考虑在N-by-M大小的网格上排列的一组点。 我正在尝试建立邻接矩阵 相邻点已连接。
例如,在带有图表的3x3网格中:
1-2-3
| | |
4-5-6
| | |
7-8-9
我们应该有相应的邻接矩阵:
+---+------------------------------------------------------+
| | 1 2 3 4 5 6 7 8 9 |
+---+------------------------------------------------------+
| 1 | 0 1 0 1 0 0 0 0 0 |
| 2 | 1 0 1 0 1 0 0 0 0 |
| 3 | 0 1 0 0 0 1 0 0 0 |
| 4 | 1 0 0 0 1 0 1 0 0 |
| 5 | 0 1 0 1 0 1 0 1 0 |
| 6 | 0 0 1 0 1 0 0 0 1 |
| 7 | 0 0 0 1 0 0 0 1 0 |
| 8 | 0 0 0 0 1 0 1 0 1 |
| 9 | 0 0 0 0 0 1 0 1 0 |
+---+------------------------------------------------------+
作为奖励,该解决方案应适用于4和8连接的相邻点,即:
o o o o
o X o vs. o X o
o o o o
这是我到目前为止的代码:
N = 3; M = 3;
adj = zeros(N*M);
for i=1:N
for j=1:M
k = sub2ind([N M],i,j);
if i>1
ii=i-1; jj=j;
adj(k,sub2ind([N M],ii,jj)) = 1;
end
if i<N
ii=i+1; jj=j;
adj(k,sub2ind([N M],ii,jj)) = 1;
end
if j>1
ii=i; jj=j-1;
adj(k,sub2ind([N M],ii,jj)) = 1;
end
if j<M
ii=i; jj=j+1;
adj(k,sub2ind([N M],ii,jj)) = 1;
end
end
end
如何改进以避免所有循环?
答案 0 :(得分:24)
如果您注意到,您正在创建的邻接矩阵有一个独特的模式。具体来说,它们是对称的banded。如果要创建稀疏矩阵,可以使用diag
函数(或spdiags
函数)轻松创建矩阵。以下是使用上面的示例矩阵作为示例为每种情况创建邻接矩阵的方法:
mat = [1 2 3; 4 5 6; 7 8 9]; % Sample matrix
[r, c] = size(mat); % Get the matrix size
diagVec1 = repmat([ones(c-1, 1); 0], r, 1); % Make the first diagonal vector
% (for horizontal connections)
diagVec1 = diagVec1(1:end-1); % Remove the last value
diagVec2 = ones(c*(r-1), 1); % Make the second diagonal vector
% (for vertical connections)
adj = diag(diagVec1, 1)+diag(diagVec2, c); % Add the diagonals to a zero matrix
adj = adj+adj.'; % Add the matrix to a transposed copy of
% itself to make it symmetric
你将获得以下矩阵:
adj =
0 1 0 1 0 0 0 0 0
1 0 1 0 1 0 0 0 0
0 1 0 0 0 1 0 0 0
1 0 0 0 1 0 1 0 0
0 1 0 1 0 1 0 1 0
0 0 1 0 1 0 0 0 1
0 0 0 1 0 0 0 1 0
0 0 0 0 1 0 1 0 1
0 0 0 0 0 1 0 1 0
mat = [1 2 3; 4 5 6; 7 8 9]; % Sample matrix
[r, c] = size(mat); % Get the matrix size
diagVec1 = repmat([ones(c-1, 1); 0], r, 1); % Make the first diagonal vector
% (for horizontal connections)
diagVec1 = diagVec1(1:end-1); % Remove the last value
diagVec2 = [0; diagVec1(1:(c*(r-1)))]; % Make the second diagonal vector
% (for anti-diagonal connections)
diagVec3 = ones(c*(r-1), 1); % Make the third diagonal vector
% (for vertical connections)
diagVec4 = diagVec2(2:end-1); % Make the fourth diagonal vector
% (for diagonal connections)
adj = diag(diagVec1, 1)+... % Add the diagonals to a zero matrix
diag(diagVec2, c-1)+...
diag(diagVec3, c)+...
