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  • 来自专栏Venyo 的专栏

    pHash的Java实现

    return 转置后的矩阵 */ private static double[][] transposingMatrix(double[][] matrix, int n) { double nMatrix [][] = new double[n][n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { nMatrix [i][j] = matrix[j][i]; } } return nMatrix; } /** * 矩阵相乘 * @author luoweifu * * @param ; j < n; j++) { t = 0; for (int k = 0; k < n; k++) { t += A[i][k] * B[k][j]; } nMatrix [i][j] = t; } } return nMatrix; } }

    2.7K70发布于 2018-03-15
  • 来自专栏我的充电站

    数值计算方法 Chapter1. 插值

    assert(len(xlist) == len(ylist)) assert(len(set(xlist)) == len(xlist)) n = len(xlist) nmatrix = [[0 for _ in range(n)] for _ in range(n)] for i in range(n): nmatrix[i][0] = ylist[i] for j in range(1, i+1): nmatrix[i][j] = (nmatrix[i][j-1] - nmatrix[i-1][j-1]) / ( xlist[i] - xlist[i-j]) nlist = [nmatrix[i][i] for i in range(n)] def fn(x):

    1.1K30编辑于 2022-05-11
  • 来自专栏深度学习

    【数据结构】数组和字符串(七):特殊矩阵的压缩存储:三元组表的转置、加法、乘法操作

    insertElement(&matrixB, 2, 1, 9); printf("Matrix A:\n"); displayMatrix(&matrixA); printf("\nMatrix displayMatrix(&matrixB); TripletTable matrixC = matrixAddition(&matrixA, &matrixB); printf("\nMatrix displayMatrix(&matrixD); TripletTable matrixE = matrixMultiplication(&matrixA, &matrixB); printf("\nMatrix insertElement(&matrixB, 2, 1, 9); printf("Matrix A:\n"); displayMatrix(&matrixA); printf("\nMatrix displayMatrix(&matrixB); TripletTable matrixC = matrixAddition(&matrixA, &matrixB); printf("\nMatrix

    53910编辑于 2024-07-30
  • 来自专栏深度学习

    【数据结构】数组和字符串(十):稀疏矩阵的链接存储:十字链表的矩阵操作(加法、乘法、转置)

    matrix2); // 矩阵加法 SparseMatrix* additionResult = matrixAddition(matrix1, matrix1); printf("\nMatrix 矩阵乘法 SparseMatrix* multiplicationResult = matrixMultiplication(matrix1, matrix2); printf("\nMatrix multiplicationResult); // 矩阵转置 SparseMatrix* transposeResult = matrixTranspose(matrix1); printf("\nMatrix matrix2); // 矩阵加法 SparseMatrix* additionResult = matrixAddition(matrix1, matrix1); printf("\nMatrix 矩阵乘法 SparseMatrix* multiplicationResult = matrixMultiplication(matrix1, matrix2); printf("\nMatrix

    61410编辑于 2024-07-30
  • 来自专栏cwl_Java

    C++经典算法题-稀疏矩阵

    +) { printf("%4d", num[i][j]); } putchar('\n'); } printf("\nmatrix

    1.1K10发布于 2020-02-13
  • 来自专栏Android知识点总结

    Flutter 绘制探索 | 来一起画箭头吧

    处理后效果如下:\n\n \n\ndart\n---->[ArrowPath#formPath]----\nPath headPath = head.formPath();\nMatrix4 headM4 head.size.width/2, 0, 0);\nheadPath = headPath.transform(headM4.storage);\n\nPath tailPath = tail.formPath();\nMatrix4 \n\n \n\ndart\nMatrix4 headM4 = Matrix4.translationValues(fixDx, fixDy, 0);\ncenter = head.position; Matrix4.translationValues(-center.dx, -center.dy, 0));\nheadPath = headPath.transform(headM4.storage);\n\nMatrix4

    98940编辑于 2022-09-08
  • 来自专栏前端达人

    WebGL基础教程:第三部分

    TransformationMatrix");  this.GL.uniformMatrix4fv(tmatrix, false, new Float32Array(TransformMatrix));  var nmatrix this.GL.getUniformLocation(this.ShaderProgram, "NormalTransformation");  this.GL.uniformMatrix4fv(nmatrix

    3.1K20发布于 2019-01-12
  • 【C++模板与泛型编程】模板定义

    1, 1) = 8; std::cout << "Matrix m1:" << std::endl; m1.print(); std::cout << "\nMatrix

    11710编辑于 2026-01-21
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