单层神经网络

Question

对于单层神经网络的实现,我有两个数据文件.

In:
    0.832 64.643
    0.818 78.843

Out:
    0 0 1
    0 0 1

以上是2个数据文件的格式.

目标输出为"1表示相应输入所属的特定类,"0表示其余2个输出.

问题如下:

您的单层神经网络将在Y = A*X + b中找到A(3乘2矩阵)和b(3乘1矢量),其中Y是[C1,C2,C3]',X是[x1,x2]' .

为了用神经网络解决上述问题,我们可以重新编写如下公式:Y = A'*X'其中A'= [A b](3乘3矩阵),X'是[x1,x2, 1]"

现在您可以使用具有三个输入节点(分别用于x1,x2和1)和三个输出(C1,C2,C3)的神经网络.

由此产生的9(因为我们在3个输入和3个输出之间有9个连接)权重将等同于A'矩阵的元素.

基本上,我试图做这样的事情,但它不起作用:

function neuralNetwork   
    load X_Q2.data
    load T_Q2.data
    x = X_Q2(:,1);
    y = X_Q2(:,2);

    learningrate = 0.2;
    max_iteration = 50;

    % initialize parameters
    count = length(x);
    weights = rand(1,3); % creates a 1-by-3 array with random weights
    globalerror = 0;
    iter = 0;
    while globalerror ~= 0 && iter <= max_iteration
        iter = iter + 1;
        globalerror = 0;
        for p = 1:count
            output = calculateOutput(weights,x(p),y(p));
            localerror = T_Q2(p) - output
            weights(1)= weights(1) + learningrate *localerror*x(p);
            weights(2)= weights(1) + learningrate *localerror*y(p);
            weights(3)= weights(1) + learningrate *localerror;
            globalerror = globalerror + (localerror*localerror);
        end 
    end 

我在其他文件中编写此函数并在之前的代码中调用它.

function result = calculateOutput (weights, x, y)
    s = x * weights(1) + y * weights(2) + weights(3);
    if s >= 0
        result = 1;
    else
        result = -1;
    end

Answer 1

我可以发现代码中的一些问题.主要问题是目标是多类(非二进制),因此您需要为每个类使用一个输出节点(称为1-of-N编码),或者使用具有不同激活功能的单个输出节点(能够超过二进制输出-1/1或0/1的东西)

在下面的解决方案中,感知器具有以下结构:

%# load your data
input = [
    0.832 64.643
    0.818 78.843
    1.776 45.049
    0.597 88.302
    1.412 63.458
];
target = [
    0 0 1
    0 0 1
    0 1 0
    0 0 1
    0 0 1
];

%# parameters of the learning algorithm
LEARNING_RATE = 0.1;
MAX_ITERATIONS = 100;
MIN_ERROR = 1e-4;

[numInst numDims] = size(input);
numClasses = size(target,2);

%# three output nodes connected to two-dimensional input nodes + biases
weights = randn(numClasses, numDims+1);

isDone = false;               %# termination flag
iter = 0;                     %# iterations counter
while ~isDone
    iter = iter + 1;

    %# for each instance
    err = zeros(numInst,numClasses);
    for i=1:numInst
        %# compute output: Y = W*X + b, then apply threshold activation
        output = ( weights * [input(i,:)';1] >= 0 );                       %#'

        %# error: err = T - Y
        err(i,:) = target(i,:)' - output;                                  %#'

        %# update weights (delta rule): delta(W) = alpha*(T-Y)*X
        weights = weights + LEARNING_RATE * err(i,:)' * [input(i,:) 1];    %#'
    end

    %# Root mean squared error
    rmse = sqrt(sum(err.^2,1)/numInst);
    fprintf(['Iteration %d: ' repmat('%f ',1,numClasses) '\n'], iter, rmse);

    %# termination criteria
    if ( iter >= MAX_ITERATIONS || all(rmse < MIN_ERROR) )
        isDone = true;
    end
end

%# plot points and one-against-all decision boundaries
[~,group] = max(target,[],2);                     %# actual class of instances
gscatter(input(:,1), input(:,2), group), hold on
xLimits = get(gca,'xlim'); yLimits = get(gca,'ylim');
for i=1:numClasses
    ezplot(sprintf('%f*x + %f*y + %f', weights(i,:)), xLimits, yLimits)
end
title('Perceptron decision boundaries')
hold off

您提供的五个样本的培训结果:

Iteration 1: 0.447214 0.632456 0.632456 
Iteration 2: 0.000000 0.447214 0.447214 
...
Iteration 49: 0.000000 0.447214 0.447214 
Iteration 50: 0.000000 0.632456 0.000000 
Iteration 51: 0.000000 0.447214 0.000000 
Iteration 52: 0.000000 0.000000 0.000000 

请注意,上例中使用的数据仅包含5个样本.如果每个班级有更多的培训实例,您将获得更有意义的结果.