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  • R语言实现人工神经网络预测实例

    R语言中很多包(package)关于神经网络,例如nnet、AMORE、neuralnet以及RSNNS。nnet提供了最常见的前馈反向传播神经网络算法。AMORE包则更进一步提供了更为丰富的控制参数,并可以增加多个隐藏层。neuralnet包的改进在于提供了弹性反向传播算法和更多的激活函数形式。RSNNS则是连接R和SNNS的工具,在R中即可直接调用SNNS的函数命令,在这方面有了极大的扩充。本文使用AMORE包来实现神经网络模型。可使用命令install.packages("AMORE")进行包的安装。

    install.packages("AMORE")

    library(AMORE)

    library(nnet)

    #输入一个11*10的矩阵,前8行用来训练,后3行用来预测
    p<-matrix(c(6977.93,24647,11356.6,9772.5,1496.92,4279.65,89.84,95.97,9194,0.6068,
    7973.37,28534,13469.77,11585.82,1618.27,5271.991,100.28,111.16,9442,0.63,
    9294.26,33272,16004.61,14076.83,1707.98,6341.86,117.78,130.22,9660,0.6314,
    10868.67,37638,18502.2,16321.46,1790.97,6849.688,134.77,125.56,9893,0.6337,
    12933.12,39436,19419.7,18052.59,1855.73,6110.941,86.04,119.81,10130,0.634,
    15623.7,44736,23014.53,20711.55,1948.06,7848.961,151.59,187.08,10441,0.6618,
    17069.2,50807,26447.38,24097.7,2006.92,9134.673,177.79,202.12,10505,0.665,
    18751.47,54095,27700.97,26519.69,2037.88,9840.205,195.18,282.05,10594,0.674,
    21169.7,60633.82,31941.45,29569.92,2211.6665,11221.01,205.5601,329.4234,10986.79,0.684065,
    23716.17,66750.29,35562.93,32993.75,2317.9223,12486.77,220.3005,398.7751,11245.69,0.694706,
    26469.74,73292.95,39458.17,36680.63,2428.5869,13849.68,235.0408,477.4204,11515.33,0.706087),11,10,byrow=T)

    #对输入矩阵进行归一化处理(0到1)
    b1=(p[,1]-min(p[,1]))/(max(p[,1])-min(p[,1]))
    b2=(p[,2]-min(p[,2]))/(max(p[,2])-min(p[,2]))
    b3=(p[,3]-min(p[,3]))/(max(p[,3])-min(p[,3]))
    b4=(p[,4]-min(p[,4]))/(max(p[,4])-min(p[,4]))
    b5=(p[,5]-min(p[,5]))/(max(p[,5])-min(p[,5]))
    b6=(p[,6]-min(p[,6]))/(max(p[,6])-min(p[,6]))
    b7=(p[,7]-min(p[,7]))/(max(p[,7])-min(p[,7]))
    b8=(p[,8]-min(p[,8]))/(max(p[,8])-min(p[,8]))
    b9=(p[,9]-min(p[,9]))/(max(p[,9])-min(p[,9]))
    b10=(p[,10]-min(p[,10]))/(max(p[,10])-min(p[,10]))
    p0=cbind(b1,b2,b3,b4,b5,b6,b7,b8,b9,b10)
    #归一化后的数据放入矩阵中

    #对应矩阵前8行的测试结果集
    t<-c(2673.5356,2991.0529,3393.0057,3504.8229,3609.4029,4060.1257,4399.0168,4619.4102)

    #第9行的实际结果
    t9=4830.1315

    #测试结果归一化
    t0=(t-min(t))/(max(t)-min(t))
    alter=1
    count=0

    #训练的结果测试第9行若误差在3%之内或者循环20次结束
    while(abs(alter)<0.03 && count<20){
    #训练网络,n.neurons表示输入的参数,以及隐藏层个数,及输出结果
    net<-newff(n.neurons = c(10,10,2,1),learning.rate.global=1e-4, momentum.global=0.05,error.criterium="LMS", Stao=NA, hidden.layer="tansig", output.layer="purelin", method="ADAPTgdwm")
    #<span style="line-height: 27.2px; font-family: 'Helvetica Neue', Helvetica, Tahoma, Arial, STXihei, 'Microsoft YaHei', 微软雅黑, sans-serif;">p0[1:8,]表示输入,t0[1:8]表示输出,show.step表示循环次数,n.shows表示满足结果的报告次数</span>
    result<-train(net,p0[1:8,],t0[1:8],error.criterium="LMS", report=TRUE, show.step=10000, n.shows=5)
    #测试第9行到11行
    y<-sim(result$net,p0[9:11,])
    #反归一化,这里反归一化感觉有问题,最后训练20次才结束并且第九行误差还是11%,当我把反归一化改成(y<-y*(t[8]-t[1])+t[1])训练一次误差就小于0.03且结束训练
    y<-y*t[8]
    #用第9行来测试训练误差,满足训练误差结束
    alter=(y[1]-t9)/t9
    count=count+1;
    }

    count
    #输出第9行到11行预测的值
    y
    #作图
    x0<-c(2013,2014,2015)
    plot(x0,y,col = "blue",pch = "+")

    注:每一行都具有实际含义,代表每一年的参数指标,预测一个值

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  • 原文地址:https://www.cnblogs.com/liquan-anran/p/9507563.html
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