Making grid triangular mesh quickly with Numpy

We could play a multi-dimensional game based on slicing and multi-dim assignment that are perfect in NumPy environment on efficiency -

def MakeFacesVectorized1(Nr,Nc):    out = np.empty((Nr-1,Nc-1,2,3),dtype=int)    r = np.arange(Nr*Nc).reshape(Nr,Nc)    out[:,:, 0,0] = r[:-1,:-1]    out[:,:, 1,0] = r[:-1,1:]    out[:,:, 0,1] = r[:-1,1:]    out[:,:, 1,1] = r[1:,1:]    out[:,:, :,2] = r[1:,:-1,None]    out.shape =(-1,3)    return out

Runtime test and verification -

In [226]: Nr,Nc = 100, 100In [227]: np.allclose(MakeFaces(Nr, Nc), MakeFacesVectorized1(Nr, Nc))Out[227]: TrueIn [228]: %timeit MakeFaces(Nr, Nc)100 loops, best of 3: 11.9 ms per loopIn [229]: %timeit MakeFacesVectorized1(Nr, Nc)10000 loops, best of 3: 133 µs per loopIn [230]: 11900/133.0Out[230]: 89.47368421052632

Around 90x speedup for Nr, Nc = 100, 100!

You can achieve a similar result without any explicit loops if you recast the problem correctly. One way would be to imagine the result as three arrays, each containing one of the vertices: first, second and third. You can then zip up or otherwise convert the arrays into whatever format you like in a fairly inexpensive operation.

You start with the actual matrix. This will make indexing and selecting elements much easier:

m = np.arange(Nr * Nc).reshape(Nr, Nc)

The first array will contain all the 90-degree corners:

c1 = np.concatenate((m[:-1, :-1].ravel(), m[1:, 1:].ravel()))

m[:-1, :-1] are the corners that are at the top, m[1:, 1:] are the corners that are at the bottom.

The second array will contain the corresponding top acute corners:

c2 = np.concatenate((m[:-1, 1:].ravel(), m[:-1, 1:].ravel()))

And the third array will contain the bottom corners:

c2 = np.concatenate((m[1:, :-1].ravel(), m[1:, :-1].ravel()))

You can now get an array like your original one back by zipping:

faces = list(zip(c1, c2, c3))

I am sure that you can find ways to improve this algorithm, but it is a start.