numpy.array.__iadd__ and repeated indices [duplicate]

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Question :

numpy.array.__iadd__ and repeated indices [duplicate]

I have an array:

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

and list of indices with repetitions:

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

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

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

The operation:

A[idx] += B

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?

Asked By: mrkwjc

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Answer #1:

for this numpy 1.8 added the at reduction:

at(a, indices, b=None)

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.

.. 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])
Answered By: jtaylor

Answer #2:

How about:

A = np.array([1, 2, 3])
idx = [0, 0, 1, 1, 2, 2]
A += np.bincount(idx, minlength=len(A))

Obviously it’s even more simple if A starts off as zeros:

A = np.bincount(idx)
Answered By: Bi Rico

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