help with sort function

I am writing my own sorting function, I've gotten the minimum value for the array and set that to the other array, and I couldn't figure out how to get rid of that minimum value in the original array and tell the program to find the next smallest minimum value. Here's my code so far

clear,clc;
x = [6 5 3 5 9 10 438 4 1 4 7 0 4 8 4 2];
len = length(x);
minval = x(1);
for n = 1:len
  for i = 1:len
      if x(i) < minval
          minval = x(i);
      end
  end
  sortval(n) = minval;
end
disp(sortval)

I was thinking maybe I could delete that value, but doing that would change the size of the array and it wouldn't work.

[John BG]

Hi Zhi

It's rewarding to be able to develop your own tools, yet sometimes, if time is of essence, on budget, perhaps you would like to consider having a look at a lost of already implemented sorting algorithms, here:

https://uk.mathworks.com/matlabcentral/fileexchange/56380-sortingalgorithms?s_tid=srchtitle

The one you want to implement, i think it's the bubble one.

Regards

John BG

<mailto:[e-mail] [e-mail]>

Answer 1

[Jan]

You can sort the vector "inplace": Move the minimum value to front and start the next iteration at the next element:

x = [6 5 3 5 9 10 438 4 1 4 7 0 4 8 4 2];
len = length(x);
for n = 1:len
  minind = n;
  minval = x(n);
  for i = n+1:len     % Start at n+1
    if x(i) < minval
        minval = x(i);
        minind = i;   % Remember index
    end
  end
  tmp  = x(n);        % Swap current with minimal element
  x(n) = minval;
  x(minind) = tmp;
end

Finally x contains the sorted array and the processing time is reduced, because the inner loop is shortend. (By the way: You can omit the last iteration)

Have fun with developping sorting algorithms. This is an interesting topic and it is worth to dig in WikiPedia.

Answer 2

[the cyclist]

I have not looked in detail at your algorithm. But one possibility that springs to mind is that you could swap NaN in place of the element you want to eliminate. That will retain the size of the array, and NaN will never be selected as less than any other element, because the logical operation

NaN < x

will return false for any value of x.

[zhi zhu]

I just tried that, added a for loop setting each value that is equal to minimum value equal to NaN and it still gave me the same result.

[the cyclist]

Can you post the new code?

[zhi zhu]

 clear,clc;
x = [6 5 3 5 9 10 438 4 1 4 7 0 4 8 4 2];
len = length(x);
minval = x(1);
for n = 1:len
    for i = 1:len
        if x(i) < minval
            minval = x(i);
        end
    end

    for j = 1:len
        if x(j) == minval
            x(j) = NaN;
        end
    end

    for r = 1:len
        if x(r) == minval
            x(r) = NaN;
        end
    end
    sortval(n) = minval;
end
disp(sortval)