我正在尝试将以前的数组(1 和 2)存储在新数组 3 中的结果可以显示数组 1 的每个结果和每个数组 2,当我分别显示每个数组的结果时,它们是可以的,但是当我尝试将它们混合在数组 3 中,我得到了数组 1 中第一个值和数组 2 中所有值的正确答案,然后出现了很多零。
这是数组 3 的代码
for (int i = 0; i < result1.GetLength(0); i++)
{
for (int j = 0; j < result1.GetLength(1); j++)
{
for (int k = 0; k < result2.GetLength(0); k++)
{
for (int m = 0; m < result2.GetLength(1); m++)
{
result3[i, j] = result1[i, j] + "," + result2[k, m];
Console.WriteLine(result3[i, j]);
counter++;
}
}
}
}
这里是完整的代码
double[,] Cranelocations = { { -12.3256, 0.5344 }, { -12.3256, -0.4656 }, { -12.3256, -1.4656 }, { -12.3256, -2.4656 } };
double[,] Picklocation = { { -0.3256, -3.4656 }, { 0.6744, -3.4656 }, { 1.6744, -3.4656 }, { 2.6744, -3.4656 }, { 3.6744, -3.4656 }, { 4.6744, -3.4656 }, { 5.6744, -3.4656 } };
double[,] Setlocation = { { 20.62, 5.03 }, { 24.28, 5.03 }, { 28.40, 5.03 }, { 32.11, 5.03 }, { 35.99, 5.26 }, { 40.18, 5.26 } };
double[] Weights = { 11.7865, 14.7335, 15.1015, 10.7465 };
double[,] result1 = new double[Weights.Length * Cranelocations.GetLength(0), Picklocation.GetLength(0)];
double[,] result2 = new double[Weights.Length * Cranelocations.GetLength(0), Setlocation.GetLength(0)];
object[,] result3 = new object[result1.GetLength(0) * result1.GetLength(1), result2.GetLength(0) * result2.GetLength(1)];
int counter = 0;
for (int m = 0; m < Weights.Length; m++)
{
for (int i = 0; i < Cranelocations.GetLength(0); i++)
{
for (int j = 0; j < Picklocation.GetLength(0); j++)
{
double x = Cranelocations[i, 0] - Picklocation[j, 0];
double y = Cranelocations[i, 1] - Picklocation[j, 1];
result1[i, j] = Weights[m] * (Math.Sqrt(Math.Pow(x, 2) + Math.Pow(y, 2)));
}
}
}
Console.WriteLine("-----------------------------------------------------------------");
for (int m = 0; m < Weights.Length; m++)
{
for (int i = 0; i < Cranelocations.GetLength(0); i++)
{
for (int j = 0; j < Setlocation.GetLength(0); j++)
{
double x = Cranelocations[i, 0] - Setlocation[j, 0];
double y = Cranelocations[i, 1] - Setlocation[j, 1];
result2[i, j] = Weights[m] * (Math.Sqrt(Math.Pow(x, 2) + Math.Pow(y, 2)));
}
}
}
for (int i = 0; i < result1.GetLength(0); i++)
{
for (int j = 0; j < result1.GetLength(1); j++)
{
for (int k = 0; k < result2.GetLength(0); k++)
{
for (int m = 0; m < result2.GetLength(1); m++)
{
result3[i, j] = result1[i, j] + "," + result2[k, m];
Console.WriteLine(result3[i, j]);
counter++;
}
}
}
}
}
最佳答案
对于每个 m
,您将 i
重置为 0 并再次从数组的开头开始写入。
当您递增m
时,您需要继续向前移动索引。
当您合并两个索引时也是如此。当您组合两个迭代以获得索引时,您通常将第一个迭代乘以第二个迭代的长度,然后将它们相加。
for (int m = 0; m < Weights.Length; m++)
{
int offset = m * Cranelocations.GetLength(0);
for (int i = 0; i < Cranelocations.GetLength(0); i++)
{
for (int j = 0; j < Picklocation.GetLength(0); j++)
{
double x = Cranelocations[i, 0] - Picklocation[j, 0];
double y = Cranelocations[i, 1] - Picklocation[j, 1];
result1[i + offset, j] = Weights[m] * (Math.Sqrt(Math.Pow(x, 2) + Math.Pow(y, 2)));
}
}
}
Console.WriteLine("-----------------------------------------------------------------");
for (int m = 0; m < Weights.Length; m++)
{
int offset = m * Cranelocations.GetLength(0);
for (int i = 0; i < Cranelocations.GetLength(0); i++)
{
for (int j = 0; j < Setlocation.GetLength(0); j++)
{
double x = Cranelocations[i, 0] - Setlocation[j, 0];
double y = Cranelocations[i, 1] - Setlocation[j, 1];
result2[i + offset, j] = Weights[m] * (Math.Sqrt(Math.Pow(x, 2) + Math.Pow(y, 2)));
}
}
}
for (int i = 0; i < result1.GetLength(0); i++)
{
int iOffset = i * result1.GetLength(1);
for (int j = 0; j < result1.GetLength(1); j++)
{
for (int k = 0; k < result2.GetLength(0); k++)
{
int kOffset = k * result2.GetLength(1);
for (int m = 0; m < result2.GetLength(1); m++)
{
result3[iOffset + j, kOffset + m] = result1[i, j] + "," + result2[k, m];
Console.WriteLine(result3[iOffset + j, kOffset + m]);
counter++;
}
}
}
}
关于c# - 将两个二维数组混合成1个二维数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55133402/