Efficiently generate a lattice of points in python

Question

Help make my code faster: My python code needs to generate a 2D lattice of points that fall inside a bounding rectangle. I kludged together some code (shown below) that generates this lattice. However, this function is called many many times and has become a serious bottleneck in my application.

I'm sure there's a faster way to do this, probably involving numpy arrays instead of lists. Any suggestions for a faster, more elegant way to do this?

Description of the function: I have two 2D vectors, v1 and v2. These vectors define a lattice. In my case, my vectors define a lattice that is almost, but not quite, hexagonal. I want to generate the set of all 2D points on this lattice that are in some bounding rectangle. In my case, one of the corners of the rectangle is at (0, 0), and the other corners are at positive coordinates.

Example: If the far corner of my bounding rectangle was at (3, 3), and my lattice vectors were:

v1 = (1.2, 0.1)
v2 = (0.2, 1.1)

I'd want my function to return the points:

(1.2, 0.1) #v1
(2.4, 0.2) #2*v1
(0.2, 1.1) #v2
(0.4, 2.2) #2*v2
(1.4, 1.2) #v1 + v2
(2.6, 1.3) #2*v1 + v2
(1.6, 2.3) #v1 + 2*v2
(2.8, 2.4) #2*v1 + 2*v2

I'm not concerned with edge cases; it's doesn't matter if the function returns (0, 0), for example.

The slow way I'm currently doing it:

import numpy, pylab

def generate_lattice( #Help me speed up this function, please!
    image_shape, lattice_vectors, center_pix='image', edge_buffer=2):

    ##Preprocessing. Not much of a bottleneck:
    if center_pix == 'image':
        center_pix = numpy.array(image_shape) // 2
    else: ##Express the center pixel in terms of the lattice vectors
        center_pix = numpy.array(center_pix) - (numpy.array(image_shape) // 2)
        lattice_components = numpy.linalg.solve(
            numpy.vstack(lattice_vectors[:2]).T,
            center_pix)
        lattice_components -= lattice_components // 1
        center_pix = (lattice_vectors[0] * lattice_components[0] +
                      lattice_vectors[1] * lattice_components[1] +
                      numpy.array(image_shape)//2)
    num_vectors = int( ##Estimate how many lattice points we need
        max(image_shape) / numpy.sqrt(lattice_vectors[0]**2).sum())
    lattice_points = []
    lower_bounds = numpy.array((edge_buffer, edge_buffer))
    upper_bounds = numpy.array(image_shape) - edge_buffer

    ##SLOW LOOP HERE. 'num_vectors' is often quite large.
    for i in range(-num_vectors, num_vectors):
        for j in range(-num_vectors, num_vectors):
            lp = i * lattice_vectors[0] + j * lattice_vectors[1] + center_pix
            if all(lower_bounds < lp) and all(lp < upper_bounds):
                lattice_points.append(lp)
    return lattice_points


##Test the function and display the output.
##No optimization needed past this point.
lattice_vectors = [
    numpy.array([-40., -1.]),
    numpy.array([ 18., -37.])]
image_shape = (1000, 1000)
spots = generate_lattice(image_shape, lattice_vectors)

fig=pylab.figure()
pylab.plot([p[1] for p in spots], [p[0] for p in spots], '.')
pylab.axis('equal')
fig.show()

Answer 1

Since lower_bounds and upper_bounds are only 2-element arrays, numpy might not be the right choice here. Try to replace

if all(lower_bounds < lp) and all(lp < upper_bounds):

with basic Python stuff:

if lower1 < lp and lower2 < lp and lp < upper1 and lp < upper2:

According to timeit, the second approach is much faster:

>>> timeit.timeit('all(lower < lp)', 'import numpy\nlp=4\nlower = numpy.array((1,5))') 
3.7948939800262451
>>> timeit.timeit('lower1 < 4 and lower2 < 4', 'lp = 4\nlower1, lower2 = 1,5') 
0.074192047119140625

From my experience, as long as you don't need to handle n-diminsional data and if you don't need double precision floats, it is generally faster to use basic Python data types and constructs instead of numpy, which is a bit overload in such cases -- have a look at this other question.

---

Another minor improvement could be to compute range(-num_vectors, num_vectors) only once and then reuse it. Additionally you might want to use a product iterator instead of a nested for loop -- though I don't think that these changes will have a significant influence on performance.

Answer 2

If you want to vectorize the whole thing, generate a square lattice and then shear it. Then chop off the edges that land outside your box.

Here is what I came up with. There are still a lot of improvements that could be made, but this is the basic idea.

def generate_lattice(image_shape, lattice_vectors) :
    center_pix = numpy.array(image_shape) // 2
    # Get the lower limit on the cell size.
    dx_cell = max(abs(lattice_vectors[0][0]), abs(lattice_vectors[1][0]))
    dy_cell = max(abs(lattice_vectors[0][1]), abs(lattice_vectors[1][1]))
    # Get an over estimate of how many cells across and up.
    nx = image_shape[0]//dx_cell
    ny = image_shape[1]//dy_cell
    # Generate a square lattice, with too many points.
    # Here I generate a factor of 4 more points than I need, which ensures 
    # coverage for highly sheared lattices.  If your lattice is not highly
    # sheared, than you can generate fewer points.
    x_sq = np.arange(-nx, nx, dtype=float)
    y_sq = np.arange(-ny, nx, dtype=float)
    x_sq.shape = x_sq.shape + (1,)
    y_sq.shape = (1,) + y_sq.shape
    # Now shear the whole thing using the lattice vectors
    x_lattice = lattice_vectors[0][0]*x_sq + lattice_vectors[1][0]*y_sq
    y_lattice = lattice_vectors[0][1]*x_sq + lattice_vectors[1][1]*y_sq
    # Trim to fit in box.
    mask = ((x_lattice < image_shape[0]/2.0)
             & (x_lattice > -image_shape[0]/2.0))
    mask = mask & ((y_lattice < image_shape[1]/2.0)
                    & (y_lattice > -image_shape[1]/2.0))
    x_lattice = x_lattice[mask]
    y_lattice = y_lattice[mask]
    # Translate to the centre pix.
    x_lattice += center_pix[0]
    y_lattice += center_pix[1]
    # Make output compatible with original version.
    out = np.empty((len(x_lattice), 2), dtype=float)
    out[:, 0] = y_lattice
    out[:, 1] = x_lattice
    return out