二维数组和

Question

我有如下所示的二维浮点数组，

{0.2,0.0,0.3,0.0,0.0}
{0.4,0.1,0.0,0.0,0.9}
{0.0,0.0,0.0,0.3,0.6}

我想得到以下输出

{0.6,0.0,0.3,0.0,0.0}
{0.6,0.1,0.0,0.0,1.5}
{0.0,0.0,0.0,0.3,1.5}

如果您进行分析，我将计算每列的非零值的总和，并用该总和值更新所有非零值。例如，在第一列中，我求和 (0.2+0.4=0.4) 并将两个值位置更新为 0.6。

我正在使用 Java，我该如何执行此操作？这是一个简单的例子，实时我有非常大的数组。

Answer 1

假设它们的长度都相同，这是可行的。特殊情况可供读者练习。

class MatTest {
    static void makeSums(float[][] floats) {
        // we wouldn't be doing any operations on these inputs anyway, so return
        if(floats == null || floats.length == 0 || floats.length == 1) return;

        // check to make sure it's retangular
        for(float[] arr : floats) {
            if(arr.length != floats[0].length) {
                throw new IllegalArgumentException("makeSums() requires rectangular array");
            }
        }

        for(int i = 0; i < floats[0].length; i++) {
            // do each column
            float sum = 0f;
            for(int j = 0; j < floats.length; j++) {
                sum += floats[j][i];
            }
            for(int j = 0; j < floats.length; j++) {
                if(floats[j][i] != 0) floats[j][i] = sum;
            }
        }
    }

    public static void main(String[] args) {
        float[][] floats = new float[3][5];
        floats[0] = new float[] {0.2f,0.0f,0.3f,0.0f,0.0f};
        floats[1] = new float[] {0.4f,0.1f,0.0f,0.0f,0.9f};
        floats[2] = new float[] {0.0f,0.0f,0.0f,0.3f,0.6f};

        makeSums(floats);

        for(int i = 0; i < floats.length; i++) {
            for(int j = 0; j < floats[0].length; j++) {
                System.out.print(floats[i][j]);
                System.out.print(" ");
            }
            System.out.println(" ");
        }
    }
}

这是它的结果：

 C:\Documents and Settings\glow\My Documents>javac MatTest.java

 C:\Documents and Settings\glow\My Documents>java MatTest
 0.6 0.0 0.3 0.0 0.0
 0.6 0.1 0.0 0.0 1.5
 0.0 0.0 0.0 0.3 1.5

Answer 2

我们将输入数组命名为 float[][] a 并将并行输出数组命名为 b 并初始化为全零。

float curSum = 0.0;   
first = true;
for(int i = 0; i < a[0].length; i++)
{
    for(int j = 0; j < a.length; j++)
    { 
         if(a[i][j] != 0)
         {
             if (first)
             {
                 for(int k = j; k < a.length; k++)
                     curSum += a[i][k];
                 first = false;
             }
             b[i][j] = curSum;
         }
     }
     curSum = 0.0;
     first = true;
}

可能有一些更好的点你必须改变，比如浮点数和其他东西的比较，但我认为这个想法就在那里，

我认为它在 O(n*m) 中运行，这看起来不太好，但我试图保留迭代尽可能短。我没有看到任何更快的方法。即使有三个 for 循环，带有 k 的循环也只会为每个 j 循环运行一次，因此渐近地它根本不会增加复杂性。