Convert to Web Mercator With Numpy

Question

My program vertically stretches a Numpy array, representing a 180 by 360 map image, so it represents a Web Mercator map image.

I wrote a function (below) that does what I want - but it is crazy slow (takes like five minutes). Is there a much faster and easier way to do this? Maybe using Numpy interpolate2d or MatPlotLib?

def row2lat(row):
  return 180.0/math.pi*(2.0*math.atan(math.exp(row*math.pi/180.0))-math.pi/2.0)

def mercator(geodetic):
    geo = np.repeat(geodetic, 2, axis=0)
    merc = np.zeros_like(geo)
    side = geo[0].size
    for row in range(side):
        lat = row2lat(180 - ((row * 1.0)/side) * 360)
        g_row = (abs(90 - lat)/180)*side
        fraction = g_row-math.floor(g_row)
        for col in range(side):
            high_row = geo[math.floor(g_row)][col] * (fraction)
            low_row = geo[math.ceil(g_row)][col] * (1-fraction)
            merc[row][col] = high_row + low_row
    return merc

Answer 1

Try to avoid the inner for loop and vectorize your functions. Numpy is highly optimized to run those things efficient. Your function would then read like

def mercator_faster(geodetic):
    geo = np.repeat(geodetic, 2, axis=0)
    merc = np.zeros_like(geo)
    side = geo[0].size
    for row in range(side):
        lat = row2lat(180 - ((row * 1.0)/side) * 360)
        g_row = (abs(90 - lat)/180)*side
        fraction = g_row-math.floor(g_row)

        # Here I optimized the code by using the numpy vector operations 
        # instead of the for loop:

        high_row = geo[math.floor(g_row), :] * (fraction)
        low_row = geo[math.ceil(g_row), :] * (1-fraction)
        merc[row, :] = high_row + low_row

    return merc

If I run it on my machine it takes less then a second:

%timeit mercator_faster(geo)
1 loops, best of 3: 727 ms per loop

And it looks like this (I had to rescale it, because it was too big for SO):

Possibly the outer for loop might be vectorized as well, but I guess this is much harder.