how to index the stating position of something?

Hi,

I was solving a question that asks to find the length of the largest section of zeros and its starting postion.

Example: x = [1 0 2 3 8 5 6 7 0 0 0 0 0 2 5 3 0 0 6 4]

LP = [5 9]

I was able to find the length, but I coudn't find a way to index the starting postion

my code:

function LP = LengthAndPosnZeros(x)
LP = [0 0];
y = [];
count = 0;
for i = 1:length(x)
    if x(i) == 0
        count = count+1;
        y = horzcat(y,count);
    else 
        count = 0;
    end
end
L = max(y)
% P = ? I don't know
LP(1,1) = L
LP(1,2) = P
end

Answer 1

[Dyuman Joshi]

The code you have requires a few modifications to work.

Try this -

x1 = [1 0 2 3 8 5 6 7 0 0 0 0 0 2 5 3 0 0 6 4];

out1 = LengthAndPosnZeros(x1)

out1 = 1×2

5 9

x2 = [1 0 0 8 5 0 0 3 3 3 3 0 0 0 0];

out2 = LengthAndPosnZeros(x2)

out2 = 1×2

4 12

x3 = [0 0 0 0 0 0 0 1 99 0];

out3 = LengthAndPosnZeros(x3)

out3 = 1×2

7 1

function LP = LengthAndPosnZeros(x)
d = diff([false,x==0,false]);
b = find(d>0);
e = find(d<0);
[m,idx] = max(e-b);
LP = [m b(idx)];
end

In case, you want to modify your current instead of using the code I provided, please let me know

[dim-ask]

I posted the almost the exact same function, so I deleted my answer. Imo this is the simplest, cleanest and probably the most scaleable (with reasonable effort put) way to implement such a function. Rule of thumb with matlab is that if you can avoid loops, then do that.

[Image Analyst]

Perhaps, unless you have the Image Processing Toolbox and can use the function specifically made for finding the longest run bwareafilt. Like I showed in my answer, the whole thing can be done in as little as 2 or even one line of code, and less cryptically/confusing.

[Nourhan]

@Dyuman Joshi I wanna understand your code, can you explain it?

[Torsten]

The best way to understand the code is to look up unknown commands (diff,max) and to remove the semicolons behind the code lines to see what comes out for d, b and e.

[dim-ask]

@Nourhan b is the start indices of each zero-segment, and e are the indices where they end (or better, one index more than their last). Then `e-b` are the lengths of each segment.

[dim-ask]

@Image Analyst Imo it depends. If you intend to check very large arrays or use the function many times in some loop, then `bwareafilt` is quite (~20x in my machine) slower. It is not optimised for 1d arrays, as its purpose is to be mainly used on general 2d arrays, thus I do not think it is an obvious case to prefer that over constructing one specific for 1d arrays. It is perfectly fine for the scale of the specific array given of course.

Answer 2

[Heisenberg]

function LP = LengthAndPosnZeros(x)
LP = [0 0];
y = [];
count = 0;
MaxInd=[];
for i = 1:length(x)
    if x(i) == 0
        count = count+1;
        if i>1 && x(i-1)~=0
            MaxInd = horzcat(MaxInd,i)
        else
            if i==1 && x(i)==0
                MaxInd = horzcat(MaxInd,1) 
            else
                MaxInd = horzcat(MaxInd,0)
            end
        end
        y = horzcat(y,count)
    else 
        count = 0;        
    end
end
try
[L,index] = max(y);

P= MaxInd(index-L+1);
% P = ? I don't know
LP(1,1) = L
LP(1,2) = P
catch
    disp('all non zeros')
end
end

This should fix your proble...cheers :)

[Nourhan]

Thank you

Answer 3

[Torsten]

x1 = [1 0 2 3 8 5 6 7 0 0 0 0 0 2 5 3 0 0 6 4];
out1 = LengthAndPosnZeros(x1)

out1 = 1×2

5 9

x2 = [1 0 0 8 5 0 0 3 3 3 3 0 0 0 0];
out2 = LengthAndPosnZeros(x2)

out2 = 1×2

4 12

x3 = [0 0 0 0 0 0 0 1 99 0];
out3 = LengthAndPosnZeros(x3)

out3 = 1×2

7 1

function LP = LengthAndPosnZeros(x)
  LP(1) = 0;
  LP(2) = 0;
  i = 1;
  while i <= length(x)
    if x(i) == 0
      start = i;
      len = 1;
      for j = i+1:length(x)
        if x(j) == 0
          len = len + 1;
        else
          break
        end
      end
      if len > LP(1)
        LP(1) = len;
        LP(2) = start;
      end
      i = j;
    else
      i = i + 1;
    end
  end
end

[Nourhan]

Thank you

Answer 4

[Image Analyst]

If you have the Image Processing Toolbox, why not use the power of the bwareafilt function to immediately extract the longest run of zeros? Everything after that is trivial.

x = [1 0 2 3 8 5 6 7 0 0 0 0 0 2 5 3 0 0 6 4];
x2 = bwareafilt(x == 0, 1) % Get longest run of zeros.  x2 will = [0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0]
zeroLength = sum(x2) % Count them.
startingIndex = find(x2, 1, 'first') % Find index of first zero
LP = [zeroLength, startingIndex] % Stitch together into a vector called "LP".

[Nourhan]

Thank you

[Image Analyst]

OK, here is it in only 2 lines.

x2 = bwareafilt(x == 0, 1) % Get longest run of zeros.  x2 will = [0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0]
LP = [sum(x2), find(x2, 1, 'first')] % Stitch together into a vector called "LP".

This is far, far simpler than the other complicated and cryptic solutions shown here because this uses the function literally meant for finding the longest run. Can't get any simpler, more obvious, and more intuitive than this.

[Nourhan]

@Image Analyst

Thank you so much. But I want to say that not only I am new to MATLAB, but I am also new to programming in general. So I am still learning how to use loops, conditions and basic functions, that's why I prefer to solve problems using them.