Finding Consecutive True Values in a Vector

2014 2 4
4 个回答
回答已采纳
更新时间：2014 2 7
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0 个投票

I want to sum consecutive 1 values given a logical input vector. An example of input and output is below. Notice that the output is the sum of the previous elements that were 1 and if a zero element is encountered, the sum starts over. I am trying to avoid a for loop here if I can. Suggestions?
Input Output
0 0
0 0
0 0
1 1
0 0
1 1
0 0
0 0
1 1
1 2
1 3
1 4
0 0
1 1
0 0
1 1
1 2
0 0

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 采纳的回答

Andrei Bobrov
Andrei Bobrov 2014-2-5
编辑：Andrei Bobrov 2014-2-5

3 个投票

a0 = a(:); % input vector
ii = strfind(a0',[1 0]);
a1 = cumsum(a0);
i1 = a1(ii);
a0(ii+1) = -[i1(1);diff(i1)];
out = cumsum(a0); % output vector

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Azzi Abdelmalek
Azzi Abdelmalek 2014-2-5
Andrei, you didn't define a0, I guess a0=a(:) in the first line. Your code is slightly faster then Roger's
Andrei Bobrov
Andrei Bobrov 2014-2-5
Thank you Azzi for reply. I corrected.
Jason Nicholson
Jason Nicholson 2014-2-7
编辑：Azzi Abdelmalek 2014-2-7
I had to work this line by line to figure out what it was doing. Very nice solution. To summarize, you add values at the consecutive 1's break to make the cumulative sum correct at the right elements. Great work!
Documented Code:
a=[0 0 0 1 0 1 0 0 1 1 1 1 0]';
a0 = a;
% Find the end of any consecutive 1's in a0
ii= strfind(a0',[1 0]);
a1 = cumsum(a);
% Cumulative sum at end of any consecutive 1's in a0
i1 = a1(ii);
% Places the amount to subtract during cumulative-sum 1-element past the
% consecutive 1's in a to produce only the cumulative sum of consecutive
% 1's in a0. If this is confusing, output a0 after this step.
a0(ii+1) = -[i1(1);diff(i1)];
a0
% output vector
out = cumsum(a0);
out
Outputs this:
a0' = 0 0 0 1 -1 1 -1 0 1 1 1 1 -4
out' =0 0 0 1 0 1 0 0 1 2 3 4 0

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更多回答（3 个）

Roger Stafford
Roger Stafford 2014-2-5

2 个投票

Let the input column vector be called x.
y = [x;0];
f = find(diff([0;y])~=0);
p = f(2:2:end);
y(p) = y(p)-p+f(1:2:end-1);
y = cumsum(y(1:end-1));
Then y is your output.
Azzi Abdelmalek
Azzi Abdelmalek 2014-2-4
编辑：Azzi Abdelmalek 2014-2-4

1 个投票

a=[0 0 0 1 0 1 0 0 1 1 1 1 0]'
ii1=strfind([0 a' 0],[0 1])
ii2=strfind([0 a' 0],[1 0])-1
out=zeros(1,numel(a));
for k=1:numel(ii1)
c1=ii1(k);
c2=ii2(k);
out(c1:c2)=1:c2-c1+1
end
out'

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Jason Nicholson
Jason Nicholson 2014-2-4
编辑：Jason Nicholson 2014-2-5
Azzi,
Thanks for the answer but I am trying to totally avoid the loop if possible. I think it can be done.
Image Analyst
Image Analyst 2014-2-5
Don't be afraid of for loops. The fear of them is way overblown, especially for more recent versions of MATLAB. Unless your vector is tens of millions of elements long, I wouldn't worry about it. I would choose the one answer from the 3 that is the most well commented, intuitive, and easy to understand , if there is any. Any speed differences are probably negligible.

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Jos (10584)
Jos (10584) 2014-2-7

0 个投票

Hide the loops ;-)
input = [1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 1 0]
[~,~,C] = logicalfind(input,1) ;
C = cellfun(@cumsum, C,'un',0) ;
output = input ;
output(output==1) = [C{:}]
LOGICALFIND can be downloaded here:
http://www.mathworks.nl/matlabcentral/fileexchange/45305-logicalfind

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