learning_rate = 0.07
# 抽出数
epoch = 1000
# パラメータの初期化
network = init_network()
# データのランダム抽出
random_datasets = np.random.choice(data_sets, epoch)
# 勾配降下の繰り返し
for dataset in random_datasets:
x, d = dataset['x'], dataset['d']
z1, y = forward(network, x)
grad = backward(x, d, z1, y)
# パラメータに勾配適用
for key in ('W1', 'W2', 'b1', 'b2'):
network[key] -= learning_rate * grad[key]
# 誤差
loss = functions.mean_squared_error(d, y)
losses.append(loss)
print("##### 結果表示 #####")
lists = range(epoch)
plt.plot(lists, losses, '.')
plt.title('loss')
# グラフの表示
plt.show()
from pathlib import Path import sys sys.path.append(Path.cwd().parent) import numpy as np from common.functions import mean_squared_error #「2」を正解とする t = [0, 0, 1, 0, 0, 0, 0, 0, 0, 0] #例1:「2」の確率が最も高い場合（0.6） y1 = [0.1, 0.05, 0.6, 0.0, 0.05, 0.1, 0.0, 0.1, 0.0, 0.0] mean_squared_error(np.array(y1), np.array(t)) #例2:「7」の確率が最も高い場合（0.6） y2 = [0.1, 0.05, 0.1, 0.0, 0.05, 0.1, 0.0, 0.6, 0.0, 0.0] mean_squared_error(np.array(y2), np.array(t))
all_loss = 0
# 時系列ループ
for t in range(binary_dim):
# 入力値
X = np.array([a_bin[ - t - 1], b_bin[ - t - 1]]).reshape(1, -1)
# 時刻tにおける正解データ
dd = np.array([d_bin[binary_dim - t - 1]])
u[:,t+1] = np.dot(X, W_in) + np.dot(z[:,t].reshape(1, -1), W)
z[:,t+1] = functions.sigmoid(u[:,t+1])
y[:,t] = functions.sigmoid(np.dot(z[:,t+1].reshape(1, -1), W_out))
#誤差
loss = functions.mean_squared_error(dd, y[:,t])
delta_out[:,t] = functions.d_mean_squared_error(dd, y[:,t]) * functions.d_sigmoid(y[:,t])
all_loss += loss
out_bin[binary_dim - t - 1] = np.round(y[:,t])
for t in range(binary_dim)[::-1]:
X = np.array([a_bin[-t-1],b_bin[-t-1]]).reshape(1, -1)
delta[:,t] = (np.dot(delta[:,t+1].T, W.T) + np.dot(delta_out[:,t].T, W_out.T)) * functions.d_sigmoid(u[:,t+1])
# 勾配更新
W_out_grad += np.dot(z[:,t+1].reshape(-1,1), delta_out[:,t].reshape(-1,1))
W_grad += np.dot(z[:,t].reshape(-1,1), delta[:,t].reshape(1,-1))
def forward(self, x, t):
self.t = t
self.y = np.tanh(x)
self.loss = mean_squared_error(self.y, self.t)
return self.loss
# 4.2.1 평균 제곱 오차(MSE) # 4.2.2 교차 엔트로피 오차(CEE) ############################################################################## import numpy as np from common.functions import mean_squared_error, cross_entropy_error # 정답은 '2' <= 2번 index 값을 1로 세팅 (one-hot encoding) t = [0, 0, 1, 0, 0, 0, 0, 0, 0, 0] ############################################################################## # 예1 : '2'일 확률이 가장 높다고 추정 <= 2번 index 값이 0.6으로 가장 높다. # - 아래 예2 와 비교해서 손실함수의 출력이 더 작다. (매개변수가 더 최적화되어있다는 의미) ############################################################################## # (1) y = [0.1, 0.05, 0.6, 0.0, 0.05, 0.1, 0.0, 0.1, 0.0, 0.0] print(mean_squared_error(np.array(y), np.array(t))) # 0.09750000000000003 # cross_entropy_error 사용하는 경우 정답(t)이 2번째 index이므로 y의 2번째만 계산한 것과 같다. print(cross_entropy_error(np.array(y), np.array(t))) # 0.510825457099338 ############################################################################## # 예2 : '7'일 확률이 가장 높다고 추정 <= 7번 index 값이 0.6으로 가장 높다. ############################################################################## y = [0.1, 0.05, 0.1, 0.0, 0.05, 0.1, 0.0, 0.6, 0.0, 0.0] print(mean_squared_error(np.array(y), np.array(t))) # 0.5975 # cross_entropy_error 사용하는 경우 정답(t)이 2번째 index이므로 y의 2번째만 계산한 것과 같다. print(cross_entropy_error(np.array(y), np.array(t))) # 2.302584092994546
import sys, os sys.path.append(os.pardir) sys.path.append(os.curdir) import numpy as np from common.functions import mean_squared_error t = [0, 0, 1, 0, 0, 0] y = [0.1, 0.05, 0.6, 0.0, 0.05, 0.1] error = mean_squared_error(np.array(y), np.array(t)) print(error)