c# - 使用 GribCS 库 C# 迭代 Grib2 值？

Question

我正在使用建议的库GribCS它没有任何示例或示例显示关于 Lat/Lon 的迭代以及它们各自的值以可读形式存储。任何想法？

Answer 1

为此，您必须像我在下面所做的那样使用坐标迭代，并且您必须设置您的要求参数条件以提取相关值。

    static void ReadGribTwoFile()
    {
        #region Grib 2 Code
        Grib2Input input = new Grib2Input(RandomAccessFile);

        if (!input.scan(false, false))
        {
            Console.WriteLine("Failed to successfully scan grib file");
            return;
        }
        Grib2Data data = new Grib2Data(RandomAccessFile);

        var records = input.Records;

        foreach (Grib2Record record in records)
        {
            IGrib2IndicatorSection iis = record.Is;
            IGrib2IdentificationSection id = record.ID;
            IGrib2ProductDefinitionSection pdsv = record.PDS;
            IGrib2GridDefinitionSection gdsv = record.GDS;

            long time = id.RefTime.AddTicks(record.PDS.ForecastTime * 3600000).Ticks;

            Console.WriteLine("Record description at " + " forecast " + new DateTime(time) + ": " + string.Format("{0} {1} {2}", iis.Discipline, pdsv.ParameterCategory, pdsv.ParameterNumber));

            float[] values = data.getData(record.getGdsOffset(), record.getPdsOffset());

            if ((iis.Discipline == 0) && (pdsv.ParameterCategory == 1) && (pdsv.ParameterNumber == 1))
            {
                // RH
                int c = 0;
                for (double lat = gdsv.La1; lat >= gdsv.La2; lat = lat - gdsv.Dy)
                {
                    for (double lon = gdsv.Lo1; lon <= gdsv.Lo2; lon = lon + gdsv.Dx)
                    {
                        Console.WriteLine("RH " + lat + "\t" + lon + "\t" + values[c]);
                        c++;
                    }
                }
            }
        }

        #endregion
    }