gpt4 book ai didi

Java:使用 lambda 在流中查找多个最小/最大属性值

转载 作者:搜寻专家 更新时间:2023-11-01 02:37:56 25 4
gpt4 key购买 nike

我正在寻找一种简洁的方法来查找一组属性值,这些属性值在给定的对象流中是最小值或最大值。

例如:

class Dimensions {
final int startX, startY, endX, endY; //Set by constructor
}

/**
* For the given dimensions, looks where the dimensions intersect.
* These coordinates define the sub-array, which is applied to the given function.
*
* @return the value returned by applying the sub-array in the given dimensions to the given function
*/
<S, T> T performOnIntersections(Function<S, T> function, S[][] inputArray, Dimensions...dimensions){

int maxStartX = Arrays.stream(dimensions).max(Comparator.comparingInt(d -> d.startX)).get().startX;
int maxStartY = Arrays.stream(dimensions).max(Comparator.comparingInt(d -> d.startY)).get().startY;
int minEndX = Arrays.stream(dimensions).min(Comparator.comparingInt(d -> d.endX)).get().endX;
int minEndY = Arrays.stream(dimensions).min(Comparator.comparingInt(d -> d.endY)).get().endY;

return applyInBetween(inputArray, function, maxStartX, maxStartY, minEndX, minEndY);
}

这是非常多余的,因为我必须为我需要的每个最小/最大属性创建一个新流。

在我的用例中,一个类似的方法是指数成本递归算法的一部分,所以有一个并发解决方案,只打开一次流会很棒。更好的解决方案是在现有流上运行而不终止(但我怀疑这是可能的)。

你知道如何改进它吗?

编辑:我忘了说,Dimension 是不可变的,这在使用 Supplier 时是相关的。

编辑 2: 使用 lambda 表达式在流上调用 collect() 而不是创建 DimensionsMinMaxCollector 的实例具有最佳的运行时性能. jessepeng 首先提到它,所以我将他的帖子标记为解决方案。我的实现现在是:

return Arrays.stream(dimensions)
.collect(() -> new int[4], (array, dimension) -> {
array[0] = Math.max(array[0], dimension.startX);
array[1] = Math.min(array[1], dimension.endX);
array[2] = Math.max(array[2], dimension.startY);
array[3] = Math.min(array[3], dimension.endY);
}, (a, b) -> {
a[0] = Math.max(a[0], b[0]);
a[1] = Math.min(a[1], b[1]);
a[2] = Math.max(a[2], b[2]);
a[3] = Math.min(a[3], b[3]);
});

最佳答案

您可以使用 collect() 将流的所有元素组合到一个包含所需值的 Dimensions 对象中。

来自 Stream 文档:

<R> R collect(Supplier<R> supplier,
BiConsumer<R, ? super T> accumulator,
BiConsumer<R, R> combiner);

Performs a mutable reduction operation on the elements of this stream. A mutable reduction is one in which the reduced value is a mutable result container, such as an ArrayList, and elements are incorporated by updating the state of the result rather than by replacing the result. This produces a result equivalent to:

 R result = supplier.get();
for (T element : this stream)
accumulator.accept(result, element);
return result;

因此,在您的情况下,您需要一个创建新 Dimension 对象的供应商,而累加器和组合器将进行比较和设置值。

Dimensions searchDimensions = Arrays.stream(dimensions).collect(Dimensions::new, (dimension, dimension2) -> {
dimension.endX = dimension.endX < dimension2.endX ? dimension.endX : dimension2.endX;
dimension.endY = dimension.endY < dimension2.endY ? dimension.endY : dimension2.endY;
dimension.startX = dimension.startX > dimension2.startX ? dimension.startX : dimension2.startX;
dimension.startY = dimension.startY > dimension2.startY ? dimension.startY : dimension2.startY;
}, (dimension, dimension2) -> {
dimension.endX = dimension.endX < dimension2.endX ? dimension.endX : dimension2.endX;
dimension.endY = dimension.endY < dimension2.endY ? dimension.endY : dimension2.endY;
dimension.startX = dimension.startX > dimension2.startX ? dimension.startX : dimension2.startX;
dimension.startY = dimension.startY > dimension2.startY ? dimension.startY : dimension2.startY;
});

return applyInBetween(inputArray, function, searchDimensions.startX, searchDimensions.startY, searchDimensions.endX, searchDimensions.endY);

编辑由于 Dimensions 是不可变的,因此不适合执行可变缩减操作。可以使用一个简单的数组来存储这四个值。

<S, T> T performOnIntersections(Function<S, T> function, S[][] inputArray, Dimensions...dimensions){

Supplier<int[]> supplier = () -> new int[]{Integer.MIN_VALUE, Integer.MIN_VALUE, Integer.MAX_VALUE, Integer.MAX_VALUE};
BiConsumer<int[], Dimensions> accumulator = (array, dim) -> {
array[0] = dim.startX > array[0] ? dim.startX : array[0];
array[1] = dim.startY > array[1] ? dim.startY : array[1];
array[2] = dim.endX < array[2] ? dim.endX : array[2];
array[3] = dim.endY < array[3] ? dim.endY : array[3];
};
BiConsumer<int[], int[]> combiner = (array1, array2) -> {
array1[0] = array1[0] > array2[0] ? array1[0] : array2[0];
array1[1] = array1[1] > array2[1] ? array1[1] : array2[1];
array1[2] = array1[2] < array2[2] ? array1[2] : array2[2];
array1[3] = array1[3] < array2[3] ? array1[3] : array2[3];
};

int[] searchDimensions = Arrays.stream(dimensions).collect(supplier, accumulator, combiner);

return applyInBetween(inputArray, function, searchDimensions[0], searchDimensions[1], searchDimensions[2], searchDimensions[3]);
}

关于Java:使用 lambda 在流中查找多个最小/最大属性值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42022406/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com