如何通过列表中的复杂逻辑获得价值 [英] how can i get value through complex logic from list

问题描述

我有过滤 personCountFilter = 3 ，并且列表如下：

I have filter personCountFilter=3, and have list as below:

Rate{ PersonCount:1, LOS:1}
Rate{ PersonCount:1, LOS:2}
Rate{ PersonCount:1, LOS:3}
Rate{ PersonCount:2, LOS:1}
Rate{ PersonCount:2, LOS:2}
Rate{ PersonCount:2, LOS:3}
Rate{ PersonCount:3, LOS:2}
Rate{ PersonCount:3, LOS:4}
Rate{ PersonCount:3, LOS:5}
Rate{ PersonCount:3, LOS:6}
Rate{ PersonCount:4, LOS:3}
Rate{ PersonCount:5, LOS:7}
Rate{ PersonCount:6, LOS:7}

过滤后我的预期：

Rate{ PersonCount:2, LOS:1}
Rate{ PersonCount:3, LOS:2}
Rate{ PersonCount:4, LOS:3}
Rate{ PersonCount:3, LOS:4}
Rate{ PersonCount:3, LOS:5}
Rate{ PersonCount:3, LOS:6}
Rate{ PersonCount:5, LOS:7}

如何在按LOS分组后得到值，如果personCount匹配过滤器获得此值，如果不匹配，则最接近 personCountFilter ，更大 personCountFilter 第一次

how can I get value after grouping by LOS, and if personCount matched filter get this one, if not matched, get closest to personCountFilter, bigger personCountFilter first

我试图使用

HashSet<Rate> testSet = rates.stream()
                .collect(Collectors.collectingAndThen(
                        Collectors.toMap(Rate::getLengthOfStayCount,
                                Function.identity(),
                                (previous, current) ->
                                {
                                    return previous.getPersonCount() > 
                                           current.getPersonCount() ? previous : current;
                                }),
                        map ->
                        {
                            HashSet<Rate> set = new HashSet<>();
                            set.addAll(map.values());
                            return set;
                        }));

但它返回

Rate{ PersonCount:2, LOS:1}
Rate{ PersonCount:3, LOS:2}
Rate{ PersonCount:4, LOS:3}
Rate{ PersonCount:3, LOS:4}
Rate{ PersonCount:3, LOS:5}
Rate{ PersonCount:3, LOS:6}
Rate{ PersonCount:6, LOS:7}

当按LOS分组后，当前获得最大人数。

Current it gets max personCount when after grouping by LOS

推荐答案

如果我理解正确，你想得到每个组的元素，它具有最近的 personCount 属性数字 3 ，如果出现平局，则优先选择较高的数字。

If I understand you correctly, you want to get the element of each group, which has the personCount property closest to the number 3, with preference to the higher number in case of a tie.

您可以使用

HashSet<Rate> testSet = rates.stream()
    .collect(Collectors.collectingAndThen(
        Collectors.toMap(Rate::getLengthOfStayCount,
                Function.identity(),
                BinaryOperator.minBy(
                   Comparator.comparingDouble(rate -> Math.abs(rate.getPersonCount()-3.1)))
        ),
        map -> new HashSet<>(map.values())
    ));

Math.abs（rate.getPersonCount（） - 3）是您的目标号 3 的距离，获得具有该值的最小值的元素将获得最接近的值。通过简单地使用 3.1 而不是 3 ，我们在平局的情况下获得更高的数字（假设你属性）有一个整数类型）。

Math.abs(rate.getPersonCount()-3) is the distance to your goal number 3, getting the element with the minimum of that value will get the closest. By simply using 3.1 instead of 3, we achieve preference to higher numbers in case of a tie (assuming that you property has an integer type).

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