读取txt文件并将数据存储在java的哈希表中

Question

我正在读取一个txt文件并将数据存储在哈希表中，但无法获得正确的输出。类似于此（部分）附加图像的txt文件 this is part of my data

我想将第1列和第2列存储为哈希表中的键（字符串类型），第3列和第4列存储为哈希表中的值（ArrayList类型）。 我的代码如下：

private Hashtable<String, ArrayList<String[]>> readData() throws Exception {
    BufferedReader br = new BufferedReader (new FileReader("MyGridWorld.txt"));
    br.readLine();

    ArrayList<String[]> value = new ArrayList<String[]>();
    String[] probDes = new String[2];
    String key = "";

    //read file line by line
    String line = null;
    while ((line = br.readLine()) != null && !line.equals(";;")) {
        //System.out.println("line ="+line);
        String source;
        String action;

        //split by tab
        String [] splited = line.split("\\t"); 
        source = splited[0];
        action = splited[1];
        key = source+","+action;

        probDes[0] = splited[2];
        probDes[1] = splited[3];

        value.add(probDes);
        hashTableForWorld.put(key, value);
        System.out.println("hash table is like this:" +hashTableForWorld);

    }

    br.close();
    return  hashTableForWorld;


}

输出如下所示： it's a very long long line

我想哈希表可能坏了，但我不知道为什么。谢谢你阅读我的问题

Answer 1

我们需要确定的第一件事是，你有一个非常明显的XY-Problem，即“你需要做什么”和“你试图如何解决它”是完全不一致的

那么，让我们回到原来的问题，首先尝试找出我们需要什么

正如我所能确定的那样，source和action是相互连接的，因为它们表示数据结构中可查询的“键”，而probability、destination和reward是数据结构中可查询的“结果”。因此，我们将首先创建对象来表示这两个概念：

public class SourceAction implements Comparable<SourceAction>{
    public final String source;
    public final String action;

    public SourceAction() {
        this("", "");
    }

    public SourceAction(String source, String action) {
        this.source = source;
        this.action = action;
    }

    public int compareTo(SourceAction sa) {
        int comp = source.compareTo(sa.source);
        if(comp != 0) return comp;
        return action.compareto(sa.action);
    }

    public boolean equals(SourceAction sa) {
        return source.equals(sa.source) && action.equals(sa.action);
    }

    public String toString() {
        return source + ',' + action;
    }
}

public class Outcome {
    public String probability; //You can use double if you've written code to parse the probability
    public String destination;
    public String reward; //you can use double if you're written code to parse the reward

    public Outcome() {
        this("", "", "");
    }

    public Outcome(String probability, String destination, String reward) {
        this.probability = probability;
        this.destination = destination;
        this.reward = reward;
    }

    public boolean equals(Outcome o) {
        return probability.equals(o.probability) && destination.equals(o.destination) && reward.equals(o.reward);

    public String toString() {
        return probability + ',' + destination + ',' + reward;
    }
}

那么，给定这些对象，如果SourceAction似乎与Outcome对象有一对多的关系，那么什么样的数据结构可以恰当地封装这些对象之间的关系呢？我的建议是Map<SourceAction, List<Outcome>>代表这种关系

private Map<SourceAction, List<Outcome>> readData() throws Exception {

可以使用哈希表（在本例中为HashMap）来包含这些对象，但我试图使代码尽可能简单，因此我们将继续使用更通用的接口

然后，我们可以重用您在原始代码中使用的逻辑，在这个数据结构中插入值，只需做一些调整

private Map<SourceAction, List<Outcome>> readData() {
    //We're using a try-with-resources block to eliminate the later call to close the reader
    try (BufferedReader br = new BufferedReader (new FileReader("MyGridWorld.txt"))) {
        br.readLine();//Skip the first line because it's just a header

        //I'm using a TreeMap because that makes the implementation simpler. If you absolutely 
        //need to use a HashMap, then make sure you implement a hash() function for SourceAction
        Map<SourceAction, List<Outcome>> dataStructure = new TreeMap<>();

