Python实现的递归神经网络简单示例 本文实例讲述了Python实现的递归神经网络。分享给大家供大家参考,具体如下: # Recurrent Neural Networks import copy, numpy as np np.random.seed(0) # compute sigmoid nonlinearity def sigmoid(x): output = 1/(1+np.exp(-x)) return output # convert output of sigmoid function to its derivative def sigmoid_output_to_derivative(output): return output*(1-output) # training dataset generation int2binary = {} binary_dim = 8 largest_number = pow(2,binary_dim) binary = np.unpackbits( np.array([range(largest_number)],dtype=np.uint8).T,axis=1) for i in range(largest_number): int2binary[i] = binary[i] # input variables alpha = 0.1 input_dim = 2 hidden_dim = 16 output_dim = 1 # initialize neural network weights synapse_0 = 2*np.random.random((input_dim,hidden_dim)) - 1 synapse_1 = 2*np.random.random((hidden_dim,output_dim)) - 1 synapse_h = 2*np.random.random((hidden_dim,hidden_dim)) - 1 synapse_0_update = np.zeros_like(synapse_0) synapse_1_update = np.zeros_like(synapse_1) synapse_h_update = np.zeros_like(synapse_h) # training logic for j in range(10000): # generate a simple addition problem (a + b = c) a_int = np.random.randint(largest_number/2) # int version a = int2binary[a_int] # binary encoding b_int = np.random.randint(largest_number/2) # int version b = int2binary[b_int] # binary encoding # true answer c_int = a_int + b_int c = int2binary[c_int] # where we'll store our best guess (binary encoded) d = np.zeros_like(c) overallError = 0 layer_2_deltas = list() layer_1_values = list() layer_1_values.append(np.zeros(hidden_dim)) # moving along the positions in the binary encoding for position in range(binary_dim): # generate input and output X = np.array([[a[binary_dim - position - 1],b[binary_dim - position - 1]]]) y = np.array([[c[binary_dim - position - 1]]]).T # hidden layer (input ~+ prev_hidden) layer_1 = sigmoid(np.dot(X,synapse_0) + np.dot(layer_1_values[-1],synapse_h)) # output layer (new binary representation) layer_2 = sigmoid(np.dot(layer_1,synapse_1)) # did we miss?... if so, by how much? layer_2_error = y - layer_2 layer_2_deltas.append((layer_2_error)*sigmoid_output_to_derivative(layer_2)) overallError += np.abs(layer_2_error[0]) # decode estimate so we can print(it out) d[binary_dim - position - 1] = np.round(layer_2[0][0]) # store hidden layer so we can use it in the next timestep layer_1_values.append(copy.deepcopy(layer_1)) future_layer_1_delta = np.zeros(hidden_dim) for position in range(binary_dim): X = np.array([[a[position],b[position]]]) layer_1 = layer_1_values[-position-1] prev_layer_1 = layer_1_values[-position-2] # error at output layer layer_2_delta = layer_2_deltas[-position-1] # error at hidden layer layer_1_delta = (future_layer_1_delta.dot(synapse_h.T) + layer_2_delta.dot(synapse_1.T)) * sigmoid_output_to_derivative(layer_1) # let's update all our weights so we can try again synapse_1_update += np.atleast_2d(layer_1).T.dot(layer_2_delta) synapse_h_update += np.atleast_2d(prev_layer_1).T.dot(layer_1_delta) synapse_0_update += X.T.dot(layer_1_delta) future_layer_1_delta = layer_1_delta synapse_0 += synapse_0_update * alpha synapse_1 += synapse_1_update * alpha synapse_h += synapse_h_update * alpha synapse_0_update *= 0 synapse_1_update *= 0 synapse_h_update *= 0 # print(out progress) if j % 1000 == 0: print("Error:" + str(overallError)) print("Pred:" + str(d)) print("True:" + str(c)) out = 0 for index,x in enumerate(reversed(d)): out += x*pow(2,index) print(str(a_int) + " + " + str(b_int) + " = " + str(out)) print("------------") 运行输出: Error:[ 3.45638663] Pred:[0 0 0 0 0 0 0 1] True:[0 1 0 0 0 1 0 1] 9 + 60 = 1 ------------ Error:[ 3.63389116] Pred:[1 1 1 1 1 1 1 1] True:[0 0 1 1 1 1 1 1] 28 + 35 = 255 ------------ Error:[ 3.91366595] Pred:[0 1 0 0 1 0 0 0] True:[1 0 1 0 0 0 0 0] 116 + 44 = 72 ------------ Error:[ 3.72191702] Pred:[1 1 0 1 1 1 1 1] True:[0 1 0 0 1 1 0 1] 4 + 73 = 223 ------------ Error:[ 3.5852713] Pred:[0 0 0 0 1 0 0 0] True:[0 1 0 1 0 0 1 0] 71 + 11 = 8 ------------ Error:[ 2.53352328] Pred:[1 0 1 0 0 0 1 0] True:[1 1 0 0 0 0 1 0] 81 + 113 = 162 ------------ Error:[ 0.57691441] Pred:[0 1 0 1 0 0 0 1] True:[0 1 0 1 0 0 0 1] 81 + 0 = 81 ------------ Error:[ 1.42589952] Pred:[1 0 0 0 0 0 0 1] True:[1 0 0 0 0 0 0 1] 4 + 125 = 129 ------------ Error:[ 0.47477457] Pred:[0 0 1 1 1 0 0 0] True:[0 0 1 1 1 0 0 0] 39 + 17 = 56 ------------ Error:[ 0.21595037] Pred:[0 0 0 0 1 1 1 0] True:[0 0 0 0 1 1 1 0] 11 + 3 = 14 ------------ 英文原文:http://iamtrask.github.io/2015/11/15/anyone-can-code-lstm/ 更多关于Python相关内容感兴趣的读者可查看本站专题:《Python数学运算技巧总结》、《Python数据结构与算法教程》、《Python函数使用技巧总结》、《Python字符串操作技巧汇总》、《Python入门与进阶经典教程》及《Python文件与目录操作技巧汇总》 希望本文所述对大家Python程序设计有所帮助。