如何在numpy中通过索引累积数组？ [重复]

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Increment Numpy array with repeated indices 4 answers

增加Numpy数组,重复索引4个答案

I have an array:

我有一个数组:

a = np.array([0,0,0,0,0,0])

I want to add some other array into each index of a, while the index can appear more than one times. I want to get the some of each index. I write:

我想在a的每个索引中添加一些其他数组,而索引可以出现多次。我想获得每个索引的一些内容。我写:

a[np.array([1,2,2,1,3])] += np.array([1,1,1,1,1])

but get a to be:

但得到一个:

array([0, 1, 1, 1, 0, 0])

But what I want is to get:

但我想要的是得到:

array([0, 2, 2, 1, 0, 0])

How to implement this in numpy without for loop?

如何在没有for循环的情况下实现这个numpy?

4 个解决方案

#1

Using pure numpy, AND avoiding a for loop:

使用纯numpy,并避免for循环:

np.add.at(a, np.array([1,2,2,1,3]), np.array([1,1,1,1,1]))

Output:

>>> a = np.array([0,0,0,0,0,0])
>>> np.add.at(a, np.array([1,2,2,1,3]), np.array([1,1,1,1,1]))
>>> a
array([0, 2, 2, 1, 0, 0])

Please note, this does in-place substitution. This is what is desired by you, but it may not be desired by future viewers. Hence the note :)

请注意,这会进行就地替换。这是您所期望的,但未来的观众可能并不希望这样。因此注意:)

#2

You could always just iterate yourself. Something like:

你总是可以自己迭代。就像是:

for i in [1,2,2,1,3]:
    a[i] += 1

#3

I don't know of a clever numpy vectorized way to do this... the best I can come up with is:

我不知道一个聪明的numpy矢量化方式来做到这一点...我能想到的最好的是:

>>> indices = np.array([1,2,2,1,3])
>>> values = np.array([1,1,1,1,1])
>>> a = np.array([0,0,0,0,0,0])
>>> for i, ix in enumerate(indices):
...   a[ix] += values[i]
... 
>>> a
array([0, 2, 2, 1, 0, 0])

#4

You can do something like (assuming for each index there is a correlated value):

您可以执行类似的操作(假设每个索引都有相关值):

a = np.array([0,0,0,0,0,0])
idxs = np.array([1,2,2,1,3])
vals = np.array([1,1,1,1,1])
for idx, val in zip(idxs,vals):
    a[idx] += val

#1