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从地图中随机选择n个项目

[英]Randomly select n items from a map

I am trying to randomly generate 'n' number of items from a HashMap where 'n' is determined by the user. 我试图从HashMap中随机生成“ n”个项目,其中“ n”由用户确定。

Here is what I have so far: 这是我到目前为止的内容:

public static void main(String []args){
    int numColors = 3;
    HashMap<String, String> map = new HashMap<String, String>();
    map.put("White","FFFFFF");
    map.put("Blank","000000");
    map.put("Red","ED0A15");
    map.put("Green","06F76C");
    map.put("Blue","0689FF");
    map.put("Sky Blue","00C2FC");
    map.put("Light Blue","08F0FC");
    map.put("Silver","C0BFC5");
    map.put("Mint","ABD3CA");
    map.put("Off White","FFEFF0");
    map.put("Purple","736FFA");
    map.put("Lavendar","DEBEEF");
    map.put("Hot Pink","F5159A");
    map.put("Pink","DB39CC");
    map.put("Light Pink","F5C2E3");
    map.put("Blush","C95FA7");
    map.put("Orange","D4361B");
    map.put("Yellow","DEF231");
    map.put("Warm White","F3E4C3");
    map.put("Turquoise","01DCA4");

    List<String> valuesList = new ArrayList<String>(map.values());
    int randomIndex = new Random().nextInt(valuesList.size());
    String randomValue = valuesList.get(randomIndex);

    System.out.printf(randomValue);
}

It prints 1 random color for me (in hex) which I want, however I am unsure of how/which loop to use in order to generate say 3 random hex colors from the map. 它为我打印了我想要的1种(十六进制)随机颜色,但是我不确定如何/要使用哪个循环才能从地图生成3种随机的十六进制颜色。 I declared numColors as 3 just to try and test this out. 我声明numColors为3只是为了尝试对此进行测试。

Here is what I ended up going with: 这就是我最终要进行的工作:

public static void main(String []args){
    int numColors = 3;
    HashMap<String, String> map = new HashMap<String, String>();
    map.put("White","FFFFFF");
    map.put("Blank","000000");
    map.put("Red","ED0A15");
    map.put("Green","06F76C");
    map.put("Blue","0689FF");
    map.put("Sky Blue","00C2FC");
    map.put("Light Blue","08F0FC");
    map.put("Silver","C0BFC5");
    map.put("Mint","ABD3CA");
    map.put("Off White","FFEFF0");
    map.put("Purple","736FFA");
    map.put("Lavendar","DEBEEF");
    map.put("Hot Pink","F5159A");
    map.put("Pink","DB39CC");
    map.put("Light Pink","F5C2E3");
    map.put("Blush","C95FA7");
    map.put("Orange","D4361B");
    map.put("Yellow","DEF231");
    map.put("Warm White","F3E4C3");
    map.put("Turquoise","01DCA4");

    List<String> keys = new ArrayList<String>(map.keySet());
    Random rand = new Random();
    for (int i = 0; i < numColors; i++) {
        String key = keys.get(rand.nextInt(keys.size()));
        System.out.println(map.get(key));
        }

 }

A simple solution is to shuffle the entire map using Collections.shuffle(map) . 一个简单的解决方案是使用Collections.shuffle(map)来随机播放整个地图。 Then just iterating over it and picking the first n elements. 然后只是对其进行迭代并选择前n个元素。

Of course this doesn't make sense if the map is huge and you only need a couple of elements. 当然,如果地图很大,而您只需要几个元素,就没有任何意义。

Edit: 编辑:

Naturally, with this solution you won't get any duplicate entries 自然,使用此解决方案,您将不会得到任何重复的条目

If I understand your question, you could do it with 如果我理解您的问题,您可以通过

List<String> keys = new ArrayList<String>(map.keySet());
Random rand = new Random();
for (int i = 0; i < numColors; i++) {
    String key = keys.get(rand.nextInt(keys.size()));
    System.out.println(map.get(key));
}

Changes are mentioned in comments 注释中提到了更改

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Random;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {

        int numColors = 3;
        HashMap<String, String> map = new HashMap<String, String>();
        map.put("White", "FFFFFF");
        map.put("Blank", "000000");
        map.put("Red", "ED0A15");
        map.put("Green", "06F76C");
        map.put("Blue", "0689FF");
        map.put("Sky Blue", "00C2FC");
        map.put("Light Blue", "08F0FC");
        map.put("Silver", "C0BFC5");
        map.put("Mint", "ABD3CA");
        map.put("Off White", "FFEFF0");
        map.put("Purple", "736FFA");
        map.put("Lavendar", "DEBEEF");
        map.put("Hot Pink", "F5159A");
        map.put("Pink", "DB39CC");
        map.put("Light Pink", "F5C2E3");
        map.put("Blush", "C95FA7");
        map.put("Orange", "D4361B");
        map.put("Yellow", "DEF231");
        map.put("Warm White", "F3E4C3");
        map.put("Turquoise", "01DCA4");

        // scanner for accepting values
        Scanner scan = new Scanner(System.in);
        System.out.println("Enter number");
        int N = scan.nextInt();

        // random object for generating random values
        Random rand = new Random();

        // converting map values to list
        List<String> valuesList = new ArrayList<String>(map.values());

        for (int i = 1; i <= N; i++) {
            // choose random value
            int randomIndex = rand.nextInt(valuesList.size());
            // get value
            String randomValue = valuesList.get(randomIndex);
            // printing
            System.out.println("Random value " + i + " : " + randomValue);
        }
    }
}

To prevent duplicates you can do something like this : 为了防止重复,您可以执行以下操作:

// random object for generating random values
Random rand = new Random();

// converting map values to list
List<String> valuesList = new ArrayList<String>(map.values());

Set<String> set = new HashSet<String>();

while (set.size() != N) {
    int randomIndex = rand.nextInt(valuesList.size());
    String randomValue = valuesList.get(randomIndex);
    set.add(randomValue);
}

System.out.println(set);

As Malt suggested, to prevent duplicates and keep code clean: 如Malt所建议,为防止重复并保持代码干净:

List<String> list = new ArrayList(map.values() );
Collections.shuffle(list);

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