NumPy数组，更改索引列表中不存在的值

I have a numpy array like:

我有一个像numpy数组：

a = np.arange(30)

I know that I can replace the values located at positions indices=[2,3,4] using for instance fancy indexing:

我知道我可以使用例如花式索引替换位置indices = [2,3,4]的值：

a[indices] = 999

But how to replace the values at the positions that are not in indices? Would be something like below?

但是如何替换不在指数中的位置的值？会是这样的吗？

a[ not in indices ] = 888

Thank you!

谢谢！

4 个解决方案

#1

I don't know of a clean way to do something like this:

我不知道干净的方法做这样的事情：

mask = np.ones(a.shape,dtype=bool) #np.ones_like(a,dtype=bool)
mask[indices] = False
a[~mask] = 999
a[mask] = 888

Of course, if you prefer to use the numpy data-type, you could use dtype=np.bool_ -- There won't be any difference in the output. it's just a matter of preference really.

当然，如果您更喜欢使用numpy数据类型，则可以使用dtype = np.bool_ - 输出中没有任何差异。这只是一个偏好的问题。

#2

Only works for 1d arrays:

仅适用于1d数组：

a = np.arange(30)
indices = [2, 3, 4]

ia = np.indices(a.shape)

not_indices = np.setxor1d(ia, indices)
a[not_indices] = 888

#3

Obviously there is no general not operator for sets. Your choices are:

显然，集合没有一般的非运算符。你的选择是：

Subtracting your indices set from a universal set of indices (depends on the shape of a), but that will be a bit difficult to implement and read.
从一组通用索引中减去索引集（取决于a的形状），但这有点难以实现和阅读。
Some kind of iteration (probably the for-loop is your best bet since you definitely want to use the fact that your indices are sorted).
某种迭代（可能for循环是你最好的选择，因为你肯定想要使用你的索引被排序的事实）。
Creating a new array filled with new value, and selectively copying indices from the old one.

创建一个填充了新值的新数组，并有选择地从旧数组中复制索引。
```
b = np.repeat(888, a.shape)
b[indices] = a[indices]
```

#4

Just overcome similar situation, solved this way:

刚刚克服类似情况，解决了这个问题：

a = np.arange(30)
indices=[2,3,4]

a[indices] = 999

not_in_indices = [x for x in range(len(a)) if x not in indices]

a[not_in_indices] = 888

#1