java 获取哈希图键作为整数数组

发布于 2024-12-04 14:45:47 字数 501 浏览 2 评论 0原文

我有一个像这样的哈希图

public HashMap <String,People> valueHashMap  = new Hashmap();

这里我的哈希图的键是以秒为单位的时间字符串，即我像这样向哈希图添加值

long timeSinceEpoch = System.currentTimeMillis()/1000;
valueHashMap.put(
                   Integer.toString((int)timeSinceEpoch)
                   , people_obj
                );

现在我想将哈希图中的所有键放入整数数组列表中。

ArrayList<Integer> intKeys = valueHashMap.keys()...

有什么办法可以做到这一点吗？

原文

I have a hashmap like this

public HashMap <String,People> valueHashMap  = new Hashmap();

Here the key to my HashMap is time in seconds as string, ie I am adding value to hashmap like this

long timeSinceEpoch = System.currentTimeMillis()/1000;
valueHashMap.put(
                   Integer.toString((int)timeSinceEpoch)
                   , people_obj
                );

Now I want to get all keys in the hashmap into an array list of integer.

ArrayList<Integer> intKeys = valueHashMap.keys()...

Is there any way to do that?

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谈情不如逗狗 2024-12-11 14:45:47

没有直接的方法将 String 列表转换为 Integer 列表：

您需要重新定义 valueHashMap像这样：

public HashMap valueHashMap = new HashMap();

....

ArrayList<整数> intKeys = new ArrayList(valueHashMap.keySet());

或者您需要循环：

ArrayList; intKeys = new ArraList();

for (String stringKey : valueHashMap.keySet())
     intKeys.add(Integer.parseInt(stringKey);

我建议您使用 Long 作为键：
```
public HashMap; valueHashMap = new HashMap();
```
那么就不会转换为 int （您可以使用上面的 (1) 和 Long 来代替）。

There is no direct way of converting a list of Strings to a list of Integers:

Either you need to redefine your valueHashMap like this:

public HashMap<Integer, People> valueHashMap  = new HashMap<Integer, People>();

....

ArrayList<Integer> intKeys = new ArrayList<Integer>(valueHashMap.keySet());

Or you need to loop:

ArrayList<Integer> intKeys = new ArraList<Integer>();

for (String stringKey : valueHashMap.keySet())
     intKeys.add(Integer.parseInt(stringKey);

I would advice you however to use the Long as key instead:
```
public HashMap<Long, People> valueHashMap  = new HashMap<Long, People>();
```
then there would be no casting to int (and you can use (1) above with Long instead).

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你是年少的欢喜 2024-12-11 14:45:47

您无法将一种类型的列表转换为另一种类型的列表，因此您必须迭代键并解析每个键。

for(String k : valueHashMap.keySet()) {
    intKeys.add(Integer.valueOf(k));
}

You can't cast a List of one type to a List of another type, so you have to iterate through the keys and parse each one.

for(String k : valueHashMap.keySet()) {
    intKeys.add(Integer.valueOf(k));
}

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雨后咖啡店 2024-12-11 14:45:47

你确实有类型问题。为什么要将长整型改为字符串以将它们存储在映射中。为什么不简单地使用 Long，它需要更少的内存并且更具描述性。那为什么要使用 Integer.toString 将 long 转换为 String 呢？通过将 long 转换为 int，您可能会面临丢失信息的风险。代码可能如下所示：

private Map<Long, People> valueHashMap = new Hashmap<Long, People>();

long timeSinceEpoch = System.currentTimeMillis()/1000;
valueHashMap.put(timeSinceEpoch, people_obj);
List<Long> longKeys = new ArrayList<Long>(valueHashMap.keySet());

You really have type problems. Why do you change the longs into Strings to store them in a map. Why not simply use Long, which needs less memory and is more descriptive. Then why use Integer.toString to transform a long into a String? By casting your long to an int, you risk loosing information by. Here's how the code should probably look like:

private Map<Long, People> valueHashMap = new Hashmap<Long, People>();

long timeSinceEpoch = System.currentTimeMillis()/1000;
valueHashMap.put(timeSinceEpoch, people_obj);
List<Long> longKeys = new ArrayList<Long>(valueHashMap.keySet());

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哑 2024-12-11 14:45:47

您可以使用 org.apache.commons.collections.Transformer 类来实现此目的，如下所示。

List<Integer> intKeys  = (List<Integer>)CollectionUtils.collect(valueHashMap.keySet(), new Transformer() {
                                    @Override
                                    public Object transform(Object key) {
                                        return Integer.valueOf(key);
                                    }
                                }, new ArrayList<Integer>());

You can use org.apache.commons.collections.Transformer class for that as follows.

List<Integer> intKeys  = (List<Integer>)CollectionUtils.collect(valueHashMap.keySet(), new Transformer() {
                                    @Override
                                    public Object transform(Object key) {
                                        return Integer.valueOf(key);
                                    }
                                }, new ArrayList<Integer>());

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