使用ImageSharp将包含Bgra32的传入缓冲区转换为ImageSharp Image<Rgba24>图像的最佳方法是什么？

using SixLabors.ImageSharp;
using SixLabors.ImageSharp.PixelFormats;
using System;

public class ImageConverter
{
    public static Image<Rgba24> ConvertBgra32ToRgba24(byte[] buffer, int width, int height)
    {
        // 创建一个Bgra32格式的图像
        using (var bgraImage = Image.LoadPixelData<Bgra32>(buffer, width, height))
        {
            // 创建一个Rgba24格式的图像
            var rgbaImage = new Image<Rgba24>(width, height);

            // 将Bgra32图像的像素数据复制到Rgba24图像中
            for (int y = 0; y < height; y++)
            {
                for (int x = 0; x < width; x++)
                {
                    var bgraPixel = bgraImage.GetPixel(x, y);
                    rgbaImage.SetPixel(x, y, new Rgba24(bgraPixel.R, bgraPixel.G, bgraPixel.B, bgraPixel.A));
                }
            }

            return rgbaImage;
        }
    }
}