zoukankan      html  css  js  c++  java
  • golang 矩阵乘法、行列式、求逆矩阵

    package matrix
    
    import (
        "math"
        "github.com/astaxie/beego"
    )
    
    type Matrix4 struct {
        Elements [16]float64 `json:"elements"`
    }
    
    type SQ struct {
        //矩阵结构
        M,N int            //m是列数,n是行数
        Data [][]float64
    }
    
    //矩阵定义
    func (this*SQ)Set(m int,n int,data []float64) {
        //m是列数,n是行数,data是矩阵数据(从左到右由上到下填充)
        this.M=m
        this.N=n
        if len(data)!=this.M*this.N {
            beego.Debug("矩阵定义失败")
            return
        }else {
            k := 0
            if this.M*this.N == len(data){
                for i := 0; i < this.N; i++ {
                    var tmpArr []float64
                    for j := 0; j < this.M; j++ {
                        tmpArr = append(tmpArr, data[k])
                        k++
                    }
                    this.Data = append(this.Data, tmpArr)
                }
            }else {
                beego.Debug("矩阵定义失败")
                return
            }
        }
    }
    //a的列数和b的行数相等
    //矩阵乘法
    func Mul(a SQ,b SQ) [][]float64{
        if a.M==b.M {
            res := [][]float64{}
            for i:=0;i<a.M;i++ {
                t := []float64{}
                for j:=0;j<b.M;j++ {
                    r := float64(0)
                    for k:=0;k<a.M;k++ {
                        r += a.Data[i][k]*b.Data[k][j]
                    }
                    t = append(t, r)
                }
                res = append(res,t)
            }
            return res
        }else {
            beego.Debug("两矩阵无法进行相乘运算")
            return [][]float64{}
        }
        /*一个应用的例子
        a := [][]int{
            {1,2},
            {3,4},
            {5,6},
        }
        b := [][]int{
            {1,2,3},
            {3,4,1},
        }
        A := SQ{
            2,3,
            a,
        }
        B := SQ{
            3,2,
            b,
        }
    
        res := mul(A,B)
         */
    }
    
    //计算n阶行列式(N=n-1)
    func Det(Matrix [][]float64,N int) float64 {
        var T0,T1,T2,Cha int
        var Num float64
        var B [][]float64
    
        if N>0 {
            Cha=0
            for i := 0; i < N; i++ {
                var tmpArr []float64
                for j := 0; j < N; j++ {
                    tmpArr = append(tmpArr, 0)
                }
                B = append(B, tmpArr)
            }
            Num=0
            for T0=0;T0<=N;T0++{    //T0循环
                for T1=1;T1<=N;T1++ {    //T1循环
                    for T2=0;T2<=N-1;T2++ {    //T2循环
                        if T2==T0 {
                            Cha = 1
                        }
                        B[T1-1][T2]=Matrix[T1][T2+Cha]
                    }//T2循环
                    Cha=0
                }//T1循环
                Num=Num+Matrix[0][T0]*Det(B,N-1)*math.Pow(-1,float64(T0))
            }//T0循环
            return Num
        }else if N==0 {
            return Matrix[0][0]
        }
        return 0
    }
    
    //矩阵求逆(N=n-1)
    func Inverse(Matrix [][]float64,N int) (MatrixC [][]float64) {
        var T0,T1,T2,T3 int
        var B [][]float64
        for i := 0; i < N; i++ {
            var tmpArr []float64
            for j := 0; j < N; j++ {
                tmpArr = append(tmpArr, 0)
            }
            B = append(B, tmpArr)
        }
        Chay := 0
        Chax := 0
        var add float64
        add = 1/Det(Matrix,N)
        for T0=0;T0<=N;T0++{
            for T3=0;T3<=N;T3++{
                for T1=0;T1<=N-1;T1++{
                    if T1<T0 {
                        Chax = 0
                    }else {
                        Chax = 1
                    }
                    for T2=0;T2<=N-1;T2++{
                        if T2<T3 {
                            Chay = 0
                        }else{
                            Chay = 1
                        }
                        B[T1][T2]=Matrix[T1+Chax][T2+Chay]
                    }//T2循环
                }//T1循环
                Det(B,N-1)
                MatrixC[T3][T0]=Det(B,N-1)*add*(math.Pow(-1, float64(T0+T3)))
            }
        }
        return MatrixC
    }
  • 相关阅读:
    jFinal基于maven简单的demo
    quartz定时任务
    cors解决跨越问题
    poi导出excel
    layui富文本编译器后台获取图片路径
    HttpClient短信接口
    js分页
    El表达式日期处理
    【传智播客】Libevent学习笔记(二):创建event_base
    【传智播客】Libevent学习笔记(一):简介和安装
  • 原文地址:https://www.cnblogs.com/zheng123/p/9648858.html
Copyright © 2011-2022 走看看