如何从一组数字计算平均值,中位数,模式和范围

Question

是否有任何函数(作为数学库的一部分)将从一组数字计算平均值,中位数,模式和范围.

Answer 1

是的,似乎确实有第3个库(Java Math中没有).两个出现的是:

http://opsresearch.com/app/

http://www.iro.umontreal.ca/~simardr/ssj/indexe.html

但是,编写自己的方法来计算平均值,中位数,模式和范围实际上并不困难.

意思

public static double mean(double[] m) {
    double sum = 0;
    for (int i = 0; i < m.length; i++) {
        sum += m[i];
    }
    return sum / m.length;
}

MEDIAN

// the array double[] m MUST BE SORTED
public static double median(double[] m) {
    int middle = m.length/2;
    if (m.length%2 == 1) {
        return m[middle];
    } else {
        return (m[middle-1] + m[middle]) / 2.0;
    }
}

模式

public static int mode(int a[]) {
    int maxValue, maxCount;

    for (int i = 0; i < a.length; ++i) {
        int count = 0;
        for (int j = 0; j < a.length; ++j) {
            if (a[j] == a[i]) ++count;
        }
        if (count > maxCount) {
            maxCount = count;
            maxValue = a[i];
        }
    }

    return maxValue;
}

UPDATE

正如Neelesh Salpe所指出的,上述内容并不适合多模式收藏.我们可以很容易地解决这个问题

public static List<Integer> mode(final int[] numbers) {
    final List<Integer> modes = new ArrayList<Integer>();
    final Map<Integer, Integer> countMap = new HashMap<Integer, Integer>();

    int max = -1;

    for (final int n : numbers) {
        int count = 0;

        if (countMap.containsKey(n)) {
            count = countMap.get(n) + 1;
        } else {
            count = 1;
        }

        countMap.put(n, count);

        if (count > max) {
            max = count;
        }
    }

    for (final Map.Entry<Integer, Integer> tuple : countMap.entrySet()) {
        if (tuple.getValue() == max) {
            modes.add(tuple.getKey());
        }
    }

    return modes;
}

加成

如果您使用的是Java 8或更高版本,则还可以确定以下模式:

public static List<Integer> getModes(final List<Integer> numbers) {
    final Map<Integer, Long> countFrequencies = numbers.stream()
            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

    final long maxFrequency = countFrequencies.values().stream()
            .mapToLong(count -> count)
            .max().orElse(-1);

    return countFrequencies.entrySet().stream()
            .filter(tuple -> tuple.getValue() == maxFrequency)
            .map(Map.Entry::getKey)
            .collect(Collectors.toList());
}

Answer 2

从apache查看公共数学.那里有很多.