在C#中是否有与F#的List.map函数相同的功能?即将函数应用于列表中的每个元素,并返回包含结果的新列表。
类似的东西:
public static IEnumerable<TResult> Map<TSource, TResult>(this IEnumerable<TSource> source, Func<TSource, TResult> funky)
{
foreach (TSource element in source)
yield return funky.Invoke(element);
}
是否已有内置方式或我应该只编写自定义扩展程序?
答案 0 :(得分:95)
这是LINQ的Select
- 即
var newSequence = originalSequence.Select(x => {translation});
或
var newSequence = from x in originalSequence
select {translation};
答案 1 :(得分:16)
ConvertAll
是内置函数:
public List<TOutput> ConvertAll<TOutput>(
Converter<T, TOutput> converter
)
自.NET 2.0版开始提供。
MSDN代码示例:
using System;
using System.Drawing;
using System.Collections.Generic;
public class Example
{
public static void Main()
{
List<PointF> lpf = new List<PointF>();
lpf.Add(new PointF(27.8F, 32.62F));
lpf.Add(new PointF(99.3F, 147.273F));
lpf.Add(new PointF(7.5F, 1412.2F));
Console.WriteLine();
foreach( PointF p in lpf )
{
Console.WriteLine(p);
}
List<Point> lp = lpf.ConvertAll(
new Converter<PointF, Point>(PointFToPoint));
Console.WriteLine();
foreach( Point p in lp )
{
Console.WriteLine(p);
}
}
public static Point PointFToPoint(PointF pf)
{
return new Point(((int) pf.X), ((int) pf.Y));
}
}
/* This code example produces the following output:
{X=27.8, Y=32.62}
{X=99.3, Y=147.273}
{X=7.5, Y=1412.2}
{X=27,Y=32}
{X=99,Y=147}
{X=7,Y=1412}
*/