Making grid triangular mesh quickly with Numpy

问题

Consider a regular matrix that represents nodes numbered as shown in the figure:

I want to make a list with all the triangles represented in the figure. Which would result in the following 2 dimensional list:[[0,1,4],[1,5,4],[1,2,5],[2,6,5],...,[11,15,14]]

Assuming that the dimensions of the matrix are (NrXNc) ((4X4) in this case), I was able to achieve this result with the following code:

def MakeFaces(Nr,Nc):
    Nfaces=(Nr-1)*(Nc-1)*2
    Faces=np.zeros((Nfaces,3),dtype=np.int32)
    for r in range(Nr-1):
        for c in range(Nc-1):
            fi=(r*(Nc-1)+c)*2
            l1=r*Nc+c
            l2=l1+1
            l3=l1+Nc
            l4=l3+1
            Faces[fi]=[l1,l2,l3]
            Faces[fi+1]=[l2,l4,l3]
    return Faces

However, the double loop operations make this approach quite slow. Is there a way of using numpy in a smart way to do this faster?

回答1:

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, 100

In [227]: np.allclose(MakeFaces(Nr, Nc), MakeFacesVectorized1(Nr, Nc))
Out[227]: True

In [228]: %timeit MakeFaces(Nr, Nc)
100 loops, best of 3: 11.9 ms per loop

In [229]: %timeit MakeFacesVectorized1(Nr, Nc)
10000 loops, best of 3: 133 µs per loop

In [230]: 11900/133.0
Out[230]: 89.47368421052632

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

回答2:

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.

来源：https://stackoverflow.com/questions/44934631/making-grid-triangular-mesh-quickly-with-numpy

标签

python

numpy

mesh

triangular