从一组点中找到近似矩形

Question

I have to extract an area from a image (can't disclose that image). With the help of this Abid K Rahman's answer I've obtained the image as Result Image I've obtained the points of that may make a rectangle but couldn't find a rectangle which is the best approximate.

    [[625, 389], [10, 385], [116, 184], [5, 35], [626, 26]]

We can assume that all four points of the rectangle will be in four different corner of image. So I have divided the set into four different group based on their location.

    [[[[41, 63]], [[613, 66]]], [[[227, 428], [25, 426], [39, 392]], [[612, 394]]]]

But I'm not able to move further. I want to extract points which approximately makes a rectangle. Better if the answer is in python.

Answer 1

So I went on to write my the code using something rayryeng suggested and come up with this.

c = [[[] for x in range(2)] for y in range (2)]
for xy in centroids:
    x_i = xy[0]*2/col
    y_i = xy[1]*2/row
    c[y_i][x_i].append(np.array(xy))

combination = []
combination = np.array([ np.array([q1,q2,q3,q4]) for q1 in c[0][0] for q2 in c[0][1] for q3 in c[1][1] for q4 in c[1][0]])

if len(combination)>0 :
    key = 0
    property = [[ 0 for j in range(4)] for i in range(len(combination))]
    for i in range(len(combination)):
        q = combination[i]
        d1,d2 = q[2]-q[0],q[1]-q[3]
        d1_len,d2_len = np.sqrt(sum(d1**2)),np.sqrt(sum(d2**2))
        angle = math.degrees(math.acos(sum(d1*d2)/(d1_len*d2_len)))
        if d1_len > d2_len:
            r,extent = int(100*d1_len/d2_len),d2_len
        else:
            r,extent = int(100*d2_len/d1_len),d1_len
        property[i] = [r,angle,extent,i]
    property.sort(key = lambda x:x[0])
    key = property[0][3]
    combination = combination[key]

Here centroids is the points I have got, which I divided in to quadrants, since in my problem the points need to be from all four qaudrants. I've made the all possible combination of four points. I found the diagonals, the angle the make at the centre. Made a list of which contain the ratio of length of diagonal(smallest to largest), the angle and the length of shortest diagonal. I used the property that the diagonal of rectangle are of equal length. Which is quite useful for now. I don't know how to use the other properties hence its kept on hold. In final answer I get the best rectangle (when there are no points to make a sqaure, also, it doesn't care about the size of rectangle).

Answer 2

Maybe Python Shapely library can help.

Example:

from shapely.geometry import MultiPoint
points = MultiPoint([(0.0, 0.0), (1.0, 1.0), ... ])

points.bounds
# A (minx, miny, maxx, maxy) tuple.

Answer 3

best thing is imcrop(I,rect); where rect is define as

rect=[xmin ymin width height]

so at first pair for example [41,63],[613,66]

xmin=41

ymin=63

width=613-41

height=66-63