numpy: efficiently summing with index arrays

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Question :

numpy: efficiently summing with index arrays

Suppose I have 2 matrices M and N (both have > 1 columns). I also have an index matrix I with 2 columns — 1 for M and one for N. The indices for N are unique, but the indices for M may appear more than once. The operation I would like to perform is,

for i,j in w:
  M[i] += N[j]

Is there a more efficient way to do this other than a for loop?

Answer #1:

For completeness, in numpy >= 1.8 you can also use np.add‘s at method:

In [8]: m, n = np.random.rand(2, 10)

In [9]: m_idx, n_idx = np.random.randint(10, size=(2, 20))

In [10]: m0 = m.copy()

In [11]:, m_idx, n[n_idx])

In [13]: m0 += np.bincount(m_idx, weights=n[n_idx], minlength=len(m))

In [14]: np.allclose(m, m0)
Out[14]: True

In [15]: %timeit, m_idx, n[n_idx])
100000 loops, best of 3: 9.49 us per loop

In [16]: %timeit np.bincount(m_idx, weights=n[n_idx], minlength=len(m))
1000000 loops, best of 3: 1.54 us per loop

Aside of the obvious performance disadvantage, it has a couple of advantages:

  1. np.bincount converts its weights to double precision floats, .at will operate with you array’s native type. This makes it the simplest option for dealing e.g. with complex numbers.
  2. np.bincount only adds weights together, you have an at method for all ufuncs, so you can repeatedly multiply, or logical_and, or whatever you feel like.

But for your use case, np.bincount is probably the way to go.

Answered By: duckworthd

Answer #2:

Using also m_ind, n_ind = w.T, just do M += np.bincount(m_ind, weights=N[n_ind], minlength=len(M))

Answered By: Jaime

Answer #3:

For clarity, let’s define

>>> m_ind, n_ind = w.T

Then the for loop

for i, j in zip(m_ind, n_ind):
    M[i] += N[j]

updates the entries M[np.unique(m_ind)]. The values that get written to it are N[n_ind], which must be grouped by m_ind. (The fact that there’s an n_ind in addition to m_ind is actually tangential to the question; you could just set N = N[n_ind].) There happens to be a SciPy class that does exactly this: scipy.sparse.csr_matrix.

Example data:

>>> m_ind, n_ind = array([[0, 0, 1, 1], [2, 3, 0, 1]])
>>> M = np.arange(2, 6)
>>> N = np.logspace(2, 5, 4)

The result of the for loop is that M becomes [110002 1103 4 5]. We get the same result with a csr_matrix as follows. As I said earlier, n_ind isn’t relevant, so we get rid of that first.

>>> N = N[n_ind]
>>> from scipy.sparse import csr_matrix
>>> update = csr_matrix((N, m_ind, [0, len(N)])).toarray()

The CSR constructor builds a matrix with the required values at the required indices; the third part of its argument is a compressed column index, meaning that the values N[0:len(N)] have the indices m_ind[0:len(N)]. Duplicates are summed:

>>> update
array([[ 110000.,    1100.]])

This has shape (1, len(np.unique(m_ind))) and can be added in directly:

>>> M[np.unique(m_ind)] += update.ravel()
>>> M
array([110002,   1103,      4,      5])
Answered By: seberg

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