计算方法（三）C#矩阵类库
测量平差以及工科中常常用到矩阵的相关运算，因此自己写了一个，同时考虑到了类库的可用性，这次又重载了比较匀运算符，修正了一些问题
using System;
using System.Collections.Generic; namespace CMath
{
[Serializable]
public class Matrix
{
public double[] element;
private int rows = ;
private int cols = ;
/// <summary>
/// 获取矩阵行数
/// </summary>
public int Rows
{
get
{
return rows;
}
}
/// <summary>
/// 获取矩阵列数
/// </summary>
public int Cols
{
get
{
return cols;
}
}
/// <summary>
/// 获取或设置第i行第j列的元素值
/// </summary>
/// <param name="i">第i行</param>
/// <param name="j">第j列</param>
/// <returns>返回第i行第j列的元素值</returns>
public double this[int i, int j]
{
get
{
if (i < Rows && j < Cols)
{
return element[i * cols + j];
}
else
{
throw new Exception("索引越界");
}
}
set
{
element[i * cols + j] = value;
}
}
/// <summary>
/// 用二维数组初始化Matrix
/// </summary>
/// <param name="m">二维数组</param>
public Matrix(double[][] m)
{
this.rows = m.GetLength();
this.cols = m.GetLength();
int count = ;
this.element=new double[Rows*Cols];
for (int i = ; i < rows; i++)
{
for (int j = ; j < cols; j++)
{
element[count++] = m[i][j];
}
}
}
public Matrix(double[,] m)
{
this.rows = m.GetLength();
this.cols = m.GetLength();
this.element = new double[this.rows * this.cols];
int count = ;
for (int i = ; i < rows; i++)
{
for (int j = ; j < cols; j++)
{
element[count++] = m[i, j];
}
}
}
public Matrix(List<List<double>> m)
{
this.rows = m.Count;
this.cols = m[].Count;
this.element = new double[Rows * Cols];
for (int i = ; i < rows; i++)
{
for (int j = ; j < cols; j++)
{
this[i, j] = m[i][j];
}
}
}
#region 矩阵数学运算
public static Matrix MAbs(Matrix a)
{
Matrix _thisCopy = a.DeepCopy();
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
_thisCopy[i, j] = Math.Abs(a[i, j]);
}
}
return _thisCopy;
}
/// <summary>
/// 矩阵相加
/// </summary>
/// <param name="a">第一个矩阵,和b矩阵必须同等大小</param>
/// <param name="b">第二个矩阵</param>
/// <returns>返回矩阵相加后的结果</returns>
public static Matrix operator +(Matrix a, Matrix b)
{
if (a.cols == b.cols && a.rows == b.rows)
{
double[,] res = new double[a.rows, a.cols];
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
res[i, j] = a[i, j] + b[i, j];
}
}
return new Matrix(res);
}
else
{
throw new Exception("两个矩阵行列不相等");
}
}
/// <summary>
/// 矩阵相减
/// </summary>
/// <param name="a">第一个矩阵,和b矩阵必须同等大小</param>
/// <param name="b">第二个矩阵</param>
/// <returns>返回矩阵相减后的结果</returns>
public static Matrix operator -(Matrix a, Matrix b)
{
if (a.cols == b.cols && a.rows == b.rows)
{
double[,] res = new double[a.rows, a.cols];
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
res[i, j] = a[i, j] - b[i, j];
}
}
return new Matrix(res);
}
else
{
throw new Exception("两个矩阵行列不相等");
}
}
/// <summary>
/// 对矩阵每个元素取相反数
/// </summary>
/// <param name="a">二维矩阵</param>
/// <returns>得到矩阵的相反数</returns>
public static Matrix operator -(Matrix a)
{
Matrix res = a;
for (int i = ; i < a.rows; i++)
{
for (int j = ; j < a.cols; j++)
{
res.element[i * a.cols + j] = -res.element[i * a.cols + j];
}
}
return res;
}
/// <summary>
/// 矩阵相乘
/// </summary>
/// <param name="a">第一个矩阵</param>
/// <param name="b">第二个矩阵,这个矩阵的行要与第一个矩阵的列相等</param>
