从分形图片用Box counting方法计算分形维数的一个例子

You can still use box count, but doing it smarter :)

Counting boxes with at least 1 white pixel from ImagePartition can be done more efficiently usingIntegral Image, a technique used by Viola-Jones (2004) in their now popular face recognition framework. For a mathematical motivation (and proof), Viola and Jones point to this source.

Actually, someone already asked about a Mathematica implementation here.

What Integral Image allows you to do is to compute efficiently the total mass of any rectangle in an image. So, you can define the following:

IntegralImage[d_] := Map[Accumulate, d, {0, 1}];
data = ImageData[ColorConvert[iEdge, "Grayscale"]]; (* iEdge: your snowflake image *)
ii = IntegralImage[data];

Then, the mass (white content) of a region, is

(* PixelCount: total mass in region delimited by two corner points, given ii, the IntegralImage *)
PixelCount[ii_, {p0x_, p0y_}, {p1x_, p1y_}] := ii[[p1x, p1y]] + ii[[p0x, p0y]] - ii[[p1x, p0y]] - ii[[p0x, p1y]];

So, instead of partitioning the image using ImagePartition, you can get a list of all the boxes of a given size by

PartitionBoxes[{rows_, cols_}, size_] := Transpose /@ Tuples[{Partition[Range[1, rows, size], 2, 1], Partition[Range[1, cols, size], 2, 1]}];

If you apply PixelCount to above, as in your algorithm, you should have the same data but calculated faster.

PixelCountsAtSize[{rows_, cols_}, ii_, size_] :=((PixelCount [ii, #1, #2] &) @@ # &) /@ PartitionBoxes[{rows, cols}, size];

Following your approach here, we should then do

fractalDimensionData = Table[{1/size, Total[Sign /@ PixelCountsAtSize[Dimensions[ii], ii, size]]}, {size, 3, Floor[Min[Dimensions[ii]]/10]}];
line = Fit[Log[fractalDimensionData], {1, x}, x]Out:= 10.4414 + 1.27104 x

which is very close to the actual fractal dimension of the snowflake (which I used as input).

Two things to note. Because this is faster, I dared to generate the table at box size 3. Also, unlike ImagePartition, my partition boxes are all of the same size and therefore, it excludes uneven boxes at the edges. So, instead of doing minSize/2 as you did, I put minSize/10 - excluding bigger and misleading values for big boxes.

Hope this helps.

Update

Just ran the algorithm starting with 2 and got this 10.4371 + 1.27008 x. And starting with 1 is 10.4332 + 1.26919 x, much better. Of course, it takes longer but still under or around 1 min for your snowflake image.

Update 2

And finally, for your image from Google Earth (eqypt2.jpg) the output is (starting at 1-pixel boxes)

12.1578 + 1.47597 x

It ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs.