How to find a chunk of a certain number of zeros inside a vector
1 次查看(过去 30 天)
显示 更早的评论
Hi all,
I have a vector of ones and zeros randomly distributed.
i.e: A = [0;1;1;0;0;0;0;1;1;1;1;0;1;]
What I want is to find the location of the first zero of the first chunk with 4 OR MORE zeros appearing in the vector.
In this example the result would be:
pos = 4;
The size of the group of zeros doesn't have to be necessarily 4, this was just an example.
I cannot find a simple way to do this but most probably there's a command for for this kind of operations that I cannot recall.
Many thanks in advance,
Pedro Cavaco
0 个评论
采纳的回答
David Young
2011-6-21
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;]
n = 4;
To find the first group of 4 or more zeros:
p = regexp(char(A.'), char(zeros(1, n)), 'once')
To find the first group of exactly 4 zeros:
zz = char(zeros(1,n));
p = regexp(char(A.'), ['(?<=^|' char(1) ')' zz '(' char(1) '|$)'], 'once')
5 个评论
David Young
2011-6-21
There are two solutions in this answer. The first of them works for the case of n or more zeros. The ?<= is a lookbehind operator to ensure that the match is at the start of the group of zeros - there is a requirement that the character before the zeros is char(1) or the start of the string. See doc regexp and follow the link to "Regular Expressions" for more details. You don't need this for the simple solution which finds groups of 4 or more zeros.
David Young
2011-6-21
By the way, in the case of n or more zeros, it's not obvious whether to use my first answer, with regexp, or Andrei's answer, with strfind. For very long strings, it may be faster to use regexp because of its 'once' option; however, strfind is simpler and will have a lower overhead.
更多回答(3 个)
Andrei Bobrov
2011-6-21
EDIT
A1 = A(:)';
out = strfind([1 A1],[1 0])-1; % all groups zeros
strfind([A1 1],[0 0 1]); % all groups two zeros
...
strfind([A1 1],[zeros(1,4) 1]); % all groups 4 zeros
6 个评论
Andrei Bobrov
2011-6-21
speed
>> A = +(rand(10000,1)<.2);
tic, zz = char(zeros(1,4));
p = regexp(char(A(:).'), ['(?<=^|' char(1) ')' zz '(' char(1) '|$)'], 'once'); toc
Elapsed time is 0.002538 seconds.
>> tic, A1 = A(:)';strfind([A1 1],[zeros(1,4) 1]);toc
Elapsed time is 0.000652 seconds.
Gerd
2011-6-21
Hi Pedro,
just programming straigforward I would use
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;];
cons = 4;
indices = find(A==0);
for ii=1:numel(indices)-cons
if (indices(ii+1)-indices(ii) == 1) && (indices(ii+2)-indices(ii+1)==1) && indices(ii+3)-indices(ii+2)==1
disp(indices(ii));
end
end
Result is 4
Gerd
3 个评论
Gerd
2011-6-21
Hi Pedro,
I tried both solution in a .m-file(David's and mine)
Please have a look at the result.
tic;
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;];
cons = 4;
indices = find(A==0);
for ii=1:numel(indices)-cons
if (indices(ii+1)-indices(ii) == 1) && (indices(ii+2)-indices(ii+1)==1) && indices(ii+3)-indices(ii+2)==1
disp(indices(ii));
end
end
t1 = toc;
tic;
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;];
n = 4;
p = regexp(char(A.'), char(zeros(1, n)), 'once');
disp(p);
t2 = toc;
With your testvector the result is really fast.
David Young
2011-6-21
Another approach to finding the first group of 4 or more zeros:
A = [0;1;1;1;0;0;0;0;1;1;1;1;0;1;0;0;0;1];
n = 4;
c = cumsum(A);
pad = zeros(n, 1)-1;
ppp = find([c; pad] == [pad; c]) - (n-1);
p = ppp(1)
EDIT Code corrected - n replaced by (n-1) to give correct offset.
3 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Characters and Strings 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!