/// <returns>返回相乘后的一个新的矩阵</returns>
public static Matrix operator *(Matrix a, Matrix b)
{
if (a.cols == b.rows)
{
double[,] res = new double[a.rows, b.cols];
for (int i = ; i < a.rows; i++)
{
for (int j = ; j < b.cols; j++)
{
for (int k = ; k < a.cols; k++)
{
res[i, j] += a[i, k] * b[k, j];
}
}
}
return new Matrix(res);
}
else
{
throw new Exception("两个矩阵行和列不等");
}
}
/// <summary>
/// 矩阵与数相乘
/// </summary>
/// <param name="a">第一个矩阵</param>
/// <param name="num">一个实数</param>
/// <returns>返回相乘后的新的矩阵</returns>
public static Matrix operator *(Matrix a, double num)
{
Matrix res = a;
for (int i = ; i < a.rows; i++)
{
for (int j = ; j < a.cols; j++)
{
res.element[i * a.cols + j] *= num;
}
}
return res;
}
/// <summary>
/// 矩阵转置
/// </summary>
/// <returns>返回当前矩阵转置后的新矩阵</returns>
public Matrix Transpose()
{
double[,] res = new double[cols, rows];
{
for (int i = ; i < cols; i++)
{
for (int j = ; j < rows; j++)
{
res[i, j] = this[j, i];
}
}
}
return new Matrix(res);
}
/// <summary>
/// 矩阵求逆
/// </summary>
/// <returns>返回求逆后的新的矩阵</returns>
public Matrix Inverse()
{
//最后原始矩阵并不变，所以需要深拷贝一份
Matrix _thisCopy = this.DeepCopy();
if (cols == rows && this.Determinant() != )
{
//初始化一个同等大小的单位阵
Matrix res = _thisCopy.EMatrix();
for (int i = ; i < rows; i++)
{
//首先找到第i列的绝对值最大的数，并将该行和第i行互换
int rowMax = i;
double max = Math.Abs(_thisCopy[i, i]);
for (int j = i; j < rows; j++)
{
if (Math.Abs(_thisCopy[j, i]) > max)
{
rowMax = j;
max = Math.Abs(_thisCopy[j, i]);
}
}
//将第i行和找到最大数那一行rowMax交换
if (rowMax != i)
{
_thisCopy.Exchange(i, rowMax);
res.Exchange(i, rowMax); }
//将第i行做初等行变换，将第一个非0元素化为1
double r = 1.0 / _thisCopy[i, i];
_thisCopy.Exchange(i, -, r);
res.Exchange(i, -, r);
//消元
for (int j = ; j < rows; j++)
{
//到本行后跳过
if (j == i)
continue;
else
{
r = -_thisCopy[j, i];
_thisCopy.Exchange(i, j, r);
res.Exchange(i, j, r);
}
}
}
return res;
}
else
{
throw new Exception("矩阵不是方阵无法求逆");
}
}
#region 重载比较运算符
public static bool operator <(Matrix a, Matrix b)
{
bool issmall = true;
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
if (a[i, j] >= b[i, j]) issmall = false;
}
}
return issmall;
}
public static bool operator >(Matrix a, Matrix b)
{
bool issmall = true;
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
if (a[i, j] <= b[i, j]) issmall = false;
}
}
return issmall;
}
public static bool operator <=(Matrix a, Matrix b)
{
bool issmall = true;
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
if (a[i, j] > b[i, j]) issmall = false;
}
}
return issmall;
}
public static bool operator >=(Matrix a, Matrix b)
{
bool issmall = true;
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
if (a[i, j] < b[i, j]) issmall = false;
}
}
return issmall;
}
public static bool operator !=(Matrix a, Matrix b)
{
bool issmall = true;
issmall = ReferenceEquals(a, b);
if (issmall) return issmall;
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
if (a[i, j] == b[i, j]) issmall = false;
}
}
return issmall;
}
public static bool operator ==(Matrix a, Matrix b)
{
bool issmall = true;
issmall = ReferenceEquals(a, b);
if (issmall) return issmall;
for (int i = ; i < a.Rows; i++)
