Find groups of 1's in array
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Sorry but ı have asked question wrong.
thanks all for great attention.
For example the data 11101111001110001111110100000000011111111 is like that and there only two groups of 1 could be because number of 0 maybe to few then we should accept them as a 1 how can we do that we have to find 1's group as a 2 there.
Examples: 1111111111100111111111111111011111111111111111= Result:1
1111111111100000000000000000111111111111111111= Result:2
11111111111000000000001111111111110000000001111111111100000000111111111 =Result :4
111111011111111010111111111101111111010011111111101001111111= Result :1
111010101111101000000011100011111 =RESULT :3 LIKE THAT
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Image Analyst
2015년 10월 15일
So ignore 3 or less. I've given you an answer that works. But you have not answered Thorsten or my questions on why you say it doesn't.
채택된 답변
Thorsten
2015년 10월 2일
Nsplit = 5; % number of consecutive zeros that split a group
Z = find(diff(x) == 1) - find(diff(x) == -1);
Ngroups = sum(Z > Nsplit) + 1;
댓글 수: 7
Thorsten
2015년 10월 8일
This file has 2337 groups of zeros. The largest group are 25 zeros, which occurs 11 times. You need a different algorithm to come up with 4 groups of 1's. How do you know that the answer is 4?
추가 답변 (3개)
Image Analyst
2015년 10월 6일
Ali, if you have the Image Processing Toolbox, this will give you exactly the results you want:
b = [1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1] % Result:1
filtered_b = ~bwareaopen(~b, 3) % Remove short stretches of 0's.
[~, numRegions] = bwlabel(filtered_b) % Count # regions of 1's
b = [1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1] % Result:2
filtered_b = ~bwareaopen(~b, 3) % Remove short stretches of 0's.
[~, numRegions] = bwlabel(filtered_b) % Count # regions of 1's
b = [1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1] % Result :4
filtered_b = ~bwareaopen(~b, 3) % Remove short stretches of 0's.
[~, numRegions] = bwlabel(filtered_b) % Count # regions of 1's
b = [1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,0,1,1,1,1,1,1,1,1,1,0,1,0,0,1,1,1,1,1,1,1] % Result :1
filtered_b = ~bwareaopen(~b, 3) % Remove short stretches of 0's.
[~, numRegions] = bwlabel(filtered_b) % Count # regions of 1's
b = [1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,0,0,0,0,0,0,1,1,1,0,0,0,1,1,1,1,1] % RESULT :3 LIKE THAT
filtered_b = ~bwareaopen(~b, 3) % Remove short stretches of 0's.
[~, numRegions] = bwlabel(filtered_b) % Count # regions of 1's
If you like my answer, please click on the "0 votes" under my avatar to vote for it.
댓글 수: 9
Image Analyst
2015년 10월 6일
Looks like you've accepted an answer that works for you. Al alternate way though is to use the Image Processing Toolbox if you have it. Then you can use bwareaopen() to get rid of small groups of zeros, then use bwlabel() to count the groups of 1's.
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