Dada la siguiente matriz numérica

import numpy as np
np_matrix = np.array([[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,2,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,2,0,0,1,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0]
                    ,[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,0]
                    ,[0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
                    ,[0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1]
                    ,[0,0,0,1,0,0,0,2,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,0]
                    ,[1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]
                    ,[2,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0]])

print(np_matrix)
print(np_matrix.shape)

Que se puede representar como una imagen.

enter image description here

A continuación, se muestran algunas suposiciones sobre la matriz de entrada:

  • La forma de la matriz puede variar. No es un tamaño fijo.
  • El número de cuadrados varía de una matriz a otra. Pero siempre hay al menos un cuadrado en la matriz.
  • Para cada punto verde hay un cuadrado alrededor y los cuadrados tienen un solo punto dentro. El punto dentro del cuadrado está marcado con el número 2
  • Los cuadrados son secuenciales en el eje X y los bordes del cuadrado nunca se superponen con los bordes de otros cuadrados. Los bordes de los cuadrados están marcados con el número 1
  • El espacio en blanco está marcado con el número 0

Tengo dos preguntas. De la manera más eficiente, ¿cómo puedo obtener una matriz con las coordenadas de todos los puntos verdes? ¿Y cómo puedo obtener una matriz con todas las esquinas de los cuadrados alrededor de un punto verde?

enter image description here

Aquí está el resultado que estoy buscando para este ejemplo:

green_dots_coordinates = [
    [0,0],  # dot 1 with coordinates x1, y1 inside square 1
    [7,12], # dot 2 with coordinates x2, y2 inside square 2
    [16,27],# dot 3 with coordinates x3, y3 inside square 3
    [29,21],# dot 4 with coordinates x4, y4 inside square 4
    [34,7], # dot 5 with coordinates x5, y5 inside square 5
]



sqaures_corners_coordinates = [
    #square nr 1
    [
        [0,6], # x1, y1
        [2,6], # x2, y2
        [0,0], # x3, y3
        [2,0], # x4, y4
    ],

    #square nr 2
    [
        [3,14], # x1, y1
        [9,14], # x2, y2
        [3,7], # x3, y3
        [9,7], # x4, y4
    ],

    #square nr 3
    [
        [12,31], # x1, y1
        [23,31], # x2, y2
        [12,24], # x3, y3
        [23,24], # x4, y4
    ],

    #square nr 4
    [
        [25,22], # x1, y1
        [32,22], # x2, y2
        [25,15], # x3, y3
        [32,15], # x4, y4
    ]

    ,#square nr 5
    [
        [33,13], # x1, y1
        [35,13], # x2, y2
        [33,0], # x3, y3
        [35,0], # x4, y4
    ],

]
3
RaduS 12 nov. 2017 a las 22:52

2 respuestas

La mejor respuesta

Su formato de salida es francamente bastante extraño y requiere invertir mucho el orden de los índices (y hace que la salida sea bastante inútil para indexar la matriz original) pero esto funciona:

def find_boxes(np_matrix):
    np_mat = np_matrix[::-1, :]  # reversed in expected output
    def find_extent(arr, val):
        xn = arr.size
        x0 = np.flatnonzero(arr == 1)
        xi = np.searchsorted(x0, val, side = 'right')
        if xi == x0.size:
            x1 = x0[xi-1]
            x2 = xn - 1
        elif xi == 0:
            x1 = 0
            x2 = x0[xi]
        else:
            x1 = x0[xi-1]
            x2 = x0[xi]
        return np.array([x1, x2])

    green = np.where(np_mat == 2)
    green = tuple(g[np.argsort(green[-1])] for g in green)
    coords = np.empty((green[0].size, 2, 4))

    for i, (x, y) in enumerate(zip(*green)):
        coords[i, 0] =   np.tile(find_extent(np_mat[x, :], y),       2)
        coords[i, 1] = np.repeat(find_extent(np_mat[:, y], x)[::-1], 2)  # reversed again
    return np.stack(green)[::-1].T, coords.swapaxes(1,2).astype(int)
    # reversed again and transposed

Pruebas:

find_boxes(np_matrix)
Out: 
(array([[ 0,  0],
        [ 7, 12],
        [16, 27],
        [29, 21],
        [34,  7]], dtype=int32), 
 array([[[ 0,  6],
         [ 2,  6],
         [ 0,  0],
         [ 2,  0]],

        [[ 3, 14],
         [ 9, 14],
         [ 3,  7],
         [ 9,  7]],

        [[12, 31],
         [23, 31],
         [12, 24],
         [23, 24]],

        [[25, 22],
         [32, 22],
         [25, 15],
         [32, 15]],

        [[33, 13],
         [35, 13],
         [33,  0],
         [35,  0]]]))
1
Daniel F 13 nov. 2017 a las 08:59

Otro método, que hace que las cuatro direcciones sean simétricas:

rm = m[::-1].T                            # (j,-i) to (x,y)
green = np.where(rm==2)                   # the costly operation
centers = np.vstack(green).T

rm[green] = 0
res = []

for x,y in centers:
    for s in (-1,1):                      # rear/front
        for t in range(2) :               # vertical/horizontal             
            v = *_,save = rm[x,y::s]            
            v[-1] = 1                     # sentinel
            res.append(y + s*v.argmax())  # find the first 1
            v[-1] = save 
            x,y,rm = y,x,rm.T             # turn 

rm[green] = 2  

coordinates = np.array(res).reshape(-1,4) 
corners = coordinates.take([[1,2],[3,2],[1,0],[3,0]],axis=1)    

Esto evita lidiar con el comportamiento incluido / excluido de los cortes de Python y administra los bordes con un sistema centinela.

print(centers);print(corners)

[[ 0  0]
 [ 7 12]
 [16 27]
 [29 21]
 [34  7]]
-----------
[[[ 0  6]
  [ 2  6]
  [ 0  0]
  [ 2  0]]

 [[ 3 14]
  [ 9 14]
  [ 3  7]
  [ 9  7]]

 [[12 31]
  [23 31]
  [12 24]
  [23 24]]

 [[25 22]
  [32 22]
  [25 15]
  [32 15]]

 [[33 13]
  [35 13]
  [33  0]
  [35  0]]]
1
B. M. 15 nov. 2017 a las 12:14