神经网络直线-GEKKO

如何解决神经网络直线-GEKKO

我是神经网络的新手。我试图创建一个神经网络，以预测使用GEKKO给出的值。但是，即使代码有效，我也无法获得准确的预测。

另外，六个数据点足以创建一个神经网络吗？

可以请人帮忙吗？代码可以在下面找到

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt  

x_m = 0.0,24.0,72.0,96.0,120.0,144.0
y_m = (0.023027367,0.02636238,0.024316255,0.001705467,-0.004823068,-0.016863735)


x = np.array(x_m)
y = np.array(y_m)
# option for fitting function

# =============================================================================

# Size with hyperbolic tangent function
nin = 1  # inputs
n1 = 2   # hidden layer 1 (linear)
n2 = 3   # hidden layer 2 (nonlinear)
n3 = 2   # hidden layer 3 (linear)
nout = 1 # outputs
# 
# =============================================================================
# Initialize gekko
train = GEKKO()
test = GEKKO()

model = [train,test]

for m in model:
    # input(s)
    m.inpt = m.param()

    # layer 1
    m.w1 = m.Array(m.FV,(nin,n1))
    m.l1 = [m.Intermediate(m.w1[0,i]*m.inpt) for i in range(n1)]

    # layer 2
    m.w2a = m.Array(m.FV,(n1,n2))
    m.w2b = m.Array(m.FV,n2))

    m.l2 = [m.Intermediate(sum([m.tanh(m.w2a[j,i]+m.w2b[j,i]*m.l1[j]) \
                                for j in range(n1)])) for i in range(n2)]

    # layer 3
    m.w3 = m.Array(m.FV,(n2,n3))
    m.l3 = [m.Intermediate(sum([m.w3[j,i]*m.l2[j] \
            for j in range(n2)])) for i in range(n3)]

    # output(s)
    m.outpt = m.CV()
    m.Equation(m.outpt==sum([m.l3[i] for i in range(n3)]))

    # flatten matrices
    m.w1 = m.w1.flatten()
    m.w2a = m.w2a.flatten()
    m.w2b = m.w2b.flatten()
    m.w3 = m.w3.flatten()

# Fit parameter weights
m = train
m.inpt.value=x
m.outpt.value=y
m.outpt.FSTATUS = 1

for i in range(len(m.w1)):
    m.w1[i].FSTATUS=1
    m.w1[i].STATUS=1
    m.w1[i].MEAS=1.0
for i in range(len(m.w2a)):
    m.w2a[i].STATUS=1
    m.w2b[i].STATUS=1
    m.w2a[i].FSTATUS=1
    m.w2b[i].FSTATUS=1
    m.w2a[i].MEAS=1.0
    m.w2b[i].MEAS=0.5
for i in range(len(m.w3)):
    m.w3[i].FSTATUS=1
    m.w3[i].STATUS=1
    m.w3[i].MEAS=1.0
m.options.IMODE = 2
m.options.soLVER = 3
m.options.EV_TYPE = 2
m.solve(disp=False)

# Test sample points
m = test
for i in range(len(m.w1)):
    m.w1[i].MEAS=train.w1[i].NEWVAL
    m.w1[i].FSTATUS = 1
    print('w1['+str(i)+']: '+str(m.w1[i].MEAS))
for i in range(len(m.w2a)):
    m.w2a[i].MEAS=train.w2a[i].NEWVAL
    m.w2b[i].MEAS=train.w2b[i].NEWVAL
    m.w2a[i].FSTATUS = 1
    m.w2b[i].FSTATUS = 1
    print('w2a['+str(i)+']: '+str(m.w2a[i].MEAS))
    print('w2b['+str(i)+']: '+str(m.w2b[i].MEAS))
for i in range(len(m.w3)):
    m.w3[i].MEAS=train.w3[i].NEWVAL
    m.w3[i].FSTATUS = 1
    print('w3['+str(i)+']: '+str(m.w3[i].MEAS))
    
m.inpt.value= np.linspace(0,140)
m.options.IMODE = 2
m.options.soLVER = 3
m.solve(disp=True)

plt.figure()
plt.plot(x,y,'bo',label = 'measured')
plt.plot(test.inpt.value,test.outpt.value,'r-',label = 'predicted')
plt.legend()
plt.show()

这是输出：

解决方法

brain module如何简化您在Gekko中用于神经网络的代码？

from gekko import brain
import numpy as np
import matplotlib.pyplot as plt  

x_m = (0.0,24.0,72.0,96.0,120.0,144.0)
y_m = (0.023027367,0.02636238,0.024316255,\
       0.001705467,-0.004823068,-0.016863735)

x = np.array(x_m)
y = np.array(y_m)

b = brain.Brain()
b.input_layer(1)
b.layer(linear=2)
b.layer(tanh=2)
b.layer(linear=2)
b.output_layer(1)

b.learn(x,y) # train
xp = np.linspace(0,144,50) 
yp = b.think(xp) # validate

plt.figure()
plt.plot(x,y,'bo')
plt.plot(xp,yp[0],'r-')
plt.show()

六个数据点不是很多。可调整的参数比数据点更多，但这说明了如何针对较大的问题进行设置。您可能需要调整每个图层的节点数才能获得合适的效果。 Machine Learning and Dynamic Optimization网站上还有其他教程。您可能还想看看Keras或PyTorch。