Java根据时间范围判断有效数据

Question

确定指定日期有效属性值的最佳方法是什么？

table 看起来像这样：

Attribute name	Value	valid from
attr1	41	01.01.2020
attr1	41,123	02.02.2021
attr1	41,456	02.02.2023

例如。如果日期是 13.06.2021，则有效值应为 41,123

第二种情况是当多一行可用时：

Attribute name	Value	valid from	valid to
attr1	41	01.01.2020	01.01.2100
attr1	41,123	02.02.2021	02.04.2021
attr1	41,456	02.02.2023	01.01.2100

在这种情况下，有效值应为 41

Answer 1

由于您只提供了数据而没有提供代码，我们假设您有一个包含 Attribute 个元素的列表，如下所示（简化）：

class Attribute {
  String name;
  double value;
  LocalDate validFrom;
  LocalDate validTo;
}

理想情况下，您可以为每个按 validFrom 排序的属性名称维护一个树图，例如Map<String, SortedMap<LocalDate, Attribute>>.

然后获取属性名称的排序映射，根据相关日期获取尾部或头部映射并迭代。假设您已按 validFrom 降序对地图进行排序，它可能如下所示：

Map<String, SortedMap<LocalDate, Attribute>> attribs = ....;

LocalDate targetDate = LocalDate.now();


//get the map for "attr1" or an empty map if there isn't one
Iterator<Attribute> itr = attribs.getOrDefault("attr1", Collections.emptySortedMap())
       .tailMap( targetDate )
       .values()
       .iterator();

//since we have descending order we're first checking the entries that have become valid last
//so just continue until you hit one that hasn't become invalid already or you run out of entries
while(itr.hasNext()) {
  Attribute attr = itr.next();
  if(!attr.validTo.isBefore(targetDate )) {
    return attr.value;
  }
}  

//no value found, could also return an empty Optional
return null;

如果您没有这样的地图，也可以使用流来完成，但对于重复查找来说效率会更低：

List<Attribute> attribs = ...;

LocalDate targetDate = LocalDate.now();

OptionalDouble value = 
       attribs.stream()
              .filter(a -> a.name.equals("attrib1")) //remove all other attribs
              .filter(a -> !a.validFrom.isAfter(targetDate)) //remove all that not yet valid
              .filter(a -> !a.validTo.isBefore(targetDate)) //remove all that aren't valid anymore
              .sorted(Comparator.comparing(a -> a.validFrom).reverseOrder()) //sort by validFrom in descending order to get the latest
              .mapToDouble(a -> a.value) //extract the value
              .findFirst(); //pick the first value if one exists

Java根据时间范围判断有效数据

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