numpy.array .__ iadd__和重复索引 [英] numpy.array.__iadd__ and repeated indices

问题描述

我有一个数组:

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

以及具有重复索引的列表:

and list of indices with repetitions:

idx = [0, 0, 1, 1, 2, 2]

和另一个我想使用上面的索引添加到A的数组:

and another array i would like to add to A using indices above:

B = np.array([1, 1, 1, 1, 1, 1])

操作:

A[idx] += B

给出结果:array([1, 1, 1])，因此显然没有对B中的值求和.结果array([2, 2, 2])的最佳获取方式是什么?我必须遍历索引吗?

Gives the result: array([1, 1, 1]), so obviously values from B were not summed up. What is the best way to get as a result array([2, 2, 2])? Do I have to iterate over indices?

推荐答案

为此numpy 1.8添加了at简化:

for this numpy 1.8 added the at reduction:

at(a，index，b = None)

at(a, indices, b=None)

对元素的操作数'a'执行无缓冲的就地操作 由索引"指定.对于加法ufunc，此方法等效 到a[indices] += b，不同之处在于元素的结果是累积的 被多次索引的索引.例如，a[[0,0]] += 1将 由于缓冲，仅将第一个元素递增一次，而 add.at(a, [0,0], 1)将第一个元素增加两次.

Performs unbuffered in place operation on operand 'a' for elements specified by 'indices'. For addition ufunc, this method is equivalent to a[indices] += b, except that results are accumulated for elements that are indexed more than once. For example, a[[0,0]] += 1 will only increment the first element once because of buffering, whereas add.at(a, [0,0], 1) will increment the first element twice.

..版本添加:: 1.8.0

.. versionadded:: 1.8.0

In [1]: A = np.array([0, 0, 0])
In [2]: B = np.array([1, 1, 1, 1, 1, 1])
In [3]: idx = [0, 0, 1, 1, 2, 2]
In [4]: np.add.at(A, idx, B)
In [5]: A
Out[5]: array([2, 2, 2])

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