动态创建未知数量的数组并将值赋给它们

时间：2016-06-01 08:55:56

标签： c# arrays

我有一个字符串格式的数据集。这些是我通过图像处理获得的一些坐标。我附加了提取这些（坐标）并存储在单个数组中的代码。从这些坐标中，我发现x值相似。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Stack
{
    class Program
    {
        static void Main(string[] args)
        {
            int count = 0;
            //String which have my coordinates
            string TextFile = "[(27, 37), (27, 38), (27, 163), (27, 164), (27, 266), (27, 267), (27, 356), (27, 357), (27, 448), (27, 449), (27, 528), (27, 529), (28, 37), (28, 38), (28, 163), (28, 164), (28, 266), (28, 267), (28, 356), (28, 357), (28, 448), (28, 449), (28, 528), (28, 529), (51, 163), (51, 164), (51, 266), (51, 267), (51, 356), (51, 357), (51, 448), (51, 449), (52, 61), (52, 62), (52, 163), (52, 164), (52, 266), (52, 267), (52, 280), (52, 305), (52, 306), (52, 356), (52, 357), (52, 448), (52, 449), (52, 504), (52, 505), (53, 61), (53, 62), (53, 280), (53, 281), (53, 305), (53, 306), (53, 504), (53, 505), (355, 61), (355, 62), (355, 280), (355, 281), (355, 305), (355, 306), (355, 504), (355, 505), (356, 61), (356, 62), (356, 280), (356, 281), (356, 305), (356, 306), (356, 504), (356, 505), (380, 37), (380, 38), (380, 528), (380, 529), (381, 37), (381, 38), (381, 528), (381, 529)]";
            string splits = TextFile.TrimStart('[');
            string[] split = TextFile.Split(')');

            string split_1 = null;
            string split_2 = null;
            string split_3 = null;

            int pos = 0;
            int lengthOfString = 0;
            int valX, valZ, valX1, valZ1 = 0;

            /*
             * Getting the count of the coordinates in the array
             */
            foreach (string x in split)
            {
                count++;
            }

            string[] stringArr = new string[count];

            /*
             * Splitting the coordinates as x,y and store in an array
             */
            foreach (string coord in split)
            {
                split_1 = coord;
                split_1 = split_1.Trim('[');
                split_1 = split_1.Trim('(');
                split_1 = split_1.Trim(',');
                split_1 = split_1.Trim(' ');
                split_1 = split_1.Trim('(');
                split_1 = split_1.TrimEnd(']');
                stringArr[pos] = split_1;
                pos++;
            }

            Console.WriteLine("Array Length " + stringArr.Length);

            /*
             * extracting simalar x coordinates
             */
            Console.WriteLine("");
            Console.WriteLine("-----------------extracting simalar x coordinates----------------");
            Console.WriteLine("");

            int indexX = 0;
            int loopRunX = 0;
            for (int a = 0; a < stringArr.Length - 1; a = indexX)
            {
                split_2 = stringArr[a];
                lengthOfString = split_2.Length;

                valX = int.Parse(split_2.Substring(0, split_2.IndexOf(',')));
                valZ = int.Parse(split_2.Substring(split_2.IndexOf(' '), (lengthOfString - split_2.IndexOf(' '))));

                int countx = 0;
                for (int x1 = indexX; x1 < stringArr.Length - 1; x1++)
                {
                    split_3 = stringArr[x1];
                    lengthOfString = split_3.Length;

                    valX1 = int.Parse(split_3.Substring(0, split_3.IndexOf(',')));
                    valZ1 = int.Parse(split_3.Substring(split_3.IndexOf(' '), (lengthOfString - split_3.IndexOf(' '))));

                    //Check for the simillar x in the text file we provide
                    if (valX == valX1)
                    {
                        countx++;
                        Console.WriteLine("X is " + valX + " and the coordinates which have simillar x ==> (" + valX1 + ", " + valZ1 + "). Index is " + x1 + " Count is " + countx);
                    }
                    else
                    {
                        break;
                    }
                }
                loopRunX++;
                indexX = indexX + countx;
                //Console.WriteLine("Next Index to check onwards : " + indexX);
                //Console.WriteLine("Looping Count : " + loopRunX);
                Console.WriteLine("");
            }
            Console.ReadLine();
        }
    }
}

enter image description here

我想将这些相似的x值存储在不同的数组中。 例如：值27出现12次，我想将这些值存储到一个单独的数组中。值28出现12次，与上面相同，它应该将所有28个值存储在一个数组中。 但是我不知道需要多少个数组。我知道一个解决方案可以动态地将这些相似的坐标存储在一个数组中吗？

1 个答案:

答案 0 :(得分：2)

