Python 이 실현 한 재 귀 신경 망 간단 한 예제

5800 단어 Python귀착 하 다
이 글 의 사례 는 Python 이 실현 하 는 재 귀 신경 망 을 서술 하 였 다.여러분 께 참고 하도록 공유 하 겠 습 니 다.구체 적 으로 는 다음 과 같 습 니 다.

# 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("------------")

출력 실행:

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
------------

영어 원문:https://iamtrask.github.io/2015/11/15/anyone-can-code-lstm/
더 많은 파 이 썬 관련 내용 에 관심 이 있 는 독자 들 은 본 사이트 의 주 제 를 볼 수 있다.
본 논문 에서 말 한 것 이 여러분 의 Python 프로 그래 밍 에 도움 이 되 기 를 바 랍 니 다.

좋은 웹페이지 즐겨찾기