Question

我有以下数组，表示我要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，我确定使用Linq或其他方法会有更好的方法。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}";
    }
}

和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来简化代码：

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

当然，与原始代码一样，这会生成一个新列表。更具性能的选择（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，而不会产生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。

对于大量项目，与其将string[]用于diseasesToRemove，不如将HashSet<string>用于存储疾病。

编辑以包含请求的单线：

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

这有点滥用，但是：-）

Answer 2

这些过滤作业可以很好地用联接表示。它们解决了重复迭代一个组以检查另一组中是否存在匹配的问题（.Contains这里是O（n），并且经常被调用...这不需要发生）。

不幸的是，linq-to-objects只能直接进行内部联接，但是我一直使用的重复使用的扩展方法简化了执行基于集合的左外部联接的工作。

如果您在疾病和diseasesToRemove之间执行从左到外的联接，则从联接的左侧集合（healthGroup.DiseaseGroups[i].Diseases中选择与右边的任何项都不匹配的项手动收集（diseasesToRemove），那么您将删除所有具有匹配项的内容。

使用扩展方法.LeftOuterJoin（下面列出），您可以按以下方式过滤数组：

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));
    }
}

如何使用linq删除项目

2 个答案: