How to measure the aspect ratio of a certain part of an image

I am working with the mnist dataset and I am trying to calculate the aspect ratio of every number. The aspect ratio is defined as the minimum ractangular area that the number covers any idea how to do that?

[Rik]

What have you tried so far? You must have had an idea.

[Adam Danz]

If the rectangle is always parallel with the axes, you could use the boundingbox property of regionprops() to get the size of the rectangle.

Answer 1

[Ameer Hamza]

Try this:

% im is the MNIST image
sum_row = sum(im, 2);
sum_col = sum(im, 1);

min_row = find(sum_row, 1);
max_row = find(sum_row, 1, 'last');
min_col = find(sum_col, 1);
max_col = find(sum_col, 1, 'last');

width = max_col-min_col+1;
height = max_row-min_row+1;

aspect_ratio = height/width;

This solution is also much faster then regionprops() and both give the same result as shown by following example

tic
sum_row = sum(im, 2);
sum_col = sum(im, 1);

min_row = find(sum_row, 1);
max_row = find(sum_row, 1, 'last');
min_col = find(sum_col, 1);
max_col = find(sum_col, 1, 'last');

width = max_col-min_col+1;
height = max_row-min_row+1;

aspect_ratio1 = height/width;
toc

tic
props = regionprops(imbinarize(im));
aspect_ratio2 = props.BoundingBox(4)/props.BoundingBox(3);
toc

Eq = isequal(aspect_ratio1, aspect_ratio2);

Result:

t1 =
   2.6254e-04

>> t2
t2 =
    0.0145

>> Eq
Eq =
  logical
   1

[Rik]

When I'm back at a computer I'll have a look if a single call to find would be even faster.

[Ameer Hamza]

Rik, thats correct. Single call to find find is indeed faster

images = data(:,2:785);

t1 = zeros(100,1);
for i=1:100
    im = reshape(images(i,:), 28, [])';
    tic
    sum_row = sum(im, 2);
    sum_col = sum(im, 1);
    min_row = find(sum_row, 1);
    max_row = find(sum_row, 1, 'last');
    min_col = find(sum_col, 1);
    max_col = find(sum_col, 1, 'last');
    width = max_col-min_col+1;
    height = max_row-min_row+1;
    aspect_ratio1 = height/width;
    t1(i) = toc;
end

t2 = zeros(100,1);
for i=1:100
    im = reshape(images(i,:), 28, [])';
    tic
    [rows, cols] = find(im);
    height2 = max(rows)-min(rows)+1;
    width2 = max(cols)-min(cols)+1;
    aspect_ratio2 = height2/width2;
    t2(i) = toc;
end

Result:

mean(t1)
ans =
   2.2505e-05

>> mean(t2)
ans =
   1.8409e-05

[Ameer Hamza]

But In long run, the first version does seem a bit faster. The following code runs 10000 iterations

t1 = zeros(10000,1);
for i=1:10000
    im = reshape(images(i,:), 28, [])';
    tic
    sum_row = sum(im, 2);
    sum_col = sum(im, 1);
    min_row = find(sum_row, 1);
    max_row = find(sum_row, 1, 'last');
    min_col = find(sum_col, 1);
    max_col = find(sum_col, 1, 'last');
    width = max_col-min_col+1;
    height = max_row-min_row+1;
    aspect_ratio1 = height/width;
    t1(i) = toc;
end

t2 = zeros(10000,1);
for i=1:10000
    im = reshape(images(i,:), 28, [])';
    tic
    [rows, cols] = find(im);
    height2 = max(rows)-min(rows)+1;
    width2 = max(cols)-min(cols)+1;
    aspect_ratio2 = height2/width2;
    t2(i) = toc;
end

Result

mean(t1)
ans =
   5.1311e-06

>> mean(t2)
ans =
   5.4781e-06

It is also confirmed by the timeit function

global im
im = reshape(images(1,:), 28, [])';

timeit(@fun1)
timeit(@fun2)

function aspect_ratio1 = fun1()
    global im
    sum_row = sum(im, 2);
    sum_col = sum(im, 1);
    min_row = find(sum_row, 1);
    max_row = find(sum_row, 1, 'last');
    min_col = find(sum_col, 1);
    max_col = find(sum_col, 1, 'last');
    width = max_col-min_col+1;
    height = max_row-min_row+1;
    aspect_ratio1 = height/width;
end

function aspect_ratio2 = fun2()
    global im
    [rows, cols] = find(im);
    height2 = max(rows)-min(rows)+1;
    width2 = max(cols)-min(cols)+1;
    aspect_ratio2 = height2/width2;
end

Result

t1 =
   6.7230e-06

t2 =
   9.7725e-06

[Adam Danz]

Note that if the numbers in the image aren't already segmented, regionprops may come in handy.

[Ameer Hamza]

Yes, this solution is just specific for the MNIST dataset. For a more general case, I will fail.