如何遍历SparseArray？

Question

有没有办法迭代Java SparseArray(适用于Android)？我sparsearray以前很容易通过索引获取值.我找不到一个.

Answer 1

似乎我找到了解决方案.我没有正确地注意到这个keyAt(index)功能.

所以我会用这样的东西:

for(int i = 0; i < sparseArray.size(); i++) {
   int key = sparseArray.keyAt(i);
   // get the object by the key.
   Object obj = sparseArray.get(key);
}

Answer 2

如果您不关心密钥,则valueAt(int)可以在迭代稀疏数组时直接访问值.

for(int i = 0, nsize = sparseArray.size(); i < nsize; i++) {
    Object obj = sparseArray.valueAt(i);
}

Answer 3

您只需创建自己的ListIterator:

public final class SparseArrayIterator<E> implements ListIterator<E> {

private final SparseArray<E> array;
private int cursor;
private boolean cursorNowhere;

/**
 * @param array
 *            to iterate over.
 * @return A ListIterator on the elements of the SparseArray. The elements
 *         are iterated in the same order as they occur in the SparseArray.
 *         {@link #nextIndex()} and {@link #previousIndex()} return a
 *         SparseArray key, not an index! To get the index, call
 *         {@link android.util.SparseArray#indexOfKey(int)}.
 */
public static <E> ListIterator<E> iterate(SparseArray<E> array) {
    return iterateAt(array, -1);
}

/**
 * @param array
 *            to iterate over.
 * @param key
 *            to start the iteration at. {@link android.util.SparseArray#indexOfKey(int)}
 *            < 0 results in the same call as {@link #iterate(android.util.SparseArray)}.
 * @return A ListIterator on the elements of the SparseArray. The elements
 *         are iterated in the same order as they occur in the SparseArray.
 *         {@link #nextIndex()} and {@link #previousIndex()} return a
 *         SparseArray key, not an index! To get the index, call
 *         {@link android.util.SparseArray#indexOfKey(int)}.
 */
public static <E> ListIterator<E> iterateAtKey(SparseArray<E> array, int key) {
    return iterateAt(array, array.indexOfKey(key));
}

/**
 * @param array
 *            to iterate over.
 * @param location
 *            to start the iteration at. Value < 0 results in the same call
 *            as {@link #iterate(android.util.SparseArray)}. Value >
 *            {@link android.util.SparseArray#size()} set to that size.
 * @return A ListIterator on the elements of the SparseArray. The elements
 *         are iterated in the same order as they occur in the SparseArray.
 *         {@link #nextIndex()} and {@link #previousIndex()} return a
 *         SparseArray key, not an index! To get the index, call
 *         {@link android.util.SparseArray#indexOfKey(int)}.
 */
public static <E> ListIterator<E> iterateAt(SparseArray<E> array, int location) {
    return new SparseArrayIterator<E>(array, location);
}

private SparseArrayIterator(SparseArray<E> array, int location) {
    this.array = array;
    if (location < 0) {
        cursor = -1;
        cursorNowhere = true;
    } else if (location < array.size()) {
        cursor = location;
        cursorNowhere = false;
    } else {
        cursor = array.size() - 1;
        cursorNowhere = true;
    }
}

@Override
public boolean hasNext() {
    return cursor < array.size() - 1;
}

@Override
public boolean hasPrevious() {
    return cursorNowhere && cursor >= 0 || cursor > 0;
}

@Override
public int nextIndex() {
    if (hasNext()) {
        return array.keyAt(cursor + 1);
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public int previousIndex() {
    if (hasPrevious()) {
        if (cursorNowhere) {
            return array.keyAt(cursor);
        } else {
            return array.keyAt(cursor - 1);
        }
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public E next() {
    if (hasNext()) {
        if (cursorNowhere) {
            cursorNowhere = false;
        }
        cursor++;
        return array.valueAt(cursor);
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public E previous() {
    if (hasPrevious()) {
        if (cursorNowhere) {
            cursorNowhere = false;
        } else {
            cursor--;
        }
        return array.valueAt(cursor);
    } else {
        throw new NoSuchElementException();
    }
}

@Override
public void add(E object) {
    throw new UnsupportedOperationException();
}

@Override
public void remove() {
    if (!cursorNowhere) {
        array.remove(array.keyAt(cursor));
        cursorNowhere = true;
        cursor--;
    } else {
        throw new IllegalStateException();
    }
}

@Override
public void set(E object) {
    if (!cursorNowhere) {
        array.setValueAt(cursor, object);
    } else {
        throw new IllegalStateException();
    }
}
}

Answer 4

像饼一样简单.只需确保在实际执行循环之前获取数组大小.

for(int i = 0, arraySize= mySparseArray.size(); i < arraySize; i++) {
   Object obj = mySparseArray.get(/* int key = */ mySparseArray.keyAt(i));
}

希望这可以帮助.

Answer 5

无论谁使用Kotlin,老实说,迭代SparseArray的最简单方法是:使用Anko或Android KTX的Kotlin扩展!(感谢Yazazzello指出Android KTX)

只需致电 forEach { i, item -> }

Answer 6

用于删除SparseArray使用上述循环导致的所有元素Exception.

