Briefly: there is a similar question and the best answer suggests using numpy.bincount
. I need the same thing, but for a matrix.
I\'ve got two arrays:>
You can still use bincount(). The trick is to convert a
and b
into a single 1D array of flat indices.
If the matrix is n
xm
, you could apply bincount()
to a * m + b
, and construct the matrix from the result.
To take the example in your question:
In [15]: a = np.array([1, 2, 1, 1, 2])
In [16]: b = np.array([2, 1, 1, 1, 1])
In [17]: cnt = np.bincount(a * 3 + b)
In [18]: cnt.resize((3, 3))
In [19]: cnt
Out[19]:
array([[0, 0, 0],
[0, 2, 1],
[0, 2, 0]])
If the shape of the array is more complicated, it might be easier to use np.ravel_multi_index() instead of computing flat indices by hand:
In [20]: cnt = np.bincount(np.ravel_multi_index(np.vstack((a, b)), (3, 3)))
In [21]: np.resize(cnt, (3, 3))
Out[21]:
array([[0, 0, 0],
[0, 2, 1],
[0, 2, 0]])
(Hat tip @Jaime for pointing out ravel_multi_index
.)