内容会更改arr的内容）。
# 请注意，如果第一个轴已经是大小为100，则不会发生任何变化
a2 = np.moveaxis(arr, arr.shape.index(100), 0)

for i, (low, high) in enumerate(values):
    a2[i, ...] *= high - low
    # 请注意，这个 `...` 在任何维度上都可以工作，甚至是1。

# Move axis whose size is 100 in 1st position (index 0) in a2 view
# (a2 is a view of arr. So modifying its contents change arr content).
# Note, if the 1st axis is already of size 100 this will do nothing
a2 = np.moveaxis(arr, arr.shape.index(100), 0) 

for i, (low, high) in enumerate(values):
    a2[i, ...] *= high - low
    # Note this `...` works whatever the dimension, even 1.

答案2

得分: 1

你可以动态创建索引。索引集合是一个元组，: 对应 slice(None)。

找到与值的长度相同的维度中的索引。然后在循环中更新索引列表中的该索引。

indexes = [slice(None)] * arr.ndims
variable_pos = arr.shape.index(len(values))
for i, (low, high) in enumerate(values):
    indexes[variable_pos] = i
    arr[tuple(indexes)] *= high - low

indexes = [slice(None)] * arr.ndims
variable_pos = arr.shape.index(len(values))
for i, (low, high) in enumerate(values):
    indexes[variable_pos] = i
    arr[tuple(indexes)] *= high - low