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：
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# Recurrent Neural Networksimport copy, numpy as npnp.random.seed(0)# compute sigmoid nonlinearitydef sigmoid(x): output = 1/(1+np.exp(-x)) return output# convert output of sigmoid function to its derivativedef sigmoid_output_to_derivative(output): return output*(1-output)# training dataset generationint2binary = {}binary_dim = 8largest_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 variablesalpha = 0.1input_dim = 2hidden_dim = 16output_dim = 1# initialize neural network weightssynapse_0 = 2*np.random.random((input_dim,hidden_dim)) - 1synapse_1 = 2*np.random.random((hidden_dim,output_dim)) - 1synapse_h = 2*np.random.random((hidden_dim,hidden_dim)) - 1synapse_0_update = np.zeros_like(synapse_0)synapse_1_update = np.zeros_like(synapse_1)synapse_h_update = np.zeros_like(synapse_h)# training logicfor 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：
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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：https://iamtrask.github.io/2015/11/15/anyone-can-code-lstm/
希望本文所述对大家Python程序设计有所帮助。
原文链接：http://www.cnblogs.com/hhh5460/p/5782539.html
