检查邻居像素Matlab [英] Check neighbour pixels Matlab
问题描述
我有一个A，它是640x1的单元格.其中每个单元格A(i,1)的值随行而变化，例如A(1,1) =[]，而A(2,1)=[1]和A(3,1)=[1,2,3].
还有另一个大小为480x640的矩阵B，其中向量A的row_index (i)对应于矩阵B的col_index.而向量A中每一行的像元值对应于矩阵B中的row_index.例如，A(2,1)= [1]表示矩阵B中的col_2 row_1，而A(3,1)= [1,2,3]表示矩阵
我想做的是针对从向量A引用的矩阵B中的每个非零值，我要检查是否还有至少4个其他邻居也从向量引用A .每个值的邻居数由值N确定.
例如，这是矩阵B的一部分，其中所有零只是为了澄清，因为实际上它们可能是非零"是N=3时像素X的邻居:
I have a A which is 640x1 cell. where the value of each cell A(i,1) varies from row to row, for example A(1,1) =[], while A(2,1)=[1] and A(3,1)=[1,2,3].
There is another matrix B of size 480x640, where the row_index (i) of vector A corresponds to the col_index of matrix B. While the cell value of each row in vector A corresponds to the row_index in matrix B. For example, A(2,1)=[1] this means col_2 row_1 in matrix B, while A(3,1)=[1,2,3] means col_3 rows 1,2&3 in matrix B.
What I'm trying to do is to for each non-zero value in matrix B that are referenced from vector A, I want to check whether there are at least 4 other neighbors that are also referenced from vector A. The number neighbors of each value are determined by a value N.
For example, this is a part of matrix B where all the zeros"just to clarify, as in fact they may be non-zeros" are the neighbors of pixel X when N=3:
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 X 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
如图所示，由于N=3，所有这些零都是像素X的邻居.因此，如果在向量A中找到了4个以上的相邻像素，则执行例如G=1的操作，如果没有，则执行G=0的操作；否则执行G=0操作.
所以，如果有人可以请指教.并请让我知道是否需要进一步澄清.
As shown, because N=3, all these zeros are pixel X's neighbors. So if more than 4 neighbor pixels are found in vector A then do something e.g G=1 if not then G=0;
So if anyone could please advise. And please let me know if any more clarification is needed.
推荐答案
我要做的第一件事是将索引为A的单元格转换为逻辑矩阵Amat.这样可以更轻松地检查A中包含多少个邻居.
The first thing I would do is to convert your cell of indices A to a logic matrix Amat. This makes it easier to check how many neighbours are included in A.
这是使用此转换的解决方案.我希望这些评论足以使它易于理解.
Here is a solution that uses this conversion. I hope the comments are enough to make it understandable.
clear all
clc
nCols = 7;
nRows = 6;
N = 3; %// Number of neighbours
M = 4; %// Minimum number of wanted connections
%// Create cell of indices A
A = cell(nCols,1);
A{1} = [];
A{2} = 1;
A{3} = [1 2 3];
A{4} = [2 5];
A{5} = 3;
A{6} = [3 5];
A{7} = [1 4 6];
%// Generate radom data B
%// (There is a 50% probability for each element of B to be zero)
Bmax = 17;
B = (randi(2,nRows,nCols)-1).*(randi(Bmax,nRows,nCols));
%// Convert the cell A to a logic matrix Amat
Amat = zeros(size(B));
for ii = 1:nCols
Amat(A{ii},ii) = 1;
end
A
B
Amat
for ii = 1:nCols
for jj = A{ii}
if B(jj,ii)>0
%// Calculate neighbour indices with a lower bound of 1
%// and an upper bound of nCols or nRows
col_lim_low = max(1,ii-N);
col_lim_high = min(nCols,ii+N);
row_lim_low = max(1,jj-N);
row_lim_high = min(nRows,jj+N);
%// Get the corresponding neighbouring-matrix from Amat
A_neighbours = ...
Amat(row_lim_low:row_lim_high,col_lim_low:col_lim_high);
%// Check the number of neighbours against the wanted number M
if sum(A_neighbours(:)) > 1 + M
%# do something
fprintf('We should do something here at (%d,%d)\n',jj,ii)
end
end
end
end
以下是一次运行代码的打印输出.
The following is a printout from one run of the code.
A =
[]
[ 1]
[1x3 double]
[1x2 double]
[ 3]
[1x2 double]
[1x3 double]
B =
1 5 0 0 11 0 16
0 13 13 0 0 0 9
0 0 0 5 0 0 0
3 8 16 16 0 2 12
0 0 5 0 9 9 0
12 13 0 6 0 15 0
Amat =
0 1 1 0 0 0 1
0 0 1 1 0 0 0
0 0 1 0 1 1 0
0 0 0 0 0 0 1
0 0 0 1 0 1 0
0 0 0 0 0 0 1
We should do something here at (1,2)
We should do something here at (2,3)
We should do something here at (5,6)
We should do something here at (4,7)
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