필터 지우기
필터 지우기

how to find locations of consecutive blocks of ones

조회 수: 25 (최근 30일)
imola
imola 2015년 3월 28일
편집: imola 2015년 3월 31일
dear all,
I have this matrix and I want to now how to find the consecutive blocks of ones in row 2 and 3, then the blocks in row 4 and 5 such that I specify their locations in their rows that block 5 is flow block 4 immediately in the next row, I think I need to use the index with some condition, but how?
A = [1 0 1 0 1
0 1 0 1 1
0 0 0 1 1
1 1 0 0 1
0 0 1 1 0
0 1 1 1 0];
I wrote this and I can't continue
for i=1:size(A,1)
tra=diff([0 A(i,:) 0]);
ss=find(tra==1);
u=find(tra==-1)-find(tra==1);
u1=[u1 u];
clear max
x=max(u1)
[I,J]=find(x)
end
by the way max command not work in my matlab R2013a.
thanks in advance to any help.
regards,
Imola
  댓글 수: 2
dpb
dpb 2015년 3월 28일
_"...by the way max command not work in my matlab R2013a."
max=max(u1)
You just aliased the builtin max by turning it into a variable max. Don't do that.
clear max
will fix the problem once you then also rename the variable to something besides max
imola
imola 2015년 3월 28일
Thanks dpb for the advice.
regards

댓글을 달려면 로그인하십시오.

채택된 답변

Image Analyst
Image Analyst 2015년 3월 28일
I don't really know how you're defining the words "find", "get", "recognize" and "better". I think what you want can be done with the function bwlabel() in the Image Processing Toolbox:
[labeledA, numRegions] = bwlabel(A, 4);
This will find regions that are "connected" regardless of how many rows they cover. Each separate region will have it's own label (ID number). You can also get the exact row,column location of the 1's in each region if you want
regionsNumberThatYouWant = 2; % Whatever region number you want to find coordinates of.
[rows, columns] = find(labeledA == regionsNumberThatYouWant);
If you want to operate on just certain rows instead of the whole matrix, then just pass it the rows you want:
[labeledA2, numRegions2] = bwlabel(A(2:3, :), 4);
[labeledA2, numRegions4] = bwlabel(A(4:5, :), 4);
I don't know what "better" means but there's a possibility that you mean that the connected region of 1's has the most number of 1's in it. If you want to find the region that has the greatest number of 1's in it, regardless of how many rows the region snakes across, and how many 1's are in it, then it's just a few lines of code more. Let me know if you need that.
There are also a whole bunch of other things you can determine about each region if you use the regionprops() function.
  댓글 수: 4
imola
imola 2015년 3월 30일
Dear,
what about my code because sofos's code doesn't give all the blocks, while mine is give all the blocks in every row. you mean trying some thing like this? for every row.
u(j) = find(g == -1) - find(g == 1)
I tried some think like this but how to start with the region and end depending on the index give me problem.
regards
imola
imola 2015년 3월 31일
편집: imola 2015년 3월 31일
Dear Image Analyst,
I did it thanks very much. I have one question, this function (bwlabel) work in my laptop (R2013a) but not in my laptop with matlab (R2012b), is it possible that the function is not their?.
regards

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Konstantinos Sofos
Konstantinos Sofos 2015년 3월 28일
편집: Konstantinos Sofos 2015년 3월 28일
Hi Imola,
If i understood well you want to build a vector which contains the maximum number of consecutive 1s values of each row of your matrix. More or less you could do this with a for loop but i want to give you as an answer a more elegant way and in my opinion is the most appropriate when you have to do with array functions.
The idea is to write a function e.g. myfunc which will be applied in each matrix row and give you back your desired result. So save the following function:
function out = myfunc(x)
out = max( diff( [0 (find( ~ (x > 0) ) ) numel(x) + 1] ) - 1);
end
Now you want to apply this function to each row of your matrix A. Have a look in the documentation of the arrayfun. Type now the following into your command and try to understand each step
applyToGivenRow = @(func, matrix) @(row) func(matrix(row, :))
applyToRows = @(func, matrix) arrayfun(applyToGivenRow(func, matrix), 1:size(matrix,1))'
A = [1 0 1 0 1
0 1 0 1 1
0 0 0 1 1
1 1 0 0 1
0 0 1 1 0
0 1 1 1 0];
f = @myfunc;
V=applyToRows(f, A);
V =
1
2
2
2
2
3
Regards

카테고리

Help CenterFile Exchange에서 Logical에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by