How many consecutive ones?

Question

Zhuoyi Chen 2017-11-9

2
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/366126-how-many-consecutive-ones

评论： Image Analyst 2019-2-22

Write a function ConsecutiveOnes to detect in a logical row array of any length the locations the length of the ones sequences. If there is a single one, the location of the one should be indicated with a length of one. For example, for the input array [0 0 1 1 0 0 1] the function ConsecutiveOnes produces the output [0 0 2 0 0 0 1]; for the input array [0,0,1,1,0,0,0,1,1,1] the function ConsecutiveOnes produces the output [0 0 2 0 0 0 0 3 0 0]

I am trying a recursion function, which should be sth like

for i = 1:a
function indicator = solve(inputArray1(i))

I know this expression is wrong, but I wonder how to display an idea like this.

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Jan 2017-11-9

编辑：Jan 2017-11-9

I do not understand the intention of these 2 lines of code. A recursive function is possible, but much more complicated then needed.

You got this homework to learn how to solve such problems. Then it would not be useful, if the solution is posted in the forum. Try it with a loop.

Zhuoyi Chen 2017-11-9

编辑：Zhuoyi Chen 2017-11-9

在 MATLAB Online 中打开

Yeah, I know that, but I did try. Also, I am here to ask for a method, not an answer. For my first attempt, I tried the following code.

    a = length(inputArray);
    indicator = zeros(1,a)
        for i = 1:(a-1)
            if (inputArray(i) == 1)&(inputArray(i+1) ==1)
                indicator(i) = indicator(i) + 1
                n = i+1
                while n<=a
                    if inputArray(n) == 0
                        break
                    end 
                        indicator(i) = indicator(i) +1
                        n = n+1
                end
                    inputArray((i+1):n) = 0;
            end
        end

It works for [0,0,1,1,0,0,0,1,1,1] actually, but it fails to [0, 1, 1, 0, 1].

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Answer 1

Andrei Bobrov 2017-11-9

3
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https://ww2.mathworks.cn/matlabcentral/answers/366126-how-many-consecutive-ones#answer_290246

编辑：Andrei Bobrov 2017-11-9

在 MATLAB Online 中打开

out = double(diff([~A(1);A(:)]) == 1);
v = accumarray(cumsum(out).*A(:)+1,1);
out(out == 1) = v(2:end);

or

out = zeros(size(A));
ii = strfind([0,A(:)'],[0 1]);
out(ii) = strfind([A(:)',0],[1 0]) - ii + 1;

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Andrei Bobrov 2017-11-9

Thank you Jan!

Image Analyst 2019-2-22

Please start a new thread on this (not here).

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Answer 2

Jan 2017-11-9

2
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https://ww2.mathworks.cn/matlabcentral/answers/366126-how-many-consecutive-ones#answer_290250

编辑：Jan 2017-11-9

在 MATLAB Online 中打开

I'd start at the end of the vector and store the value. Then run a loop from length(a)-1 to 1. If the corresponding element is 0, the value of c is written to the element on the right and c is set to 0. If it is 1, c is increased by 1. After the loop the last value of c is written to the first element of the output.

a = [0,0,1,1,0,0,0,1,1,1];
c = a(end);
for k = length(a)-1:-1:1
  if a(k) == 0
    a(k+1) = c;
    c      = 0;
  else
    a(k+1) = 0;
    c      = c + 1;
  end
end
a(1) = c;

Or the same with multiplications instead of the IF branches:

a = [0,0,1,1,0,0,0,1,1,1];
c = a(end);
for k = length(a)-1:-1:1
  a(k+1) = c * (1 - a(k));
  c      = a(k) * (c + 1);
end
a(1) = c;

Well, this is ugly. Without good comments, it will take a while until its intention gets clear.

In productive code a vectorized method would be the best idea:

k    = find([true, diff(a) ~= 0]);      % Indices of changed values
r    = zeros(size(a));
r(k) = a(k) .* diff([k, length(a)+1]);  % Distance of changes

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Jan 2017-11-9

编辑：Jan 2017-11-9

在 MATLAB Online 中打开

As usual some timings:

function neatoTest
a = randi([0,1], 1, 1e6);
tic; for k = 1:20, r = Accum(a); end; toc
tic; for k = 1:20, r = Strfind1(a); end; toc
tic; for k = 1:20, r = Strfind2(a); end; toc
tic; for k = 1:20, r = Loop1(a); end; toc
tic; for k = 1:20, r = Loop2(a); end; toc
tic; for k = 1:20, r = Find1(a); end; toc
end
function out = Accum(a)
out = double(diff([~a(1);a(:)]) == 1);
v = accumarray(cumsum(out).*a(:)+1,1);
out(out == 1) = v(2:end);
end
function out = Strfind1(a)
out = zeros(size(a));
ii = strfind([0,a(:)'],[0 1]);
out(ii) = strfind([a(:)',0],[1 0]) - ii + 1;
end
function out = Strfind2(a)
out     = zeros(size(a));
aa      = [0,a,0];
ii      = strfind(aa, [0 1]);
out(ii) = strfind(aa, [1 0]) - ii;
end
function a = Loop1(a)
c = a(end);
for k = length(a)-1:-1:1
  if a(k) == 0
    a(k+1) = c;
    c      = 0;
  else
    a(k+1) = 0;
    c      = c + 1;
  end
end
a(1) = c;
end
function a = Loop2(a)
c = a(end);
for k = length(a)-1:-1:1
  a(k+1) = c * (1 - a(k));
  c      = a(k) * (c + 1);
end
a(1) = c;
end
function r = Find1(a)
k    = find([true, diff(a) ~= 0]);      % Indices of changed values
r    = zeros(size(a));
r(k) = a(k) .* diff([k, length(a)+1]);  % Distance of changes
end

Results (R2009a, Win64):

 Elapsed time is 0.873261 seconds.  Andrei Accumarray
 Elapsed time is 0.601190 seconds.  Andrei strfind
 Elapsed time is 0.532654 seconds.  Andrei strfind + [0,a,0]
 Elapsed time is 1.053366 seconds.  Jan loop
 Elapsed time is 0.956481 seconds.  Jan ugly loop
 Elapsed time is 0.709494 seconds.  Jan vectorized

The best vectorized solution is two times faster than the loop. But loops are not as bad as the ancient rumors claim.

Andrei Bobrov 2017-11-9

+1

Jan 2017-11-10

@Zhuoyi Chen: Flags are thought to call admins or editors to care about contents, which conflict with the terms of use, e.g. if they are offending. You can use a comment to mention, that a solution is working.

@Andrei: Thanks. It is interesting, that your STRFIND solution gets faster, if the input contains less 1s, while the loops do not profit.

I think, although we have posted solutions of a homework, the overkill of 6 different methods has supported the OP to learn something new about Matlab. :-)

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Answer 3

Image Analyst 2017-11-10

1
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https://ww2.mathworks.cn/matlabcentral/answers/366126-how-many-consecutive-ones#answer_290306

在 MATLAB Online 中打开

Interested in the method I first thought of?

v =  [0,0,1,1,0,0,0,1,1,1] 
v2 = zeros(size(v)); % Initialize vector of same length.
props = regionprops(logical(v), 'Area', 'PixelIdxList');
for k = 1 : length(props)
  v2(props(k).PixelIdxList(1)) = props(k).Area;
end

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Andrei Bobrov 2017-11-10

+1

Alexander Cranney 2018-3-15

bwconncomp may offer some speed improvements over regionprops. But I didn't know of either of these remarkable functions before seeing this, so thank you!

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How many consecutive ones?

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