我有以下代码:

public class Info
{
  public string Name;
  public string Num;
}

string s1 = "a,b";
string s2 = "1,2";

IEnumerable<Info> InfoSrc =
    from name in s1.Split(',')
    from num in s2.Split(',')
    select new Info()
    {
        Name = name,
        Num = num
    };

List<Info> listSrc = InfoSrc.ToList();

我希望我的listSrc结果包含两个Info项,它们的NameNum属性是:
a, 1
b, 2

但是,上面显示的代码有四个结果:
a, 1
a, 2
b, 1
b, 2

最佳答案

您可以使用Enumerable.Zip

IEnumerable<Info> InfoSrc = s1.Split(',')
    .Zip(s2.Split(','), (name, num) => new Info(){ Name = name, Num = num });

如果需要将两个以上的集合映射到属性,可以将多个Zip与包含第二个和第三个的匿名类型链接在一起:
IEnumerable<Info> InfoSrc = s1.Split(',')
    .Zip(s2.Split(',').Zip(s3.Split(','), (second, third) => new { second, third }),
        (first, x) => new Info { Name = first, Num = x.second, Prop3 = x.third });

下面是一个更具可读性的版本:
var arrays = new List<string[]> { s1.Split(','), s2.Split(','), s3.Split(',') };
int minLength = arrays.Min(arr => arr.Length);  // to be safe, same behaviour as Zip
IEnumerable<Info> InfoSrc = Enumerable.Range(0, minLength)
 .Select(i => new Info
 {
     Name = arrays[0][i],
     Num = arrays[1][i],
     Prop3 = arrays[2][i]
 });

07-24 22:34