Python 更改不在索引列表中的 NumPy 数组的值
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Change the values of a NumPy array that are NOT in a list of indices
提问by Saullo G. P. Castro
I have a NumPy array like:
我有一个 NumPy 数组，如：
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indices=[2,3,4]使用例如花式索引替换位于位置的值：
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 中的位置的值indices？会像下面这样吗？
a[ not in indices ] = 888
采纳答案by mgilson
I don't know of a clean way to do something like this:
我不知道做这样的事情的干净方法：
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_-- 输出不会有任何差异。这真的只是一个偏好问题。
回答by kirelagin
Obviously there is no general notoperator for sets. Your choices are:
显然not，集合没有通用运算符。您的选择是：
- Subtracting your
indicesset from a universal set of indices (depends on the shape ofa), but that will be a bit difficult to implement and read. - 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). Creating a new array filled with new value, and selectively copying indices from the old one.
b = np.repeat(888, a.shape) b[indices] = a[indices]
indices从一组通用索引中减去您的集合（取决于 的形状a），但这将有点难以实现和阅读。- 某种迭代（可能
for-loop 是您最好的选择，因为您肯定想使用索引已排序的事实）。 创建一个填充新值的新数组，并有选择地复制旧数组的索引。
b = np.repeat(888, a.shape) b[indices] = a[indices]
回答by aaren
Only works for 1d arrays:
仅适用于一维数组：
a = np.arange(30)
indices = [2, 3, 4]
ia = np.indices(a.shape)
not_indices = np.setxor1d(ia, indices)
a[not_indices] = 888
回答by Alexey Trofimov
Just overcome similar situation, solved this way:
刚刚克服了类似的情况，这样解决了：
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
