如何合并2 List< T>删除C#中的重复值 [英] how to merge 2 List<T> with removing duplicate values in C#
问题描述
有点难以解释,所以让我举个例子代码看起来像我想要的结果,在样例我使用int类型不是ResultAnalysisFileSql类。
first_list = [1,12,12,5] p>
second_list = [12,5,7,9,1]
组合这两个列表的结果应该是在此列表中:
due_list = [1,12,5,7,9]
您会注意到结果有第一个列表,其中包括两个12值,而在second_list中还有一个额外的12,1和5值。
ResultAnalysisFileSql类
[Serializable]
public partial class ResultAnalysisFileSql
{
public string FileSql {get;组; }
public string PathFileSql {get;组; }
public List< ErrorAnalysisSql>错误{get;组; }
public List< WarningAnalysisSql>警告{get;组; }
public ResultAnalysisFileSql()
{
}
public ResultAnalysisFileSql(string fileSql)
{
if(string.IsNullOrEmpty(fileSql)
|| fileSql.Trim()。Length == 0)
{
throw new ArgumentNullException(fileSql,fileSql is null);
}
if(!fileSql.EndsWith(Utility.ExtensionFicherosErrorYWarning))
{
throw new ArgumentOutOfRangeException(fileSql,Ruta de fichero Sql no tieneextensión + Utility.ExtensionFicherosErrorYWarning);
}
PathFileSql = fileSql;
FileSql = ObtenerNombreFicheroSql(fileSql);
Errors = new List< ErrorAnalysisSql>();
Warnings = new List< WarningAnalysisSql>();
}
私有字符串ObtenerNombreFicheroSql(string fileSql)
{
var f = Path.GetFileName(fileSql);
return f.Substring(0,f.IndexOf(Utility.ExtensionFicherosErrorYWarning));
}
public override bool Equals(object obj)
{
if(obj == null)
return false;
if(!(obj是ResultAnalysisFileSql))
return false;
var t = obj as ResultAnalysisFileSql;
return t.FileSql == this.FileSql
&&& t.PathFileSql == this.PathFileSql
&&& t.Errors.Count == this.Errors.Count
&& t.Warnings.Count == this.Warnings.Count;
}
}
任何示例代码为了合并和删除重复项目?
你有看过 Enumerable.Union
此方法从返回集合中排除重复项。这对于Concat
方法是不同的
行为,它返回输入序列中所有元素
,包括
重复。
列表< int> list1 = new List< int> {1,12,12,5};
列表< int> list2 = new List< int> {12,5,7,9,1};
列表< int> ulist = list1.Union(list2).ToList();
I have two lists List that i need to combine and removing duplicate values of both lists
A bit hard to explain, so let me show an example of what the code looks like, and what i want as a result, in sample I use int type not ResultAnalysisFileSql class.
first_list = [1, 12, 12, 5]
second_list = [12, 5, 7, 9, 1]
The result of combining the two lists should result in this list: resulting_list = [1, 12, 5, 7, 9]
You'll notice that the result has the first list, including its two "12" values, and in second_list has an additional 12, 1 and 5 value.
ResultAnalysisFileSql class
[Serializable]
public partial class ResultAnalysisFileSql
{
public string FileSql { get; set; }
public string PathFileSql { get; set; }
public List<ErrorAnalysisSql> Errors { get; set; }
public List<WarningAnalysisSql> Warnings{ get; set; }
public ResultAnalysisFileSql()
{
}
public ResultAnalysisFileSql(string fileSql)
{
if (string.IsNullOrEmpty(fileSql)
|| fileSql.Trim().Length == 0)
{
throw new ArgumentNullException("fileSql", "fileSql is null");
}
if (!fileSql.EndsWith(Utility.ExtensionFicherosErrorYWarning))
{
throw new ArgumentOutOfRangeException("fileSql", "Ruta de fichero Sql no tiene extensión " + Utility.ExtensionFicherosErrorYWarning);
}
PathFileSql = fileSql;
FileSql = ObtenerNombreFicheroSql(fileSql);
Errors = new List<ErrorAnalysisSql>();
Warnings= new List<WarningAnalysisSql>();
}
private string ObtenerNombreFicheroSql(string fileSql)
{
var f = Path.GetFileName(fileSql);
return f.Substring(0, f.IndexOf(Utility.ExtensionFicherosErrorYWarning));
}
public override bool Equals(object obj)
{
if (obj == null)
return false;
if (!(obj is ResultAnalysisFileSql))
return false;
var t = obj as ResultAnalysisFileSql;
return t.FileSql== this.FileSql
&& t.PathFileSql == this.PathFileSql
&& t.Errors.Count == this.Errors.Count
&& t.Warnings.Count == this.Warnings.Count;
}
}
Any sample code for combine and removing duplicates ?
Have you had a look at Enumerable.Union
This method excludes duplicates from the return set. This is different behavior to the Concat method, which returns all the elements in the input sequences including duplicates.
List<int> list1 = new List<int> { 1, 12, 12, 5};
List<int> list2 = new List<int> { 12, 5, 7, 9, 1 };
List<int> ulist = list1.Union(list2).ToList();
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