Question

我将简要解释函数正在做什么/应该做什么：

参数：所需设施清单
返回：满足50％以上设施的原型列表是必需的
原型具有X个设施（存储在DB中）。
用户提供他想要的设施清单（参数）

函数查找哪个原型最适合所有原型提供的设施。

public List<ArchetypesLokaal> GetArchetypeUitFaciliteiten(List<FaciliteitenLokaalFixed> faciliteiten)
{
    var archetypes = new List<ArchetypesLokaal>();
    var sortedArchetypes = new List<ArchetypesLokaal>();


    //Iterate over all the archetypes
    foreach(var archetype in db.ArchetypesLokaals)
    {
        //Get all the facilities that are present in archetype
        var temp = GetFaciliteitenVoorArchetype(archetype);

        //Create a new list to add all the facilities that are matched
        var temp2 = new List<FaciliteitenLokaalFixed>();

        //Iterate over the facilities provided as a parameter, these are the facilities that have to be present in the archetype
        foreach(var faciliteit in faciliteiten)
        {
            //CHeck if facility is present in an archetype
            var aanwezig = temp.FirstOrDefault(t => t.Naam.Equals(faciliteit.Naam));
            if (aanwezig!=null)
            {
                //Add if it's present
                temp2.Add(faciliteit);
            }
        }
        var needed = faciliteiten.Count;
        var present = temp2.Count;

        //Compare how many facilities were met
        if (present / needed >= .5)
        {
            archetypes.Add(archetype);
        }
    }

   //Todo: sort archetypes by % facilities that were met
    return archetypes;
}

所以现在我想根据当前/需要对列表进行排序，因此我可以先显示最佳结果。我也不确定我所做的是否是处理此问题的最有效方法，但我确信我的代码现在正常工作。如果您有任何建议，请务必使用。

谢谢。

Answer 1

如果我理解正确，您需要根据present = temp2.Count局部变量对输出进行排序。您可以将结果包装在某个容器中，例如Tuple：

var archetypes = new List<Tuple<int, ArchetypesLokaal>>();

//Iterate over all the archetypes
foreach(var archetype in db.ArchetypesLokaals)
{
    //Get all the facilities that are present in archetype
    var temp = GetFaciliteitenVoorArchetype(archetype);

    //Create a new list to add all the facilities that are matched
    var temp2 = new List<FaciliteitenLokaalFixed>();

    //Iterate over the facilities provided as a parameter, these are the facilities that have to be present in the archetype
    foreach(var faciliteit in faciliteiten)
    {
        //CHeck if facility is present in an archetype
        var aanwezig = temp.FirstOrDefault(t => t.Naam.Equals(faciliteit.Naam));
        if (aanwezig!=null)
        {
            //Add if it's present
            temp2.Add(faciliteit);
        }
    }
    var needed = faciliteiten.Count;
    var present = temp2.Count;

    //Compare how many facilities were met
    if (present / needed >= .5)
    {
        archetypes.Add(new Tuple<int, ArchetypesLokaal>(present, archetype));
    }
}

return archetypes.OrderByDescending(q => q.Item1).Select(q => q.Item2).ToList();

Answer 2

所以这不是一个完整的解决方案，但假设你在Archetype中有一个Facilteits的嵌套对象

 var matchedArchetypes = archetypes
            .Select(x => new { arch = x, cnt = x.Faciliteits.Join(faciliteitNaams, a => a.Naam, b => b.Naam, (a, b) => a).Count() })
            .ToList();

将为您提供原型和匹配设施的数量

Answer 3

正如其他人所述：我首先将其映射到原型/计数列表中，然后将其填入您需要的列表中（并根据此信息进行排序）。像这样：

 var archetypeUitFaciliteiten = db.ArchetypesLokaals.Select(archetypeLokaal => { 
    var faciliteitenVoorArchetype = GetFaciliteitenVoorArchetype(archetypeLokaal);
    return new {
        ArcheType = archetypeLokaal,
        Present = faciliteiten.Count(f => faciliteitenVoorArchetype.FirstOrDefault(t => t.Naam.Equals(f.Naam)) != null),
        Needed = faciliteiten.Count
    };
});

现在，您可以根据示例中的“当前/需要”比率将此列表（Select）映射到列表中，或者根据这些属性对此列表进行排序（OrderBy）。

（对不起，如果我屠杀了这种语言 - 我不会说荷兰语）

C＃使用外部标准对列表进行排序的方法

3 个答案: