比较hashmap以在Excel中下载数据

我首先建议您将次优数据表示规范化为如下所示。然后你只需要维护两张地图。然后很容易在它们之间迭代并以您想要的方式显示。

如果你能用 google-guava 然后,使用multiset对库进行分组比使用id进行分组更容易。

下面是我的方法的高级细节。您可以使用“reportbuilder.build(plan1,plan2)”的返回类型,并根据其他人的建议使用apache poi创建Excel。

• 运营商提供一个或多个计划
• 每个计划都有和ID以及1个或多个功能

• 每个功能都有id、decsription和value

  公共类主 { 私有静态地图>计划1映射;

  私有静态地图>计划1其他详细信息地图;

  私人静态地图>计划2映射;

  私有静态地图>平面2其他详细信息地图;

  私人静态平面图1;

  私人静态平面图2;

  公共静态void main(final string[]args) { initiaizedata(); normalizedata(); system.out.println(计划1); system.out.println(计划2); plancomaprisonreportbuilder reportbuilder=新建plancomaprisonreportbuilder(); system.out.println(reportbuilder.build(plan1,plan2)); }

  私有静态void normalizedata()。 { plan1=建筑平面图(plan1map,plan1additionaldetailsmap);

  plan2 = buildPlan(plan2Map, plan2AdditionalDetailsMap);
  

  }

  private static Plan buildPlan(final Map<String, Map<String, String>> planMap,
      final Map<String, Map<String, String>> planAdditionalDetailsMap)
  

  { string planid=iterables.getOnlyElement(planmap.keyset());

  Plan plan = new Plan(planId);
  
  
  Map<String, String> planDetails = planMap.get(planId);
  Iterator<Entry<String, String>> features = planDetails.entrySet().iterator();
  
  
  Map<String, String> additionalDetails = planAdditionalDetailsMap.get(planId);
  
  
  while (features.hasNext())
  {
      Entry<String, String> entry = features.next();
  
  
  String tagId = entry.getKey();
  String tagDescription = entry.getValue();
  String tagValue = additionalDetails.get(tagId);
  
  
  plan.addFeature(new Feature(tagId, tagDescription, tagValue));
  
  
  }
  return plan;
  

  }

  私有静态void initiazedata()。 { plan1Map=maps.newHashMap(); map1value=maps.newtreemap(); map1value.put(“id_1”,“plan features”); MAP1VALUE.PUT(“A”,“免赔额”); map1value.put(“b”,“个人”); map1value.put(“c”,“family”); map1value.put(“id_4”,“个人脱离网络”); map1value.put(“id_2”,“out-of-pocked annual maximum”); map1value.put(“d”,“个人”); map1value.put(“e”,“family”); map1value.put(“u”,“最长寿命”); 计划1资产净值(“343”,地图1价值);

  plan1AdditionalDetailsMap = Maps.newHashMap();
  Map<String, String> policy1ExtensionValue = Maps.newTreeMap();
  policy1ExtensionValue.put("id_1", "");
  policy1ExtensionValue.put("a", "Calendar Year");
  policy1ExtensionValue.put("b", "5000");
  policy1ExtensionValue.put("c", "10000");
  policy1ExtensionValue.put("id_4", "15000");
  policy1ExtensionValue.put("id_2", "");
  policy1ExtensionValue.put("d", "5000");
  policy1ExtensionValue.put("e", "10000");
  policy1ExtensionValue.put("u", "Unlimited");
  plan1AdditionalDetailsMap.put("343", policy1ExtensionValue);
  
  
  plan2Map = Maps.newHashMap();
  Map<String, String> policy2Value = Maps.newTreeMap();
  policy2Value.put("id_1", "Plan Features");
  policy2Value.put("a", "Deductible");
  policy2Value.put("b", "Individual");
  policy2Value.put("id_5", "Individual Out-of-network");
  policy2Value.put("c", "Family");
  policy2Value.put("id_4", "Family Out-of-network");
  policy2Value.put("id_2", "Out-of-pocket Annual Maximum");
  policy2Value.put("d", "Individual");
  policy2Value.put("e", "Family");
  policy2Value.put("u", "Life Time Maximum");
  plan2Map.put("347", policy2Value);
  
  
  plan2AdditionalDetailsMap = Maps.newHashMap();
  Map<String, String> policy2ExtensionValue = Maps.newTreeMap();
  policy2ExtensionValue.put("id_1", "");
  policy2ExtensionValue.put("a", "Calendar Year");
  policy2ExtensionValue.put("b", "7000");
  policy2ExtensionValue.put("id_5", "9000");
  policy2ExtensionValue.put("c", "12000");
  policy2ExtensionValue.put("id_4", "14000");
  policy2ExtensionValue.put("id_2", "");
  policy2ExtensionValue.put("d", "6000");
  policy2ExtensionValue.put("e", "15000");
  policy2ExtensionValue.put("u", "Unlimited");
  plan2AdditionalDetailsMap.put("347", policy2ExtensionValue);
  

