Question

我试图在2D二进制矩阵中计算岛屿的数量（一组连接的1形成一个岛）。

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]
]

在上面的矩阵中有5个岛，它们是：

First: (0,0), (0,1), (1,1), (2,0)
Second: (1,4), (2,3), (2,4)
Third: (4,0)
Fourth: (4,2)
Fifth: (4,4)

为了计算2D矩阵中岛的数量，我假设矩阵为图形，然后我使用DFS类算法来计算岛屿。

我正在跟踪DFS（递归函数）调用的数量，因为图中有很多组件。

以下是我为此目的编写的代码：

# 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输出错误。我得到7，但有5个群集。

我尝试调试任何逻辑错误的代码。但我无法找出问题所在。

Answer 1

大锤方法，供参考

必须添加structure参数np.ones((3,3))才能添加对角线连接

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

你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，因为您希望继续计数，但至少有一个坐标与您的中心不同。

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

另一种更直观的方法是编写以下内容

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

在2D二进制矩阵中找到岛的数量

2 个答案: