Question

我有两个ArrayLists。

日期列表
各个数据的列表。

两者都是同步。我有时在同一日期有多个数据。我需要创建两个列表：唯一日期和数据（平均值）。到目前为止，我已经尝试了以下方法

int i = 1;
for(int it =0; it < predatetime.size() - 1; it++){
   //Compare each element with the next one
    if(predatetime.get(it+1) == predatetime.get(it)){
        i++;
        weight = preweight.get(it+1) + weight;
        //If equal, add weights and increment a divisor for averaging
    } 
    else {    //if not equal, add to the new lists
        if(it == predatetime.size() - 2){  //if the last element is not equal to its previous one, just add it to the list
            newDateTime.add(predatetime.get(it+1));
            newWeight.add(preweight.get(it+1));
            break;
        }
        weight = weight / i;
        newDateTime.add(predatetime.get(it));
        newWeight.add(weight);

        weight = preweight.get(it+1);    //re-initialize variables
        i = 1;
    }
    if(it == predatetime.size() - 2){
        weight = weight / i;
        newDateTime.add(predatetime.get(it));
        newWeight.add(weight);
    }
}

此代码存在很多问题。

如果列表只有一个元素，则失败。（我知道我可以再写两行来照顾这个）。有更好的方法吗？

我知道这个网站上有类似的问题，但我仍然无法解决问题。

Answer 1

如果您尝试使列表元素唯一，为什么不尝试将列表转换为集合集

Set<Foo> foo = new HashSet<Foo>(myList);

Answer 2

为什么不以日期作为键创建地图，并将值作为列表。这样您就可以保持日期的唯一性，同时允许您将数据作为列表。

Map<String, ArrayList<myData>> myMap = new HashMap<String, ArrayList<myData>>();

然后你可以找到你的密钥是否存在，如果它确实通过使用密钥将其添加到数组列表来识别正确的列表。如果它不存在，请将其添加到地图

Answer 3

感谢@Rambler和@JulianGurung，我创建了一个HashMap并且它可以工作

        HashMap<Integer, Float> hm = new HashMap<Integer,Float>();  
        int occurance = 0;
        float weight = 0;
        hm.put(predatetime.get(0), 0f); //initialize with the first value

        for(Map.Entry m : hm.entrySet()){  
            for( int it = 0; it < predatetime.size(); it++){
                if(m.getKey() == predatetime.get(it)){
                    weight = (Float) m.getValue() + preweight.get(it);  //Sum all the same data in order to avg later
                    hm.put(predatetime.get(it), weight);
                    occurance++;
                }
                else{   //if it is not equal, add the new element to the map
                    hm.put(predatetime.get(it), preweight.get(it));
                }
            }
            weight = weight / occurance;
            hm.put((Integer) m.getKey(), weight);
            weight = 0;
            occurance = 0;
        }

Answer 4

这是完整的解决方案

import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.LinkedList;
import java.util.List;

public class CustomList {

public static void main(String[] args) {

    ArrayList<String> date = new ArrayList<>();
    date.add("1");
    date.add("2");
    date.add("2");
    date.add("3");
    System.out.println(date);

    ArrayList<Integer> value = new ArrayList<>();
    value.add(1);
    value.add(2);
    value.add(4);
    value.add(3);
    System.out.println(value);

    new MyCls().createList(date, value);

}

}

类MyCls {

ArrayList uniqueDate = new ArrayList<String>();
ArrayList averageValue = new ArrayList<Integer>();
LinkedHashMap store = new LinkedHashMap<String, CountEntry>();

class CountEntry {
    int value;
    int count;

    CountEntry() {
    }

    CountEntry(int v, int c) {
        value = v;
        count = c;
    }

    public int getValue() {
        return value;
    }

    public void setValue(int value) {
        this.value = value;
    }

    public int getCount() {
        return count;
    }

    public void setCount(int count) {
        this.count = count;
    }
}

public void createList(ArrayList<String> date, ArrayList<Integer> value) {
    for (int i = 0; i < date.size(); i++) {
        CountEntry tmp = (CountEntry) store.get(date.get(i));

        if (tmp == null) {
            store.put(date.get(i), new CountEntry(value.get(i), 1));
        } else {
            int tmpVal = tmp.getValue();
            int tmpCount = tmp.getCount();
            store.put(date.get(i), new CountEntry(value.get(i) + tmpVal, ++tmpCount));
        }
    }

    ArrayList<String> uniqueDate = new ArrayList<String>(store.keySet());
    ArrayList<CountEntry> tempAvgList = new ArrayList<CountEntry>(store.values());

    for (CountEntry ce : tempAvgList) {

        averageValue.add(ce.getValue() / ce.getCount());

    }
    System.out.println("Output");
    System.out.println(uniqueDate);
    System.out.println(averageValue);
}

} / *

OUTPUT Snap：

[1, 2, 2, 3]
[1, 2, 4, 3]
Output
[1, 2, 3]
[1, 3, 3]

* /

Java：从现有List创建唯一List，保持索引号不变

4 个答案:

OUTPUT Snap：