  } }

公共班级计划 { 私有最终字符串ID;

private final Set<Feature> features = Sets.newHashSet();

public Plan(final String id)
{
    this.id = id;
}

public String getId()
{
    return id;
}

public void addFeature(final Feature f)
{
    features.add(f);
}

public Set<Feature> getFeatures()
{
    return Collections.unmodifiableSet(features);
}

@Override
public int hashCode()
{
    final int prime = 31;
    int result = 1;
    result = prime * result + ((id == null) ? 0 : id.hashCode());
    return result;
}

@Override
public boolean equals(final Object obj)
{
    if (this == obj)
    {
        return true;
    }
    if (obj == null)
    {
        return false;
    }
    if (getClass() != obj.getClass())
    {
        return false;
    }
    Plan other = (Plan) obj;
    if (id == null)
    {
        if (other.id != null)
        {
            return false;
        }
    }
    else if (!id.equals(other.id))
    {
        return false;
    }
    return true;
}

@Override
public String toString()
{
    return "Plan [features=" + features + ", id=" + id + "]";
}

}

    public class Feature
{
    private final String id;

    private final String description;

    private final String value;

    public Feature(final String id, final String description, final String value)
    {
        this.id = id;
        this.description = description;
        this.value = value;
    }

    public String getId()
    {
        return id;
    }

    public String getDescription()
    {
        return description;
    }

    public String getValue()
    {
        return value;
    }

    @Override
    public int hashCode()
    {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((id == null) ? 0 : id.hashCode());
        return result;
    }

    @Override
    public boolean equals(final Object obj)
    {
        if (this == obj)
        {
            return true;
        }
        if (obj == null)
        {
            return false;
        }
        if (getClass() != obj.getClass())
        {
            return false;
        }
        Feature other = (Feature) obj;
        if (id == null)
        {
            if (other.id != null)
            {
                return false;
            }
        }
        else if (!id.equals(other.id))
        {
            return false;
        }
        return true;
    }

    @Override
    public String toString()
    {
        return "Attribute [description=" + description + ", id=" + id + ", value=" + value + "]";
    }

    }

    public class PlanComaprisionReportBuilder
    {

    Multimap<String, String> build(final Plan... plans)
    {
        Multimap<String, String> rows = ArrayListMultimap.create(100, plans.length);

        for (Plan p : plans)
        {
            for (Feature f : p.getFeatures())
            {
                rows.put(f.getDescription(), f.getValue() != null ? f.getValue() : "");
                // if (!rows.containsKey(f.getDescription()))
                // {

                // }
                // else
                // {
                // existing row needs separate handling
                // }
            }
        }
        return rows;
    }
   }

所以在一个Excel文件中有两组要比较的哈希图,两组数据中的信息可能相同,也可能不同…(假设他们有相同的信息,他们在地图中可能有或没有相同的键)

我不确定我是否知道你到底遇到了什么问题,但下面是我如何快速解决的。

我会用一个小类来保存行标签(“PlanFeatures”,例如),并保存要比较的两个映射的值(例如vala、valb),这样:

class ThisIsARow {
    String label, valA, valB;
}

我将把这两组哈希图组合在一个结果中 HashMap<String, ThisIsARow> 对于它,键将是标签本身。

然后,我将遍历第一组哈希图,创建thisisarow的新实例,为每个实例设置标签和vala值。

然后,我将遍历第二组哈希映射,首先检查结果中是否已经有thisisarow实例 hashmap<string,thisisarow> 对于每个标签,如果该标签还没有新标签,则创建并添加一个新标签(设置其标签和值),否则只设置现有ThisisArrow实例的值。

然后我会用 Apache POI 在Excel文件中写下所有内容。(然后,您只需遍历生成的哈希图,每行打印一个thisisarow实例。)

label    valA    valB
label    valA    valB
label    valA    valB
label    valA    valB
...

我希望这有帮助。如果你需要澄清或我偏离轨道,请告诉我!