您可以使用LINQ GroupBy方法。

notifyDataSetChanged

结果：

    class Program
    {
        static void Main(string[] args)
        {
            int count = 0;
            //String which have my coordinates
            string TextFile = "[(27, 37), (27, 38), (27, 163), (27, 164), (27, 266), (27, 267), (27, 356), (27, 357), (27, 448), (27, 449), (27, 528), (27, 529), (28, 37), (28, 38), (28, 163), (28, 164), (28, 266), (28, 267), (28, 356), (28, 357), (28, 448), (28, 449), (28, 528), (28, 529), (51, 163), (51, 164), (51, 266), (51, 267), (51, 356), (51, 357), (51, 448), (51, 449), (52, 61), (52, 62), (52, 163), (52, 164), (52, 266), (52, 267), (52, 280), (52, 305), (52, 306), (52, 356), (52, 357), (52, 448), (52, 449), (52, 504), (52, 505), (53, 61), (53, 62), (53, 280), (53, 281), (53, 305), (53, 306), (53, 504), (53, 505), (355, 61), (355, 62), (355, 280), (355, 281), (355, 305), (355, 306), (355, 504), (355, 505), (356, 61), (356, 62), (356, 280), (356, 281), (356, 305), (356, 306), (356, 504), (356, 505), (380, 37), (380, 38), (380, 528), (380, 529), (381, 37), (381, 38), (381, 528), (381, 529)]";
            string splits = TextFile.TrimStart('[');
            string[] split = TextFile.Split(')');

            string split_1 = null;
            string split_2 = null;
            string split_3 = null;

            int pos = 0;
            int lengthOfString = 0;
            int valX, valZ, valX1, valZ1 = 0;

            /*
             * Getting the count of the coordinates in the array
             */
            foreach (string x in split)
            {
                count++;
            }

            string[] stringArr = new string[count];

            /*
             * Splitting the coordinates as x,y and store in an array
             */
            foreach (string coord in split)
            {
                split_1 = coord;
                split_1 = split_1.Trim('[');
                split_1 = split_1.Trim('(');
                split_1 = split_1.Trim(',');
                split_1 = split_1.Trim(' ');
                split_1 = split_1.Trim('(');
                split_1 = split_1.TrimEnd(']');
                stringArr[pos] = split_1;
                pos++;
            }

            Console.WriteLine("Array Length " + stringArr.Length);

            //It is better to convert to Point type but to be short I've used your string
            var groupedPoints = stringArr.GroupBy(
                str => str.Split(',')[0]);

            foreach (var group in groupedPoints)
            {
                Console.WriteLine("Group {0}", group.Key);
                foreach (var point in group)
                    Console.WriteLine("\tPoint {0}", point);
            }

            Console.ReadLine();
        }
    }
}

此示例基于您的代码。但我建议用Array Length 80 Group 27 Point 27, 37 Point 27, 38 Point 27, 163 Point 27, 164 Point 27, 266 Point 27, 267 Point 27, 356 Point 27, 357 Point 27, 448 Point 27, 449 Point 27, 528 Point 27, 529 Group 28 Point 28, 37 Point 28, 38 Point 28, 163 Point 28, 164 Point 28, 266 Point 28, 267 Point 28, 356 Point 28, 357 Point 28, 448 Point 28, 449 Point 28, 528 Point 28, 529 Group 51 Point 51, 163 Point 51, 164 Point 51, 266 Point 51, 267 Point 51, 356 Point 51, 357 Point 51, 448 Point 51, 449 Group 52 Point 52, 61 Point 52, 62 Point 52, 163 Point 52, 164 Point 52, 266 Point 52, 267 Point 52, 280 Point 52, 305 Point 52, 306 Point 52, 356 Point 52, 357 Point 52, 448 Point 52, 449 Point 52, 504 Point 52, 505 Group 53 Point 53, 61 Point 53, 62 Point 53, 280 Point 53, 281 Point 53, 305 Point 53, 306 Point 53, 504 Point 53, 505 Group 355 Point 355, 61 Point 355, 62 Point 355, 280 Point 355, 281 Point 355, 305 Point 355, 306 Point 355, 504 Point 355, 505 Group 356 Point 356, 61 Point 356, 62 Point 356, 280 Point 356, 281 Point 356, 305 Point 356, 306 Point 356, 504 Point 356, 505 Group 380 Point 380, 37 Point 380, 38 Point 380, 528 Point 380, 529 Group 381 Point 381, 37 Point 381, 38 Point 381, 528 Point 381, 529 方法编写自己的MyPoint2d class \ struct。

因此，您将能够转换

  

TryParse - ＆gt; string TextFile

然后按MyPoint2d[] arrPts

对此数组进行分组

此外，您不必使用静态大小数组（MyPoint2d.x）。 string[]将为您提供更多自由（在分配数组之前无需计算大小）。

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