Median in timeseries array

If you don't copy the array then you can't sort it. If you don't sort it then you can't pinpoint the median value. pretty straightforward for me. If there was a standard functions then it would have to do it this way too AFAIK.

I understand what you are doing but I don't know HOW you do it.

Of course, I can research how to "copy the array" and how to "sort the copy".

But if you could explain how you do it, it would make the task a bit easier.

Thank you, Helmut

It's pretty easy if you use the standard functions:

https://docs.mql4.com/array

the copy and sort is straight forward but on reflection, the selection of median isn't. What I have done by simply dividing array length by 2 is good enough for me because I have enough elements to not worry about it, but my method is technically incorrect. if my array length was 5, I need to select element 2 but if it was 6, I need to find the mean of elements 2 & 3. To accurately find the median,imho, is quite convoluted and my method is technically inaccurate. Hence, I hoped there might be a function to help... by the looks of it there isn't.... I might try and build it. Anyway, abridged code below is my current status...

Thanks for the responses

V

   int   len=50;
   double   array[len],copy[len];
   static int   val;
   
   if (New_Bar)
      {
      val++;
      array[0]=val;
      ArrayCopy(copy,array,0,0,WHOLE_ARRAY);
      ArraySort(copy,WHOLE_ARRAY,0,MODE_DESCEND);
      median=copy[len/2];                                         
      for(i=len-1;i>=1;i--)
          {
          array[i]=array[i-1];
          }
      }

Set len to be an odd number and you will always get correct median.

For sake of completion, here is the function I wrote...

V

[edited as per following discussion]

   double   ArrayMedian(double array[])
      {
      double median;
      double copy[];
      int len=ArraySize(array);
      ArrayResize(copy,len);
      ArrayCopy(copy,array,0,0,WHOLE_ARRAY);
      ArraySort(copy,WHOLE_ARRAY,0,MODE_DESCEND);
      if (len%2==0) // it's even
         {
         median=(copy[len/2]+copy[(len/2)-1])/2.0;
         }
      else        // it's odd
         {
         median=copy[len/2];
         }
      return(median);
      }          

median=(copy[len/2]+copy[(len/2)+1])/2; in this case median may result in a value not present in the copy array. It will be a mere median value. If that's your intention you

can simply use ArrayMinimum() and ArrayMaximum() and find the median like that: median = (max+min)/2 without sorting the array. But if you want the find the median element

in the array use simply median = copy[len/2];

The median is not (max+min)/2 except for the special case where the array has only 2 elements.

http://www.purplemath.com/modules/meanmode.htm

Median is explicitly described as the middle element in a rank-sorted series. If the series contains an even number of elements then the median is explicitly described as being the mean value of the middle two values.

Right, sorry I mixed the terms.

V make sure the typecast is double (not integer) when you compute the mean which require you to divide by 2.0 for those cases where the values indexed by len/2 and len/2-1 are integers.

median=(copy[len/2]+copy[(len/2)-1])/2.; //