是否有任何函数（作为数学库的一部分）将从一组数字计算mean，中位数，模式和范围。
是的，似乎确实有第3个库（Java Math中没有）。两个出现的是：
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());
}
看看commons math from apache。那里有很多。
MODE算法不考虑具有多种模式的情况（双峰，三峰，......） - 当出现的次数与maxCount的次数相同时会发生。考虑到这一点，它应该返回一个数组而不是一个int值。
public class Mode {
public static void main(String[] args) {
int[] unsortedArr = new int[] { 3, 1, 5, 2, 4, 1, 3, 4, 3, 2, 1, 3, 4, 1 ,-1,-1,-1,-1,-1};
Map<Integer, Integer> countMap = new HashMap<Integer, Integer>();
for (int i = 0; i < unsortedArr.length; i++) {
Integer value = countMap.get(unsortedArr[i]);
if (value == null) {
countMap.put(unsortedArr[i], 0);
} else {
int intval = value.intValue();
intval++;
countMap.put(unsortedArr[i], intval);
}
}
System.out.println(countMap.toString());
int max = getMaxFreq(countMap.values());
List<Integer> modes = new ArrayList<Integer>();
for (Entry<Integer, Integer> entry : countMap.entrySet()) {
int value = entry.getValue();
if (value == max)
modes.add(entry.getKey());
}
System.out.println(modes);
}
public static int getMaxFreq(Collection<Integer> valueSet) {
int max = 0;
boolean setFirstTime = false;
for (Iterator iterator = valueSet.iterator(); iterator.hasNext();) {
Integer integer = (Integer) iterator.next();
if (!setFirstTime) {
max = integer;
setFirstTime = true;
}
if (max < integer) {
max = integer;
}
}
return max;
}
}
测试数据
模式{1,3}为{3,1,5,2,4,1,3,4,3,2,1,3,4,1}; 模式{-1}为{3,1,5,2,4,1,3,4,3,2,1,3,4,1，-1，-1，-1，-1，-1} ;
public static Set<Double> getMode(double[] data) {
if (data.length == 0) {
return new TreeSet<>();
}
TreeMap<Double, Integer> map = new TreeMap<>(); //Map Keys are array values and Map Values are how many times each key appears in the array
for (int index = 0; index != data.length; ++index) {
double value = data[index];
if (!map.containsKey(value)) {
map.put(value, 1); //first time, put one
}
else {
map.put(value, map.get(value) + 1); //seen it again increment count
}
}
Set<Double> modes = new TreeSet<>(); //result set of modes, min to max sorted
int maxCount = 1;
Iterator<Integer> modeApperance = map.values().iterator();
while (modeApperance.hasNext()) {
maxCount = Math.max(maxCount, modeApperance.next()); //go through all the value counts
}
for (double key : map.keySet()) {
if (map.get(key) == maxCount) { //if this key's value is max
modes.add(key); //get it
}
}
return modes;
}
//std dev function for good measure
public static double getStandardDeviation(double[] data) {
final double mean = getMean(data);
double sum = 0;
for (int index = 0; index != data.length; ++index) {
sum += Math.pow(Math.abs(mean - data[index]), 2);
}
return Math.sqrt(sum / data.length);
}
public static double getMean(double[] data) {
if (data.length == 0) {
return 0;
}
double sum = 0.0;
for (int index = 0; index != data.length; ++index) {
sum += data[index];
}
return sum / data.length;
}
//by creating a copy array and sorting it, this function can take any data.
public static double getMedian(double[] data) {
double[] copy = Arrays.copyOf(data, data.length);
Arrays.sort(copy);
return (copy.length % 2 != 0) ? copy[copy.length / 2] : (copy[copy.length / 2] + copy[(copy.length / 2) - 1]) / 2;
}
如果您只关心单峰分布，请考虑某事。像这样。
public static Optional<Integer> mode(Stream<Integer> stream) {
Map<Integer, Long> frequencies = stream
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
return frequencies.entrySet().stream()
.max(Comparator.comparingLong(Map.Entry::getValue))
.map(Map.Entry::getKey);
}
正如Nico Huysamen已经指出的那样，在Java 1.8中找到多个模式可以如下交替进行。
import java.util.ArrayList;
import java.util.List;
import java.util.HashMap;
import java.util.Map;
public static void mode(List<Integer> numArr) {
Map<Integer, Integer> freq = new HashMap<Integer, Integer>();;
Map<Integer, List<Integer>> mode = new HashMap<Integer, List<Integer>>();
int modeFreq = 1; //record the highest frequence
for(int x=0; x<numArr.size(); x++) { //1st for loop to record mode
Integer curr = numArr.get(x); //O(1)
freq.merge(curr, 1, (a, b) -> a + b); //increment the frequency for existing element, O(1)
int currFreq = freq.get(curr); //get frequency for current element, O(1)
//lazy instantiate a list if no existing list, then
//record mapping of frequency to element (frequency, element), overall O(1)
mode.computeIfAbsent(currFreq, k -> new ArrayList<>()).add(curr);
if(modeFreq < currFreq) modeFreq = currFreq; //update highest frequency
}
mode.get(modeFreq).forEach(x -> System.out.println("Mode = " + x)); //pretty print the result //another for loop to return result
}
快乐的编码！