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Home/ Questions/Q 282195
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T05:16:05+00:00 2026-05-12T05:16:05+00:00

I use the following lines to sort a LinkedHashMap, but not all items are

  • 0

I use the following lines to sort a LinkedHashMap, but not all items are sorted, anything wrong ?

LinkedHashMap<String,PatternData> statisticsMap;
// fill in the map ...

LinkedHashMap<String,PatternData> sortedStatisticsMap=new LinkedHashMap<String,PatternData>();       // Sort it by patternData's average

ArrayList<PatternData> statisticsMapValues=new ArrayList<PatternData>(statisticsMap.values());
Collections.sort(statisticsMapValues,Collections.reverseOrder());                // Sorting it (in reverse order)

patternData last_i=null;
for (PatternData i : statisticsMapValues)                                       // Now, for each value
{
  if (last_i==i) continue;                                                         // Without dublicates
  last_i=i;

  for (String s : statisticsMap.keySet())                                         // Get all hash keys
    if (statisticsMap.get(s)==i)                                                  // Which have this value
    {
      sortedStatisticsMap.put(s,i);
    }
}


class PatternData implements Comparable<PatternData>
{
  float sum=0,average;
  int totalCount=0;
  Vector<String> records=new Vector<String>();

  public PatternData() { }

  public void add(float data)
  {
    sum+=data;
    totalCount++;
    average=sum/totalCount;
  }

  public void add(float data,String record)
  {
    add(data);
    records.add(record);
  }

  float getAverage() { return average; }

  public int compareTo(patternData o) { return (int)(average-o.average); }
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T05:16:06+00:00Added an answer on May 12, 2026 at 5:16 am

    When you return int, the range when average-o.average is between -1 and 1 will always return 0.

    One solution is simply change your compareTo function to:

    return Float.compare(average, o.average);
    
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