【DL--22】实现神经网络算法NeuralNetwork以及手写数字识别

1.NeuralNetwork.py

#coding:utf-8import numpy as np#定义双曲函数和他们的导数def tanh(x):    return np.tanh(x)def tanh_deriv(x):    return 1.0 - np.tanh(x)**2def logistic(x):    return 1/(1 + np.exp(-x))def logistic_derivative(x):    return logistic(x)*(1-logistic(x))#定义NeuralNetwork 神经网络算法class NeuralNetwork:    #初始化，layes表示的是一个list，eg[10,10,3]表示第一层10个神经元，第二层10个神经元，第三层3个神经元    def __init__(self, layers, activation='tanh'):        """        :param layers: A list containing the number of units in each layer.        Should be at least two values        :param activation: The activation function to be used. Can be        "logistic" or "tanh"        """        if activation == 'logistic':            self.activation = logistic            self.activation_deriv = logistic_derivative        elif activation == 'tanh':            self.activation = tanh            self.activation_deriv = tanh_deriv        self.weights = []        #循环从1开始，相当于以第二层为基准，进行权重的初始化        for i in range(1, len(layers) - 1):            #对当前神经节点的前驱赋值            self.weights.append((2*np.random.random((layers[i - 1] + 1, layers[i] + 1))-1)*0.25)            #对当前神经节点的后继赋值            self.weights.append((2*np.random.random((layers[i] + 1, layers[i + 1]))-1)*0.25)            #训练函数   ，X矩阵，每行是一个实例 ，y是每个实例对应的结果，learning_rate 学习率，    # epochs，表示抽样的方法对神经网络进行更新的最大次数    def fit(self, X, y, learning_rate=0.2, epochs=10000):        X = np.atleast_2d(X) #确定X至少是二维的数据        temp = np.ones([X.shape[0], X.shape[1]+1]) #初始化矩阵        temp[:, 0:-1] = X  # adding the bias unit to the input layer        X = temp        y = np.array(y) #把list转换成array的形式        for k in range(epochs):            #随机选取一行，对神经网络进行更新            i = np.random.randint(X.shape[0])            a = [X[i]]            #完成所有正向的更新            for l in range(len(self.weights)):                a.append(self.activation(np.dot(a[l], self.weights[l])))                #            error = y[i] - a[-1]            deltas = [error * self.activation_deriv(a[-1])]            #开始反向计算误差，更新权重            for l in range(len(a) - 2, 0, -1): # we need to begin at the second to last layer                deltas.append(deltas[-1].dot(self.weights[l].T)*self.activation_deriv(a[l]))            deltas.reverse()            for i in range(len(self.weights)):                layer = np.atleast_2d(a[i])                delta = np.atleast_2d(deltas[i])                self.weights[i] += learning_rate * layer.T.dot(delta)                #预测函数    def predict(self, x):        x = np.array(x)        temp = np.ones(x.shape[0]+1)        temp[0:-1] = x        a = temp        for l in range(0, len(self.weights)):            a = self.activation(np.dot(a, self.weights[l]))        return a

2、基于NeuralNetwork的手写数字识别

#-*-coding:utf-8-*-import sysreload(sys)sys.setdefaultencoding('utf-8')import numpy as npfrom sklearn.datasets import load_digitsfrom sklearn.metrics import confusion_matrix, classification_reportfrom sklearn.preprocessing import LabelBinarizerfrom NeuralNetwork import NeuralNetworkfrom sklearn.cross_validation import train_test_splitdigits = load_digits()X = digits.datay = digits.targetX -= X.min() # normalize the values to bring them into the range 0-1X /= X.max()###############################训练模型########################nn = NeuralNetwork([64,100,10],'logistic')X_train, X_test, y_train, y_test = train_test_split(X, y)labels_train = LabelBinarizer().fit_transform(y_train)labels_test = LabelBinarizer().fit_transform(y_test)print "start fitting"nn.fit(X_train,labels_train,epochs=3000)###############预测结果###############################predictions = []for i in range(X_test.shape[0]):    o = nn.predict(X_test[i] )    predictions.append(np.argmax(o))###############混淆矩阵#####################################print confusion_matrix(y_test,predictions)print classification_report(y_test,predictions)#################打印预测结果###################### for each in predictions:#     print each# for each in y_test:#     print each

3、运行结果：

start fitting[[44  0  0  0  0  0  0  0  0  0] [ 0 44  0  0  0  1  0  0  2  0] [ 0  1 39  0  0  0  0  0  0  0] [ 0  1  0 49  0  0  0  2  2  0] [ 0  2  0  0 34  0  0  2  1  0] [ 0  2  0  0  1 44  1  0  0  3] [ 1  2  0  0  0  0 43  0  0  0] [ 0  0  0  0  0  0  0 41  0  0] [ 0  4  0  0  0  1  0  1 31  2] [ 0  4  0  0  0  0  0  1  1 43]]             precision    recall  f1-score   support          0       0.98      1.00      0.99        44          1       0.73      0.94      0.82        47          2       1.00      0.97      0.99        40          3       1.00      0.91      0.95        54          4       0.97      0.87      0.92        39          5       0.96      0.86      0.91        51          6       0.98      0.93      0.96        46          7       0.87      1.00      0.93        41          8       0.84      0.79      0.82        39          9       0.90      0.88      0.89        49avg / total       0.92      0.92      0.92       450Process finished with exit code 0