如何使用 linq 删除项目

Question

我有以下数组，它代表我试图从某个数据结构实例中删除的项目：

string[] diseasesToRemove = new string[] { "D1", "D3", "D5" };

我可以使用以下方法做到这一点：

for (int i = 0; i < healthGroup.DiseaseGroups.Length; i++)
{
    var dgDiseases = new List<Disease>(healthGroup.DiseaseGroups[i].Diseases);
    for (int j = 0; j < dgDiseases.Count; j++)
    {
        if (diseasesToRemove.Contains(dgDiseases[j].Name))
        {                 
            dgDiseases.RemoveAt(j);
            j--;
        }
    }
    healthGroup.DiseaseGroups[i].Diseases = dgDiseases.ToArray();
}

但是，我确信使用 Linq 或其他东西有更好的方法。有吗？

以下是课程：

public class HealthGroup
{
    public DiseaseGroup[] DiseaseGroups { get; set; }

    public HealthGroup(DiseaseGroup[] diseaseGroups)
    {
        DiseaseGroups = diseaseGroups;
    }
}

public class DiseaseGroup
{
    public string Name { get; set; }
    public Disease[] Diseases;

    public DiseaseGroup(string name, Disease[] diseases)
    {
        Name = name;
        Diseases = diseases;
    }
}

public class Disease
{
    public string Name { get; set; } = "My Disease";
    public int Risk { get; set; } = 7;

    public Disease(string name, int risk)
    {
        Name = name;
        Risk = risk;
    }

    public override string ToString()
    {
        return $"{Name} with risk {Risk}";
    }
}

还有一些用于生成实例的样板：

Disease d1 = new Disease("D1", 1);
Disease d2 = new Disease("D2", 2);
Disease d3 = new Disease("D3", 3);
Disease d4 = new Disease("D4", 4);
Disease d5 = new Disease("D5", 5);
Disease d6 = new Disease("D6", 6);
Disease d7 = new Disease("D7", 7);

DiseaseGroup dg1 = new DiseaseGroup("DG1", new Disease[] { d1, d2 });
DiseaseGroup dg2 = new DiseaseGroup("DG2", new Disease[] { d3, d4, d5 });
DiseaseGroup dg3 = new DiseaseGroup("DG3", new Disease[] { d6, d7 });

HealthGroup healthGroup = new HealthGroup(new DiseaseGroup[] { dg1, dg2, dg3 });

Answer 1

您可以使用Where 将代码简化为：

foreach (var diseaseGroup in healthGroup.DiseaseGroups)
{
    diseaseGroup.Diseases 
        = diseaseGroup.Diseases.Where(g => !diseasesToRemove.Contains(g.Name)).ToArray();
}

当然，就像原始代码一样，这会生成一个新列表。一个性能更高的选项（尽管不是 LINQ）可能是使 Diseases 成为 List<Disease>：

public class DiseaseGroup
{
    public string Name { get; set; }
    public List<Disease> Diseases;

    public DiseaseGroup(string name, Disease[] diseases)
    {
        Name = name;
        Diseases = new List<Disease>(diseases);
    }
}

然后您可以使用类似的代码，而无需生成新列表和新数组的开销：

for (int i = 0; i < healthGroup.DiseaseGroups.Length; i++)
{
    for (int j = healthGroup.DiseaseGroups[i].Diseases.Count - 1; j >= 0; --j)
    {
        if (diseasesToRemove.Contains(healthGroup.DiseaseGroups[i].Diseases[j].Name))
        {                 
            healthGroup.DiseaseGroups[i].Diseases.RemoveAt(j);
        }
    }
}

我还修改了您的 for 循环以向后工作，以解决您在 --j 部分遇到的问题。

而且，对于大量的项目，与其使用string[] 来存储diseasesToRemove，不如使用HashSet<string> 来存储疾病。

---

编辑以包含请求的单行：

healthGroup.DiseaseGroups = healthGroup.DiseaseGroups.Select(g => { return g.Diseases = g.Diseases.Where(g => !diseasesToRemove.Contains(g.Name)).ToArray(); }).ToArray();

这有点滥用选择，但是:-)

Answer 2

这些类型的过滤作业可以很好地用连接来表示。他们解决了重复迭代一个组以检查另一组中的匹配项的问题（.Contains 这里是 O(n) 并且被频繁调用......这不需要发生）。

不幸的是，linq-to-objects 仅执行开箱即用的内连接，但我保留的一种重复使用较多的扩展方法简化了执行基于集合的左外连接的工作。

如果您在疾病和diseasesToRemove 之间执行左外连接，则您从连接的左侧集合 (healthGroup.DiseaseGroups[i].Diseases) 中选择与右侧集合中的任何内容都不匹配的项目 ( diseasesToRemove)，那么您将删除任何匹配的内容。

使用扩展方法.LeftOuterJoin（如下所列），您可以按如下方式过滤您的数组：

for (int i = 0; i < healthGroup.DiseaseGroups.Length; i++)
{
    healthGroup.DiseaseGroups[i].Diseases =
        healthGroup.DiseaseGroups[i].Diseases
            .LeftOuterJoin(
                diseasesToRemove,
                d => d.Name,
                dr => dr,
                (d, dr) => ( d, dr ))
            .Where(x => x.dr == null)
            .Select(x => x.d)
            .ToArray();

}

左外连接扩展方法：

public static class JoinExtensions
{
    public static IEnumerable<TResult> LeftOuterJoin<TLeft, TRight, TKey, TResult>(
        this IEnumerable<TLeft> leftSeq,
        IEnumerable<TRight> rightSeq,
        Func<TLeft, TKey> keySelectorLeft,
        Func<TRight, TKey> keySelectorRight,
        Func<TLeft, TRight, TResult> projectionSelector)
    {
        return leftSeq
            .GroupJoin(
                rightSeq,
                keySelectorLeft,
                keySelectorRight,
                (leftItem, rightItems) => new { leftItem, rightItems })
            .SelectMany(
                x => x.rightItems.DefaultIfEmpty(),
                (x, rightItem) => projectionSelector(x.leftItem, rightItem));
    }
}