{
for (int j = ; j < a.Cols; j++)
{
if (a[i, j] != b[i, j]) issmall = false;
}
}
return issmall;
}
public override bool Equals(object obj)
{
Matrix b = obj as Matrix;
return this == b;
}
public override int GetHashCode()
{
return base.GetHashCode();
}
#endregion
public double Determinant()
{
if (cols == rows)
{
Matrix _thisCopy = this.DeepCopy();
//行列式每次交换行，都需要乘以-1
double res = ;
for (int i = ; i < rows; i++)
{
//首先找到第i列的绝对值最大的数
int rowMax = i;
double max = Math.Abs(_thisCopy[i, i]);
for (int j = i; j < rows; j++)
{
if (Math.Abs(_thisCopy[j, i]) > max)
{
rowMax = j;
max = Math.Abs(_thisCopy[j, i]);
}
}
//将第i行和找到最大数那一行rowMax交换,同时将单位阵做相同初等变换
if (rowMax != i)
{
_thisCopy.Exchange(i, rowMax);
res *= -;
}
//消元
for (int j = i + ; j < rows; j++)
{
double r = -_thisCopy[j, i] / _thisCopy[i, i];
_thisCopy.Exchange(i, j, r);
}
}
//计算对角线乘积
for (int i = ; i < rows; i++)
{
res *= _thisCopy[i, i];
}
return res;
}
else
{
throw new Exception("不是行列式");
}
}
#endregion
#region 初等变换
/// <summary>
/// 初等变换：交换第r1和第r2行
/// </summary>
/// <param name="r1">第r1行</param>
/// <param name="r2">第r2行</param>
/// <returns>返回交换两行后的新的矩阵</returns>
public Matrix Exchange(int r1, int r2)
{
if (Math.Min(r2, r1) >= && Math.Max(r1, r2) < rows)
{
for (int j = ; j < cols; j++)
{
double temp = this[r1, j];
this[r1, j] = this[r2, j];
this[r2, j] = temp;
}
return this;
}
else
{
throw new Exception("超出索引");
}
}
/// <summary>
/// 初等变换：将r1行乘以某个数加到r2行
/// </summary>
/// <param name="r1">第r1行乘以num</param>
/// <param name="r2">加到第r2行，若第r2行为负，则直接将r1乘以num并返回</param>
/// <param name="num">某行放大的倍数</param>
/// <returns></returns>
public Matrix Exchange(int r1, int r2, double num)
{
if (Math.Min(r2, r1) >= && Math.Max(r1, r2) < rows)
{
for (int j = ; j < cols; j++)
{
this[r2, j] += this[r1, j] * num;
}
return this;
}
else if (r2 < )
{
for (int j = ; j < cols; j++)
{
this[r1, j] *= num;
}
return this;
}
else
{
throw new Exception("超出索引");
}
}
/// <summary>
/// 得到一个同等大小的单位矩阵
/// </summary>
/// <returns>返回一个同等大小的单位矩阵</returns>
public Matrix EMatrix()
{
if (rows == cols)
{
double[,] res = new double[rows, cols];
for (int i = ; i < rows; i++)
{
for (int j = ; j < cols; j++)
{
if (i == j)
res[i, j] = ;
else
res[i, j] = ;
}
}
return new Matrix(res);
}
else
throw new Exception("不是方阵，无法得到单位矩阵");
}
#endregion
/// <summary>
/// 深拷贝，仅仅将值拷贝给一个新的对象
/// </summary>
/// <returns>返回深拷贝后的新对象</returns>
public Matrix DeepCopy()
{
double[,] ele = new double[rows, cols];
for (int i = ; i < rows; i++)
{
for (int j = ; j < cols; j++)
{
ele[i, j] = this[i, j];
}
}
return new Matrix(ele);
} public override string ToString()
{
string str = "";
for (int i = ; i < Rows; i++)
{
for (int j = ; j < Cols; j++)
{
str += this[i, j].ToString();
if (j != Cols - )
str += " ";
else if (i != Rows - )
str += Environment.NewLine;
}
}
return str;
}
}
}
矩阵的求秩过几天补上。
计算方法（三）C#矩阵类库的更多相关文章
- P4888 三去矩阵
P4888 三去矩阵 给出一个字符矩阵, 多次询问求以 \((x, y)\) 为中心的最长回文串长度(即横竖两种) \(l, q <= 2000\) Solution 数据范围小直接模拟即可 C ...
- 3D数学基础（三）矩阵
3D引擎中对于矩阵的使用非常多,介绍这些知识也是为了告诉开发者原理,更有助于开发者编写逻辑. (1)固定流水线 各种坐标系之间的转化是通过矩阵相乘得到的,这里面就涉及到了3D固定流水线.作为3D游戏开 ...
