计算Java中的平均数

Question

Below is example of time series data:

time - 3 sec, value  - 6 
time - 10 sec, value - 7
time - 12 sec, value - 8
time - 17 sec, value - 9
time - 28 sec, value - 93
time - 29 sec, value - 94

I would like to calculate the average value for the last 30 seconds. Is there any simple standard algorithm for that ? Would be nice to see a link to Java implementation as well

Answer 1

I wrote(with a friend) a code that calculates the average number:

package dingen;
import java.util.Scanner;


public class Gemiddelde {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Scanner sc = new Scanner(System.in);

    float maxCount = 0;
    float avgCount = 0;

    System.out.println("How many numbers do you want");

    int n = sc.nextInt();

    for(int i = 0; i < n; i++) {
        System.out.println("Number: ");
        float number = sc.nextInt();
        maxCount = maxCount + number;

    }

    avgCount = maxCount / n;
    System.out.println("maxCount = " + maxCount);
    System.out.println("avgCount = " + avgCount);
}

}

the only thing you have to do is replace your class and package.

you will get the message: How many numbers do you want?:

and then it wil ask you the amount of numbers you inserted.

example:

How many numbers do you want?:6

Number:6

Number:7

Number:8

Number:9

Number:93

Number:94

maxCount = 217.0

avgCount = 36.166668

I have hoped I helped you with your problem :)

Answer 2

I'd use a TreeMap<Integer, Integer> , with the key being the seconds and the value, well, your value ;)

TreeMap has useful methods, such as lastKey() and tailMap() , which would fit your use case:

TreeMap<Integer, Integer> map = new TreeMap<>();

// add all your entries

Integer lastSecond = map.lastKey(); // returns the highest key

Integer thirtySecondsAgo = lastSecond - 30;

SortedMap<Integer, Integer> range = map.tailMap(thirtySecondsAgo);

int count = 0;
int sum = 0;
for (Integer value : range.values()) {
    sum += value;
    count++;
}

double average = (double) sum / (double) count;

tailMap() docs:

Returns a view of the portion of this map whose keys are greater than or equal to fromKey

For your problem, fromKey would be thirtySecondsAgo .