Matlab：保持非零矩阵元素彼此相邻并忽略单独的元素
[英]Matlab: keeping non-zero matrix elements adjacent to each other and ignoring lone elements
问题描述
Here is an example matrix (but the result shouldn't be constrained to only working on this): 
这是一个示例矩阵（但结果不应仅限于此处）：
a=zeros(7,7);
a(5,3:6)=1;
a(2,2)=1;
a(2,4)=1;
a(7,1:2)=1
a=
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
I want to get rid of all the 1's that are alone (the noise), such that I only have the line of 1's on the fifth row. 我想摆脱所有单独的1（噪音），这样我在第五行只有1行。
rules: -the 1's are in 'connected lines' if there are adjacent 1's (including diagonally) eg: 规则： - 如果有相邻的1（包括对角线），则1在'连接线'中，例如：
0 0 0 1 0 0 1 0 1
1 1 1 0 1 0 0 1 0
0 0 0 0 0 1 0 0 0
(The connected lines are what I want to keep. I want to get rid of all the 1's that are not in connected lines, the connected lines can intersect each other) （连接线是我想要保留的。我想摆脱所有不在连接线上的1，连接线可以相互交叉）
- the 'connected lines need to be at least 3 elements long. '连接线需要至少3个元素长。 So in the 7x7 example, there would only be one line that matches this criteria. 因此，在7x7示例中，只有一行符合此条件。 If a(7,3) was set to 1, then there would be a connected line at the bottom left also 如果a（7,3）设置为1，那么左下角也会有一条连线
I am currently looking at this through a column by column approach, and here is the first draft of my code so far: 我目前正在通过逐列方法来看这个，这是我的代码到目前为止的第一个草稿：
for nnn=2:6
rowPoss=find(a(:,nnn)==1);
rowPoss2=find(a(:,nnn+1)==1);
for nn=1:length(rowPoss)
if myResult(rowPoss(nn)-1:rowPoss(nn)+1,n-1)==0 %
%then?
end
end
end
My difficulty is, during this column by column process, I'd have to enable a way to recognise the beginning of the connected line, the middle of the connected line, and when a connected line ends. 我的困难在于，在此列的逐列过程中，我必须启用一种方法来识别连接线的开始，连接线的中间以及连接线的结束时间。 The same rules for this, when applied to noise (the lone 1's), would just ignore the lone 1's. 当应用于噪声（单独的1）时，相同的规则将忽略单独的1。
The output I want is basically: 我想要的输出基本上是：
b=
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 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
1 个解决方案
解决方案1
3 2015-04-27 06:34:57
If you have image processing toolbox, try bwareaopen 如果您有图像处理工具箱，请尝试bwareaopen
b = bwareaopen(a, 3);
Sample Run #1: 示例运行＃1：
>> a
a =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
>> b
b =
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 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Sample Run #2: 示例运行＃2：
>> a
a =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
>> b
b =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
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