        //read file line by line
        String line = null;
        while ((line = br.readLine()) != null && !line.equals(";;")) {
            //split by tab
            String [] splited = line.split("\\t"); 
            SourceAction sourceAction = new SourceAction(splited[0], splited[1]);

            Outcome outcome = new Outcome(splited[2], splited[3], splited[4]);

            if(dataStructure.contains(sourceAction)) {
                //Entry already found; we're just going to add this outcome to the already
                //existing list.
                dataStructure.get(sourceAction).add(outcome);
            } else {
                List<Outcome> outcomes = new ArrayList<>();
                outcomes.add(outcome);
                dataStructure.put(sourceAction, outcomes);
            }
        }
    } catch (IOException e) {//Do whatever, or rethrow the exception}
    return dataStructure;
}

然后，如果要查询与给定源+操作关联的所有结果，只需构造一个SourceAction对象并查询映射即可

Map<SourceAction, List<Outcome>> actionMap = readData();
List<Outcome> outcomes = actionMap.get(new SourceAction("(1,1)", "Up"));
assert(outcomes != null);
assert(outcomes.size() == 3);
assert(outcomes.get(0).equals(new Outcome("0.8", "(1,2)", "-0.04")));
assert(outcomes.get(1).equals(new Outcome("0.1", "(2,1)", "-0.04")));
assert(outcomes.get(2).equals(new Outcome("0.1", "(1,1)", "-0.04")));

这将为您的问题提供所需的功能

Answer 2

Hashtable和ArrayList（以及其他集合）不复制键和值，因此，您存储的所有值都是在开始时分配的相同的probDes数组（请注意，String[]以一种神秘的形式出现是正常的，您必须让它看起来很漂亮，但您仍然可以看到它一直都是非常神秘的东西）
可以确定的是，您应该为循环中的每个元素分配一个新的probDes
根据您的数据，您可以使用数组作为值在我看来，ArrayList没有实际用途 这同样适用于value，遇到新的key时必须单独分配：

private Hashtable<String, ArrayList<String[]>> readData() throws Exception {
    try(BufferedReader br=new BufferedReader(new FileReader("MyGridWorld.txt"))) {
        br.readLine();

        Hashtable<String, ArrayList<String[]>> hashTableForWorld=new Hashtable<>();

        //read file line by line
        String line = null;
        while ((line = br.readLine()) != null && !line.equals(";;")) {
            //System.out.println("line ="+line);
            String source;
            String action;

            //split by tab
            String[] split = line.split("\\t"); 
            source = split[0];
            action = split[1];
            String key = source+","+action;

            String[] probDesRew = new String[3];
            probDesRew[0] = split[2];
            probDesRew[1] = split[3];
            probDesRew[2] = split[4];

            ArrayList<String[]> value = hashTableForWorld.get(key);
            if(value == null){
                value = new ArrayList<>();
                hashTableForWorld.put(key, value);
            }
            value.add(probDesRew);
        }
        return hashTableForWorld;
    }
}

除了将变量重新定位到实际使用的位置之外，还将在本地创建返回值，并且将读取器包装到try-with-resource构造中，该构造确保即使发生异常也会关闭它（请参见官方教程here）

Answer 3

您应该更改添加到哈希表的逻辑，以检查您创建的键。如果键存在，则获取它映射到的数组的数组列表，并将数组添加到其中。当前您将覆盖数据

试试这个

if(hashTableForWorld.containsKey(key))
{
    value = hashTableForWorld.get(key);
    value.add(probDes);
    hashTableForWorld.put(key, value);
}
else
{
    value = new ArrayList<String[]>();
    value.add(probDes);
    hashTableForWorld.put(key, value);
}

然后，要打印内容，请尝试以下方法

for (Map.Entry<String, ArrayList<String[]>> entry : hashTableForWorld.entrySet()) {
    String key = entry.getKey();
    ArrayList<String[]> value = entry.getValue();

    System.out.println ("Key: " + key + " Value: ");
    for(int i = 0; i < value.size(); i++)
    {
        System.out.print("Array " + i + ": ");
        for(String val : value.get(i))
            System.out.print(val + " :: ")
        System.out.println();
    }
}