在c#中将XML文件转换为csv文件格式

Question

<?xml version="1.0"?>
<ArrayOfSequence xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <Sequence>

    <SourcePath>
      <Point>
        <X>261</X>
        <Y>210</Y>
      </Point>
      <Point>
        <X>261</X>
        <Y>214</Y>
      </Point>
      <Point>
        <X>261</X>
        <Y>227</Y>
      </Point>
      <Point>
        <X>261</X>
        <Y>229</Y>
      </Point>
      <Point>
        <X>261</X>
        <Y>231</Y>
      </Point>
      <Point>
        <X>261</X>
        <Y>234</Y>
      </Point>
      <Point>
        <X>261</X>
        <Y>237</Y>
      </Point>
</Sequence>
</ArrayOfSequence>

我正在使用accord.net鼠标手势识别示例应用程序,它以上面的xml格式保存文件.我需要帮助将上面的xml转换为CSV格式,这样我就可以使用accord.net动态时间扭曲进行机器学习.我无法弄清楚如何转换为csv文件.例如:261,210,261,214,261,229,261,231

Answer 1

using System.IO;
using System.Xml.Serialization;

你可以这样做:

public class Sequence
{
    public Point[] SourcePath { get; set; }
}

using (FileStream fs = new FileStream(@"D:\youXMLFile.xml", FileMode.Open))
{
    XmlSerializer serializer = new XmlSerializer(typeof(Sequence[]));
    var data=(Sequence[]) serializer.Deserialize(fs);
    List<string> list = new List<string>();
    foreach(var item in data)
    {
        List<string> ss = new List<string>();
        foreach (var point in item.SourcePath) ss.Add(point.X + "," + point.Y);
        list.Add(string.Join(",", ss));
    }
    File.WriteAllLines("D:\\csvFile.csv", list);
}