Increment given indices in a matrix
numpy, python
Solution
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`x`m`, 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`.)
Problem
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: ``` array([1, 2, 1, 1, 2]) array([2, 1, 1, 1, 1]) ``` together they make indices that should be incremented: ``` >>> np.array([a, b]).T array([[1, 2], [2, 1], [1, 1], [1, 1], [2, 1]]) ``` I want to get this matrix: ``` array([[0, 0, 0], [0, 2, 1], # (1,1) twice, (1,2) once [0, 2, 0]]) # (2,1) twice ``` The matrix will be small (like, 5×5), and the number of indices will be large (somewhere near 10^3 or 10^5). So, is there anything better (faster) than a `for`-loop?