Converting a List<int> to a comma separated string
C#CollectionsC# Problem Overview
Is there a way to take a List
I know I can just loop and build it, but somehow I think some of you guys a more cool way of doing it?
I really want to learn these types of 'tricks', so please explain or link to the docs on the method you use.
C# Solutions
Solution 1 - C#
List<int> list = ...;
string.Join(",", list.Select(n => n.ToString()).ToArray())
Solution 2 - C#
Simple solution is
List<int> list = new List<int>() {1,2,3};
string.Join<int>(",", list)
I used it just now in my code, working funtastic.
Solution 3 - C#
List<int> list = new List<int> { 1, 2, 3 };
Console.WriteLine(String.Join(",", list.Select(i => i.ToString()).ToArray()));
Solution 4 - C#
For approximately one gazillion solutions to a slightly more complicated version of this problem -- many of which are slow, buggy, or don't even compile -- see the comments to my article on this subject:
https://docs.microsoft.com/en-us/archive/blogs/ericlippert/comma-quibbling
and the StackOverflow commentary:
https://stackoverflow.com/questions/788535/eric-lipperts-challenge-comma-quibbling-best-answer
Solution 5 - C#
For extra coolness I would make this an extension method on IEnumerable<T> so that it works on any IEnumerable:
public static class IEnumerableExtensions {
public static string BuildString<T>(this IEnumerable<T> self, string delim = ",") {
return string.Join(delim, self)
}
}
Use it as follows:
List<int> list = new List<int> { 1, 2, 3 };
Console.WriteLine(list.BuildString(", "));
Solution 6 - C#
My "clever" entry:
List<int> list = new List<int> { 1, 2, 3 };
StringBuilder sb = new StringBuilder();
var y = list.Skip(1).Aggregate(sb.Append(x.ToString()),
(sb1, x) => sb1.AppendFormat(",{0}",x));
// A lot of mess to remove initial comma
Console.WriteLine(y.ToString().Substring(1,y.Length - 1));
Just haven't figured how to conditionally add the comma.
Solution 7 - C#
Seems reasonablly fast.
IList<int> listItem = Enumerable.Range(0, 100000).ToList();
var result = listItem.Aggregate<int, StringBuilder, string>(new StringBuilder(), (strBuild, intVal) => { strBuild.Append(intVal); strBuild.Append(","); return strBuild; }, (strBuild) => strBuild.ToString(0, strBuild.Length - 1));
Solution 8 - C#
you can use, the System.Linq library; It is more efficient:
using System.Linq;
string str =string.Join(",", MyList.Select(x => x.NombreAtributo));