将数字拆分为不同的范围-Splitthenumbersinadifferentrange

0  

SortedSet keys = new TreeSet(map.keySet());
Map values = new HashMap();
Integer total = null;
Integer current = null;
Long point = null;
for (Long key : keys) {
    System.out.print(key + " : ");
    current = map.get(key);
    if(key >= 1 && key <= 20) {
        point = 1;
    } // Do Other Comparisons also and change point 2, 3, 4, 5, 6

    total = values.get(point);
    if(total == null) {
        total = 0;
    }
    total += current;
    values.put(point, total);
    System.out.println();
}

Now if you loop throw the keySet of values

现在,如果你循环抛出值的keySet

Point 1 will be How many calls(X number) came back in between 1 and 20 ms

第1点将是在1到20毫秒之间回来的多少次呼叫(X号码)

0  

You may try:

你可以尝试:

SortedSet keys = new TreeSet(map.keySet());

int group1To20=0;
int group20To40=0;
int group40To60=0;
int group60To80=0;
int group80To100=0;
int groupAbove100=0;
for (Long key : keys) {
 if(key>=0 && key<=20){
  group1To20=group1To20+map.get(key);
  }elseif(key>20 && key<=40){
   group20To40=group20To40+map.get(key);
  }
 //Similarly do as above for other range of group

}//end of loop


System.out.print("group 1-20 contains " +  group1To20);
//Now print the group range and values here

}

I have tried for your solution. I may misunderstand your question. If so, then clearify the question for me.

我试过你的解决方案。我可能会误解你的问题。如果是这样,那么请为我澄清问题。

0  

You could use a NavigableMap which allows you to query a range of numbers (head, tail). A thread safe implementation is ConcurrentSkipListMap.

您可以使用NavigableMap来查询一系列数字(头部,尾部)。一个线程安全的实现是ConcurrentSkipListMap。

In particular look the the methods NavigableMap headMap(K toKey, boolean inclusive), NavigableMap tailMap(K fromKey, boolean inclusive) and SortedMap subMap(K fromKey, K toKey)

特别是查看方法NavigableMap headMap(K toKey,boolean inclusive),NavigableMap tailMap(K fromKey,boolean inclusive)和SortedMap subMap(K fromKey,K toKey)

Example

//your existing concurrent map changed to concurrent navigable map
NavigableMap throughputCounter = new ConcurrentSkipListMap();
            // this prints for inclusive values - 1 and 20 are both included
            System.out.println("How many calls(X number) came back in between 1 and 20 ms:" + calcThroughput(throughputCounter.subMap(1L, true, 20L, true)));
            System.out.println("How many calls(X number) came back in between 20 and 40 ms:" + calcThroughput(throughputCounter.subMap(20L, true, 40L, true)));
            System.out.println("How many calls(X number) came back in between 40 and 60 ms:" + calcThroughput(throughputCounter.subMap(40L, true, 60L, true)));
            System.out.println("How many calls(X number) came back in between 60 and 80 ms:" + calcThroughput(throughputCounter.subMap(60L, true, 80L, true)));
            System.out.println("How many calls(X number) came back in between 80 and 100 ms:" + calcThroughput(throughputCounter.subMap(80L, true, 100L, true)));
            System.out.println("How many calls(X number) came back in after 100 ms:" + calcThroughput(throughputCounter.tailMap(100L)));

    private Long calcThroughput(NavigableMap subMap) {
        Long sumOfARange = new Long(0);
        for (Long value : subMap.values()) {
            sumOfARange += value;
        }
        return sumOfARange;
    }

0  

You don't need a map at all. You could simply divide the time by 20 (ms) and increment a counter in an array.

你根本不需要地图。您可以简单地将时间除以20(ms)并递增数组中的计数器。

public static void main( String[] args) {
    int counter[] = new int[6];
    for ( int i = 0 ; i <100 ; i++ ) {
        int time = (int) ( Math.random() * 200 );
        System.out.println( "" + time  );
        // add sample
        int idx = time / 20;
        idx = Math.min( idx, counter.length-1);
        counter[idx]++;
    }

    for ( int i = 0 ; i 

13 came back in between 0 and 20 ms
10 came back in between 20 and 40 ms
13 came back in between 40 and 60 ms
10 came back in between 60 and 80 ms
11 came back in between 80 and 100 ms
43 came back in between 100 and 120 ms

for ( int i = 0 ; i