diag(diagVec4, c+1);
adj = adj+adj.'; % Add the matrix to a transposed copy of
% itself to make it symmetric
你将获得以下矩阵:
adj =
0 1 0 1 1 0 0 0 0
1 0 1 1 1 1 0 0 0
0 1 0 0 1 1 0 0 0
1 1 0 0 1 0 1 1 0
1 1 1 1 0 1 1 1 1
0 1 1 0 1 0 0 1 1
0 0 0 1 1 0 0 1 0
0 0 0 1 1 1 1 0 1
0 0 0 0 1 1 0 1 0
答案 1 :(得分:14)
只是为了好玩,这里有一个解决方案,通过计算网格上所有点对之间的距离来构建邻接矩阵(显然不是最有效的方式)
N = 3; M = 3; %# grid size
CONNECTED = 8; %# 4-/8- connected points
%# which distance function
if CONNECTED == 4, distFunc = 'cityblock';
elseif CONNECTED == 8, distFunc = 'chebychev'; end
%# compute adjacency matrix
[X Y] = meshgrid(1:N,1:M);
X = X(:); Y = Y(:);
adj = squareform( pdist([X Y], distFunc) == 1 );
这里有一些代码可视化邻接矩阵和连接点图:
%# plot adjacency matrix
subplot(121), spy(adj)
%# plot connected points on grid
[xx yy] = gplot(adj, [X Y]);
subplot(122), plot(xx, yy, 'ks-', 'MarkerFaceColor','r')
axis([0 N+1 0 M+1])
%# add labels
[X Y] = meshgrid(1:N,1:M);
X = reshape(X',[],1) + 0.1; Y = reshape(Y',[],1) + 0.1;
text(X, Y(end:-1:1), cellstr(num2str((1:N*M)')) )
答案 2 :(得分:4)
我在搜索同样的问题时发现了这个问题。但是,由于需要使用稀疏矩阵类型的问题大小,所提供的解决方案都不适用于我。这是我的解决方案,适用于大型实例:
function W = getAdjacencyMatrix(I)
[m, n] = size(I);
I_size = m*n;
% 1-off diagonal elements
V = repmat([ones(m-1,1); 0],n, 1);
V = V(1:end-1); % remove last zero
% n-off diagonal elements
U = ones(m*(n-1), 1);
% get the upper triangular part of the matrix
W = sparse(1:(I_size-1), 2:I_size, V, I_size, I_size)...
+ sparse(1:(I_size-m),(m+1):I_size, U, I_size, I_size);
% finally make W symmetric
W = W + W';
答案 3 :(得分:2)
您当前的代码似乎并不那么糟糕。您需要以这种或那种方式迭代所有邻居对。如果你真的需要优化代码,我建议:
1 <= i <= (N*M)
[i-M, i+1, i+M, i-1]
请注意,要获取所有邻居节点对:
i % M != 0
的“正确”邻居(即水平边缘)(因为Matlab不是基于0而是基于1)i > M
这将导致单个循环(但N * M次迭代次数相同),不调用sub2ind(),并且循环中只有两个if语句。
答案 4 :(得分:2)
刚刚遇到这个问题。我有一个很好的工作m函数(链接:sparse_adj_matrix.m
)非常通用。
它可以处理4连接网格(根据L1规范的半径1),8连接网格(根据L_infty规范的半径1)。
它还可以支持3D(和任意更高的内部网格)
该函数还可以连接半径= 1以外的节点。
这是函数的签名:
% Construct sparse adjacency matrix (provides ii and jj indices into the
% matrix)
%
% Usage:
% [ii jj] = sparse_adj_matrix(sz, r, p)
%
% inputs:
% sz - grid size (determine the number of variables n=prod(sz), and the
% geometry/dimensionality)
% r - the radius around each point for which edges are formed
% p - in what p-norm to measure the r-ball, can be 1,2 or 'inf'
%
% outputs
% ii, jj - linear indices into adjacency matrix (for each pair (m,n)
% there is also the pair (n,m))
%
% How to construct the adjacency matrix?
% >> A = sparse(ii, jj, ones(1,numel(ii)), prod(sz), prod(sz));
%
%
% Example:
% >> [ii jj] = sparse_adj_matrix([10 20], 1, inf);
% construct indices for 200x200 adjacency matrix for 8-connect graph over a
% grid of 10x20 nodes.
% To visualize the graph:
% >> [r c]=ndgrid(1:10,1:20);
% >> A = sparse(ii, jj, 1, 200, 200);;
% >> gplot(A, [r(:) c(:)]);
答案 5 :(得分:0)
对于图表中的每个节点,向右添加一个连接,向下添加一个连接。检查您是否没有超出您的网格。考虑以下构建邻接矩阵的函数。
function adj = AdjMatrixLattice4( N, M )
% Size of adjacency matrix
MN = M*N;
adj = zeros(MN,MN);
% number nodes as such
% [1]---[2]-- .. --[M]
% | | |
% [M+1]-[M+2]- .. -[2*M]
% : : :
% [] [] .. [M*N]
for i=1:N
for j=1:N
A = M*(i-1)+j; %Node # for (i,j) node
if(j<N)
B = M*(i-1)+j+1; %Node # for node to the right
adj(A,B) = 1;
adj(B,A) = 1;
end
if(i<M)
B = M*i+j; %Node # for node below
adj(A,B) = 1;
adj(B,A) = 1;
end
end
end
end
以上示例AdjMatrixLattice4(3,3)=
0 1 0 1 0 0 0 0 0
1 0 1 0 1 0 0 0 0
0 1 0 0 0 1 0 0 0
1 0 0 0 1 0 1 0 0
0 1 0 1 0 1 0 1 0
0 0 1 0 1 0 0 0 1
0 0 0 1 0 0 0 1 0
0 0 0 0 1 0 1 0 1
0 0 0 0 0 1 0 1 0