Subract specific areas from array

1 次查看（过去 30 天）
Hugo
Hugo 2014-1-6
评论： Hugo 2014-1-7
I would like to find and define specific areas in an array. For example for the next array;
Array = [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0]
i would like to define the area's with connected 2's and divide them in two groups (the area of 2's which which is limited by two 1's, and the 2's which aren't closed by two ones, but for example two zeros or a one and a zero).
In the end i would like to keep the area which is closed in by two ones and dump the rest of my array
What would be the best way to do this?
  2 个评论
Jos (10584)
Jos (10584) 2014-1-6
编辑：Jos (10584) 2014-1-6
What do you mean by "divide in two groups" and "dump the rest"?
It would help if you'd also given an example of the expected outcome, like:
ArrayIn = [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0]
ArrayOut = [0 0 0 0 0 0 1 1 1 2 2 2 2 2 2 2 2 1 1 1 0 0 0 0 0 0 0 0] %?
Hugo
Hugo 2014-1-6
Hi Jos,
The outcome i want is something like that, but i would like to create two arrays, so something like this:
ArrayIn = [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0];
ArrayOut1 = [0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0];
ArrayOut2 = [0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 2 2 0 0 0]; %
So i want to have one array with the 2's which are locked in between two 1's and i want one array with al 2's which do not meet this condition.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2014-1-6
编辑：Azzi Abdelmalek 2014-1-7
A= [1 2 2 1 1 1 2 2 2 2 2 1 1 2 2 0 2 2 0 0 0 1 2 2 2 1 0 1 ];
ArrayOut1=zeros(size(A));
ArrayOut2=ArrayOut1;
idx1=union(strfind(A,[1 2]),strfind(A,[0 2]));
idx2=union(strfind(A,[2 1]),strfind(A,[2 0]))+1;
v=all(A([idx1' idx2']),2);
ArrayOut1(cell2mat(arrayfun(@(x) idx1(x)+1:idx2(x)-1,find(v)','un',0)))=2;
ArrayOut2(cell2mat(arrayfun(@(x) idx1(x)+1:idx2(x)-1,find(~v)','un',0)))=2;
Or in case A start (or/and) ends with 2
A= [2 2 1 1 1 2 2 2 2 2 1 1 2 2 0 2 2 0 0 0 1 2 2 2 1 0 1 2 ];
ArrayOut1=zeros(size(A));
ArrayOut2=ArrayOut1;
A=[0 A 0];
idx1=union(strfind(A,[1 2]),strfind(A,[0 2]));
idx2=union(strfind(A,[2 1]),strfind(A,[2 0]))+1;
v=all(A([idx1' idx2']),2);
ArrayOut1(cell2mat(arrayfun(@(x) idx1(x):idx2(x)-2,find(v)','un',0)))=2
ArrayOut2(cell2mat(arrayfun(@(x) idx1(x):idx2(x)-2,find(~v)','un',0)))=2

更多回答（3 个）

Walter Roberson
Walter Roberson 2014-1-6
nonz = [0 (ArrayIn ~= 0) 0];
begin_groups = strfind(nonz, [0 1]);
end_groups = strfind(nonz, [1 0]);
Now look at the locations indicated by begin_groups and see if [2 2] starts there; likewise look in the corresponding end_groups and see if it ends with [2 2]
Azzi Abdelmalek
Azzi Abdelmalek 2014-1-6
ArrayIn= [0 0 0 0 2 2 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 0 2 2 0 0 0 ]
B=num2str(ArrayIn);
B=strrep(B,' ','');
[ii1,ii2]=regexp(B,'(?<=1)2+(?=1)','start','end');
[jj1,jj2]=regexp(B,'(?<=0)2+(?=0)|(?<=0)2+(?=1)|(?<=1)2+(?=0) ','start','end');
ArrayOut1=zeros(size(ArrayIn));
ArrayOut2=ArrayOut1;
ArrayOut1(cell2mat(arrayfun(@(x,y) x:y,ii1,ii2,'un',0)))=2
ArrayOut2(cell2mat(arrayfun(@(x,y) x:y,jj1,jj2,'un',0)))=2
Andrei Bobrov
Andrei Bobrov 2014-1-6
编辑：Andrei Bobrov 2014-1-6
[a,b] = regexp(num2str(A(:))','(?<=1)2*(?=1)');
t = false(size(A));
for jj = 1:numel(a)
t(a(jj):b(jj)) = true;
end
out = A.*(A==2);
out1=out.*t;
out2 = out.*~t;
or without num2str and regexp
t = [true;diff(A(:))~=0];
n = A(t);
ii = find(t);
m = bsxfun(@plus,strfind(n(:)',[1 2 1]),(1:2)');
tt = false(size(A));
for jj = 1:size(m,2), tt(ii(m(1,jj)):ii(m(2,jj))-1) = true; end
out = A.*(A==2);
out1 = out.*tt;
out2 = out.*~tt;

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by