在二维二元矩阵中求岛的个数

Question

我试图在一个二维二进制矩阵中计算岛的数目（一组相连的1形成一个岛）。在

示例：

[
[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[1, 0, 1, 0, 1]
]

在上述矩阵中，有5个岛屿，分别是：

^{pr2}$

为了计算二维矩阵中的岛数，我假设矩阵是一个图，然后使用DFS算法来计算岛数。在

我跟踪DFS（递归函数）调用的数量，因为图中会有许多组件。在

下面是我为此编写的代码：

# count the 1's in the island
def count_houses(mat, visited, i, j):
    # base case
    if i < 0 or i >= len(mat) or j < 0 or j >= len(mat[0]) or\
            visited[i][j] is True or mat[i][j] == 0:
        return 0
    # marking visited at i, j
    visited[i][j] = True
    # cnt is initialized to 1 coz 1 is found
    cnt = 1
    # now go in all possible directions (i.e. form 8 branches)
    # starting from the left upper corner of i,j and going down to right bottom
    # corner of i,j
    for r in xrange(i-1, i+2, 1):
        for c in xrange(j-1, j+2, 1):
            # print 'r:', r
            # print 'c:', c
            # don't call for i, j
            if r != i and c != j:
                cnt += count_houses(mat, visited, r, c)
    return cnt


def island_count(mat):
    houses = list()
    clusters = 0
    row = len(mat)
    col = len(mat[0])
    # initialize the visited matrix
    visited = [[False for i in xrange(col)] for j in xrange(row)]
    # run over matrix, search for 1 and then do dfs when found 1
    for i in xrange(row):
        for j in xrange(col):
            # see if value at i, j is 1 in mat and val at i, j is False in
            # visited
            if mat[i][j] == 1 and visited[i][j] is False:
                clusters += 1
                h = count_houses(mat, visited, i, j)
                houses.append(h)
    print 'clusters:', clusters
    return houses


if __name__ == '__main__':
    mat = [
        [1, 1, 0, 0, 0],
        [0, 1, 0, 0, 1],
        [1, 0, 0, 1, 1],
        [0, 0, 0, 0, 0],
        [1, 0, 1, 0, 1]
    ]
    houses = island_count(mat)
    print houses
    # print 'maximum houses:', max(houses)

我在参数中传递的矩阵的输出错误。我得到了7，但是有{}簇。在

我试着调试代码中的任何逻辑错误。但我找不到问题出在哪里。在

Answer 1

大锤法，供参考

必须添加structure参数np.ones((3,3))来添加对角线连接

import numpy as np
from scipy import ndimage

ary = np.array([
[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[1, 0, 1, 0, 1]
])

labeled_array, num_features = ndimage.label(ary, np.ones((3,3)))

labeled_array, num_features
Out[183]: 
(array([[1, 1, 0, 0, 0],
        [0, 1, 0, 0, 2],
        [1, 0, 0, 2, 2],
        [0, 0, 0, 0, 0],
        [3, 0, 4, 0, 5]]), 5)

Answer 2

您的算法几乎是正确的，除了第21行：

if r != i and c != j:
    cnt += count_houses(mat, visited, r, c)

相反，您希望使用or继续计数，前提是至少有一个坐标与您的中心不相同。在

^{pr2}$

另一种更直观的方法是

if (r, c) != (i, j):
    cnt += count_houses(mat, visited, r, c)