要避免这种情况请按照以下代码删除SparseArray使用普通循环的所有元素

private void getValues(){      
    for(int i=0; i<sparseArray.size(); i++){
          int key = sparseArray.keyAt(i);
          Log.d("Element at "+key, " is "+sparseArray.get(key));
          sparseArray.remove(key);
          i=-1;
    }
}

Answer 7

这是简单Iterator<T>而Iterable<T>实现的SparseArray<T>:

public class SparseArrayIterator<T> implements Iterator<T> {
    private final SparseArray<T> array;
    private int index;

    public SparseArrayIterator(SparseArray<T> array) {
        this.array = array;
    }

    @Override
    public boolean hasNext() {
        return array.size() > index;
    }

    @Override
    public T next() {
        return array.valueAt(index++);
    }

    @Override
    public void remove() {
        array.removeAt(index);
    }

}

public class SparseArrayIterable<T> implements Iterable<T> {
    private final SparseArray<T> sparseArray;

    public SparseArrayIterable(SparseArray<T> sparseArray) {
        this.sparseArray = sparseArray;
    }

    @Override
    public Iterator<T> iterator() {
        return new SparseArrayIterator<>(sparseArray);
    }
}

如果您不仅要迭代值而且要迭代密钥:

public class SparseKeyValue<T> {
    private final int key;
    private final T value;

    public SparseKeyValue(int key, T value) {
        this.key = key;
        this.value = value;
    }

    public int getKey() {
        return key;
    }

    public T getValue() {
        return value;
    }
}

public class SparseArrayKeyValueIterator<T> implements Iterator<SparseKeyValue<T>> {
    private final SparseArray<T> array;
    private int index;

    public SparseArrayKeyValueIterator(SparseArray<T> array) {
        this.array = array;
    }

    @Override
    public boolean hasNext() {
        return array.size() > index;
    }

    @Override
    public SparseKeyValue<T> next() {
        SparseKeyValue<T> keyValue = new SparseKeyValue<>(array.keyAt(index), array.valueAt(index));
        index++;
        return keyValue;
    }

    @Override
    public void remove() {
        array.removeAt(index);
    }

}

public class SparseArrayKeyValueIterable<T> implements Iterable<SparseKeyValue<T>> {
    private final SparseArray<T> sparseArray;

    public SparseArrayKeyValueIterable(SparseArray<T> sparseArray) {
        this.sparseArray = sparseArray;
    }

    @Override
    public Iterator<SparseKeyValue<T>> iterator() {
        return new SparseArrayKeyValueIterator<T>(sparseArray);
    }
}

创建返回的实用程序方法很有用,Iterable<T>并且Iterable<SparseKeyValue<T>>:

public abstract class SparseArrayUtils {
    public static <T> Iterable<SparseKeyValue<T>> keyValueIterable(SparseArray<T> sparseArray) {
        return new SparseArrayKeyValueIterable<>(sparseArray);
    }

    public static <T> Iterable<T> iterable(SparseArray<T> sparseArray) {
        return new SparseArrayIterable<>(sparseArray);
    }
}

现在你可以迭代SparseArray<T>:

SparseArray<String> a = ...;

for (String s: SparseArrayUtils.iterable(a)) {
   // ...
}

for (SparseKeyValue<String> s: SparseArrayUtils.keyValueIterable(a)) {
  // ...
}

Answer 8

如果您使用 Kotlin，则可以使用扩展函数，例如：

fun <T> LongSparseArray<T>.valuesIterator(): Iterator<T> {
    val nSize = this.size()
    return object : Iterator<T> {
        var i = 0
        override fun hasNext(): Boolean = i < nSize
        override fun next(): T = valueAt(i++)
    }
}

fun <T> LongSparseArray<T>.keysIterator(): Iterator<Long> {
    val nSize = this.size()
    return object : Iterator<Long> {
        var i = 0
        override fun hasNext(): Boolean = i < nSize
        override fun next(): Long = keyAt(i++)
    }
}

fun <T> LongSparseArray<T>.entriesIterator(): Iterator<Pair<Long, T>> {
    val nSize = this.size()
    return object : Iterator<Pair<Long, T>> {
        var i = 0
        override fun hasNext(): Boolean = i < nSize
        override fun next() = Pair(keyAt(i), valueAt(i++))
    }
}

如果您愿意，也可以转换为列表。例子：

sparseArray.keysIterator().asSequence().toList()

我认为它甚至可能是安全删除使用项目remove的LongSparseArray本身（而不是在迭代器），因为它是按升序排列。

编辑：似乎有一种更简单的方法，通过使用collection-ktx (example here )。实际上，它的实现方式与我所写的非常相似。

Gradle 要求：

implementation 'androidx.core:core-ktx:#'
implementation 'androidx.collection:collection-ktx:#'

这是 LongSparseArray 的用法：

    val sparse= LongSparseArray<String>()
    for (key in sparse.keyIterator()) {
    }
    for (value in sparse.valueIterator()) {
    }
    sparse.forEach { key, value -> 
    }

例如，对于那些使用 Java 的人，您可以使用LongSparseArrayKt.keyIterator,LongSparseArrayKt.valueIterator和LongSparseArrayKt.forEach。其他情况也一样。