Count the "areas" of zeros and the "areas" of ones in a vector?

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Giorgos Papakonstantinou
Giorgos Papakonstantinou 2013 年 6 月 21 日
I have a vector:
a=[0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1 ];
in which I would like to count the "areas" of zeros and the "areas" of ones. In the example above I have
4 areas of zeros and 4 areas of ones. Thank you!

採用された回答

Andrei Bobrov
Andrei Bobrov 2013 年 6 月 21 日
a=[0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1 ];
lc = [true;diff(a(:))~=0];
x = a(lc);
zerosareas = sum(~x);
onesareas = sum(x);

その他の回答 (2 件)

Jan
Jan 2013 年 6 月 21 日
Another approach which might be faster for large data sets: FEX: RunLength:
a = [0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1 ];
b = RunLength(a);
nZero = sum(b == 0);
nOne = length(b) - nZero; % Or: sum(b == 1)
  2 件のコメント
Giorgos Papakonstantinou
Giorgos Papakonstantinou 2013 年 6 月 22 日
Thank you Jan!
Jan
Jan 2013 年 6 月 22 日
編集済み: Jan 2013 年 6 月 22 日
For 100'000 elements, the RunLength approach is twice as fast as the DIFF method. But as long as both methods need only some milliseconds, this might not really matter.

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Image Analyst
Image Analyst 2013 年 6 月 23 日
I know you already accepted an answer, but if you have the Image Processing Toolbox there is a straightforward one-liner solution to get either of those numbers:
a=[0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1];
[~, numberOf1Regions] = bwlabel(a) % Get # of "1" regions.
[~, numberOf0Regions] = bwlabel(~a) % Get # of "0" regions.

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