python - 如何教keras神经网络解决sqrt

Question

我正在使用 python 和 keras 学习机器学习。我创建了一个神经网络，从 {1, 4, 9, 16, 25, 36, ..., 100} 范围内的偶数整数中预测平方根。我已经编写了代码来做到这一点，但结果远非如此（无论我向网络提供什么数字，它都预测它是 1.0）。

我试过改变层数、每层神经元的数量、激活函数，但没有任何帮助。

这是我到目前为止编写的代码：

from numpy import loadtxt
from keras.models import Sequential
from keras.layers import Dense
from keras import optimizers

# laod dataset
# dataset = loadtxt('pima-indians-diabetes.csv', delimiter=',')
dataset = loadtxt('sqrt.csv', delimiter=',')

# split into input (X) and output (y) variables
X = dataset[:,0:1] * 1.0
y = dataset[:,1] * 1.0

# define the keras model
model = Sequential()
model.add(Dense(6, input_dim=1, activation='relu'))
model.add(Dense(1, activation='linear'))

# compile the keras model
opt = optimizers.adam(lr=0.01)
model.compile(loss='mean_squared_error', optimizer=opt, metrics=['accuracy'])

# fit the keras model on the dataset (CPU)
model.fit(X, y, epochs=150, batch_size=10, verbose=0)

# evaluate the keras model
_, accuracy = model.evaluate(X, y, verbose=0)
print('Accuracy: %.2f' % (accuracy*100))

# make class predictions with the model
predicitions = model.predict_classes(X)

# summarize the first 10 cases
for i in range(10):
    print('%s => %.2f (expected %.2f)' % (X[i].tolist(), predicitions[i], y[i]))

这是数据集：

1,1
4,2
9,3
16,4
25,5
36,6
49,7
64,8
81,9
100,10

当我运行这个网络时，我得到以下结果：

[1.0] => 0.00 (expected 1.00)
[4.0] => 0.00 (expected 2.00)
[9.0] => 1.00 (expected 3.00)
[16.0] => 1.00 (expected 4.00)
[25.0] => 1.00 (expected 5.00)
[36.0] => 1.00 (expected 6.00)
[49.0] => 1.00 (expected 7.00)
[64.0] => 1.00 (expected 8.00)
[81.0] => 1.00 (expected 9.00)
[100.0] => 1.00 (expected 10.00)

我究竟做错了什么？

标签： pythontensorflowmachine-learningkerasneural-network

Answer 1

这是一个回归问题。所以你应该使用model.predict()而不是model.predict_classes().

数据集也不够大。但是，您可以使用以下代码获得一些明智的预测。

from numpy import loadtxt
from keras.models import Sequential
from keras.layers import Dense
from keras import optimizers

# laod dataset
# dataset = loadtxt('pima-indians-diabetes.csv', delimiter=',')
dataset = loadtxt('sqrt.csv', delimiter=',')

# split into input (X) and output (y) variables
X = dataset[:,0:1] * 1.0
y = dataset[:,1] * 1.0

# define the keras model
model = Sequential()
model.add(Dense(6, input_dim=1, activation='relu'))
model.add(Dense(10, activation='relu'))
model.add(Dense(1))

# compile the keras model
opt = optimizers.adam(lr=0.001)
model.compile(loss='mean_squared_error', optimizer=opt)

# fit the keras model on the dataset (CPU)
model.fit(X, y, epochs=1500, batch_size=10, verbose=0)

# evaluate the keras model
_, accuracy = model.evaluate(X, y, verbose=0)
print('Accuracy: %.2f' % (accuracy*100))

# make class predictions with the model
predicitions = model.predict(X)

# summarize the first 10 cases
for i in range(10):
    print('%s => %.2f (expected %.2f)' % (X[i].tolist(), predicitions[i], y[i]))

输出：

[1.0] => 1.00 (expected 1.00)
[4.0] => 2.00 (expected 2.00)
[9.0] => 3.32 (expected 3.00)
[16.0] => 3.89 (expected 4.00)
[25.0] => 4.61 (expected 5.00)
[36.0] => 5.49 (expected 6.00)
[49.0] => 6.52 (expected 7.00)
[64.0] => 7.72 (expected 8.00)
[81.0] => 9.07 (expected 9.00)
[100.0] => 10.58 (expected 10.00)

编辑：

正如@desertnaut 在评论中指出的那样，该指标accuracy在回归任务中没有意义。R_squared因此，通常使用自定义值（AKA 确定系数）作为度量。R_squared值表示回归模型中的拟合优度。以下是计算的代码R_squared。

def r_squared(y_true, y_pred):
    from keras import backend as K
    SS_res =  K.sum(K.square(y_true - y_pred)) 
    SS_tot = K.sum(K.square(y_true - K.mean(y_true))) 
    return ( 1 - SS_res/(SS_tot + K.epsilon()) )

现在，您可以使用以下命令编译模型；

model.compile(loss='mean_squared_error', optimizer=opt, metrics=[r_squared])