- python 矩阵分成上三角下三角和对角三个矩阵
diagonal Return specified diagonals. diagflat Create a 2-D array with the flattened input as a diago ...
- hihoCoder #1162 : 骨牌覆盖问题·三 （矩阵快速幂，DP）
题意:有一个k*n的棋盘,要求用1*2的骨牌来铺满,有多少种方案?(k<8,n<100000001) 思路: 由于k是比较小,但是又不那么小,可以专门构造这样的一个矩阵M,使得只要我们有一 ...
- Azure KeyVault（三）通过 Microsoft.Azure.KeyVault 类库在 .NET Core 上获取 Secrets
一,引言 上一篇文章,我们介绍了 Azure Key Vault 在实际项目中的用途,Azure Key Vault 作为密钥管理的服务,我们可以很轻松的利用它创建和控制用于加密的密钥,和管理证书和机 ...
- 机器学习中的矩阵方法04：SVD 分解
前面我们讲了 QR 分解有一些优良的特性,但是 QR 分解仅仅是对矩阵的行进行操作(左乘一个酉矩阵),可以得到列空间.这一小节的 SVD 分解则是将行与列同等看待,既左乘酉矩阵,又右乘酉矩阵,可以得出 ...
- [WebGL入门]五，矩阵的基础知识
注:文章译自http://wgld.org/,原作者杉本雅広(doxas),文章中假设有我的额外说明.我会加上［lufy:］,另外.鄙人webgl研究还不够深入,一些专业词语,假设翻译有误,欢迎大家指 ...
- 从矩阵（matrix）角度讨论PCA（Principal Component Analysis 主成分分析）、SVD（Singular Value Decomposition 奇异值分解）相关原理
0. 引言 本文主要的目的在于讨论PAC降维和SVD特征提取原理,围绕这一主题,在文章的开头从涉及的相关矩阵原理切入,逐步深入讨论,希望能够学习这一领域问题的读者朋友有帮助. 这里推荐Mit的Gilb ...
- 机器学习 | SVD矩阵分解算法，对矩阵做拆分，然后呢？
本文始发于个人公众号:TechFlow,原创不易,求个关注 今天是机器学习专题第28篇文章,我们来聊聊SVD算法. SVD的英文全称是Singular Value Decomposition,翻译过来 ...
随机推荐
- 使用C++11 实现的线程池
最近打算做一个服务器端程序,每来一个客户端请求新开一个线程进行处理.在网上查了一些资料后,准备使用线程池来做这个东西.使用C++11新的库处理想线程问题比以前简单了许多,在网上找到一份线程池的实现,h ...
- Java获取昨天的时间
Calendar cal = Calendar.getInstance(); cal.add(Calendar.DATE, -1); String yesterday = new ...
- 【转】Hibernate和ibatis的比较
1. 简介 Hibernate是当前最流行的O/R mapping框架.它出身于sf.net,现在已经成为Jboss的一部分了.iBATIS是另外一种优秀的O/R mapping框架,现已改名叫myB ...
- Spring4.0学习笔记(7) —— 通过FactoryBean配置Bean
1.实现Spring 提供的FactoryBean接口 package com.spring.facoryBean; import org.springframework.beans.factory. ...
- Linux socket编程 DNS查询IP地址
本来是一次计算机网络的实验,但是还没有完全写好,DNS的响应请求报文的冗余信息太多了,不只有IP地址.所以这次的实验主要就是解析DNS报文.同时也需要正确的填充请求报文.如果代码有什么bug,欢迎指正 ...
- 多项式求和 AC 杭电
多项式求和 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Subm ...
- ASP.NET MVC的Action Filter
一年前写了一篇短文ASP.NET MVC Action Filters,整理了Action Filter方面的资源,本篇文章详细的描述Action Filter.Action Filter作为一个可以 ...
- JS之路——字符串函数
JS自带函数concat将两个或多个字符的文本组合起来,返回一个新的字符串.var a = "hello";var b = ",world";var c = a ...
- 转：7个鲜为人知却超实用的PHP函数
PHP have lots of built-in functions, and most developers know many of them. But a few functions are ...
- moveToThread的最简单用法（依葫芦画瓢即可）（使得线程也更偏向于信号槽的使用方法）
/*! * \file main.cpp * * Copyright (C) 2010, dbzhang800 * All rights reserved. * */ #include <QtC ...