我正在尝试Python和scikit-learn。我无法让MLPRegressor接近数据。哪里出错了?
from sklearn.neural_network import MLPRegressor
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(0.0, 1, 0.01).reshape(-1, 1)
y = np.sin(2 * np.pi * x).ravel()
reg = MLPRegressor(hidden_layer_sizes=(10,), activation='relu', solver='adam', alpha=0.001,batch_size='auto',
learning_rate='constant', learning_rate_init=0.01, power_t=0.5, max_iter=1000, shuffle=True,
random_state=None, tol=0.0001, verbose=False, warm_start=False, momentum=0.9,
nesterovs_momentum=True, early_stopping=False, validation_fraction=0.1, beta_1=0.9, beta_2=0.999,
epsilon=1e-08)
reg = reg.fit(x, y)
test_x = np.arange(0.0, 1, 0.05).reshape(-1, 1)
test_y = reg.predict(test_x)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.scatter(x, y, s=10, c='b', marker="s", label='real')
ax1.scatter(test_x,test_y, s=10, c='r', marker="o", label='NN Prediction')
plt.show()
结果不是很好: 谢谢。
答案 0 :(得分:10)
这个非非线性模型的点数太少,因此拟合对种子很敏感。一颗好种子有帮助,但它不是先验的。您还可以添加更多数据点。
通过迭代各种种子,我确定printf()
能够很好地运作。当然还有其他人。
random_state=9
以下是种子整数from sklearn.neural_network import MLPRegressor
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(0.0, 1, 0.01).reshape(-1, 1)
y = np.sin(2 * np.pi * x).ravel()
nn = MLPRegressor(
hidden_layer_sizes=(10,), activation='relu', solver='adam', alpha=0.001, batch_size='auto',
learning_rate='constant', learning_rate_init=0.01, power_t=0.5, max_iter=1000, shuffle=True,
random_state=9, tol=0.0001, verbose=False, warm_start=False, momentum=0.9, nesterovs_momentum=True,
early_stopping=False, validation_fraction=0.1, beta_1=0.9, beta_2=0.999, epsilon=1e-08)
n = nn.fit(x, y)
test_x = np.arange(0.0, 1, 0.05).reshape(-1, 1)
test_y = nn.predict(test_x)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.scatter(x, y, s=1, c='b', marker="s", label='real')
ax1.scatter(test_x,test_y, s=10, c='r', marker="o", label='NN Prediction')
plt.show()
的拟合的绝对误差:
i = 0..9
产生:
print(i, sum(abs(test_y - np.sin(2 * np.pi * test_x).ravel())))
现在,我们仍然可以通过将目标点的数量从100增加到1000以及将隐藏层的大小从10增加到100来改善即使0 13.0874999193
1 7.2879574143
2 6.81003360188
3 5.73859777885
4 12.7245375367
5 7.43361211586
6 7.04137436733
7 7.42966661997
8 7.35516939164
9 2.87247035261
的拟合:
random_state=0
顺便说一下,from sklearn.neural_network import MLPRegressor
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(0.0, 1, 0.001).reshape(-1, 1)
y = np.sin(2 * np.pi * x).ravel()
nn = MLPRegressor(
hidden_layer_sizes=(100,), activation='relu', solver='adam', alpha=0.001, batch_size='auto',
learning_rate='constant', learning_rate_init=0.01, power_t=0.5, max_iter=1000, shuffle=True,
random_state=0, tol=0.0001, verbose=False, warm_start=False, momentum=0.9, nesterovs_momentum=True,
early_stopping=False, validation_fraction=0.1, beta_1=0.9, beta_2=0.999, epsilon=1e-08)
n = nn.fit(x, y)
test_x = np.arange(0.0, 1, 0.05).reshape(-1, 1)
test_y = nn.predict(test_x)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.scatter(x, y, s=1, c='b', marker="s", label='real')
ax1.scatter(test_x,test_y, s=10, c='r', marker="o", label='NN Prediction')
plt.show()
中不需要某些参数,例如MLPRegressor()
,momentum
等。检查文档。此外,它有助于为您的示例添加种子以确保结果可重现;)
答案 1 :(得分:10)
你只需要
'lbfgs'
。默认'adam'
是一种类似SGD的方法,对大型和大型有效。凌乱的数据,但对于这种平滑的&小数据。tanh
。 relu
几乎是线性的,不适合学习这种简单的非线性函数。这里是result和完整代码。即使只有3个隐藏的神经元也可以达到很高的准确度。
from sklearn.neural_network import MLPRegressor
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(0.0, 1, 0.01).reshape(-1, 1)
y = np.sin(2 * np.pi * x).ravel()
nn = MLPRegressor(hidden_layer_sizes=(3),
activation='tanh', solver='lbfgs')
n = nn.fit(x, y)
test_x = np.arange(-0.1, 1.1, 0.01).reshape(-1, 1)
test_y = nn.predict(test_x)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.scatter(x, y, s=5, c='b', marker="o", label='real')
ax1.plot(test_x,test_y, c='r', label='NN Prediction')
plt.legend()
plt.show()