Question

感谢您阅读本文。

我尝试使用theano实现多标签逻辑回归：

import numpy
import theano
import theano.tensor as T
rng = numpy.random

examples = 5
features = 10
labels = 2
D = (rng.randn(examples, labels, features), rng.randint(size=(labels, examples), low=0, high=2))
training_steps = 10000

# Declare Theano symbolic variables
x = T.matrix("x")
y = T.vector("y")
w = theano.shared(rng.randn(1 , labels ,features), name="w")
b = theano.shared(0., name="b")
print "Initial model:"
print w.get_value(), b.get_value()

# Construct Theano expression graph
p_1 = 1 / (1 + T.exp(-T.dot(x, w) - b))   # Probability that target = 1
prediction = p_1 > 0.5                    # The prediction thresholded
xent = -y * T.log(p_1) - (1-y) * T.log(1-p_1) # Cross-entropy loss function
cost = xent.mean() + 0.01 * (w ** 2).sum()# The cost to minimize
gw, gb = T.grad(cost, [w, b])             # Compute the gradient of the cost
                                          # (we shall return to this in a
                                          # following section of this tutorial)

# Compile
train = theano.function(
          inputs=[x,y],
          outputs=[prediction, xent],
          updates=((w, w - 0.1 * gw), (b, b - 0.1 * gb)),
          name='train')
predict = theano.function(inputs=[x], outputs=prediction , name='predict')

# Train
for i in range(training_steps):
    pred, err = train(D[0], D[1])

print "Final model:"
print w.get_value(), b.get_value()
print "target values for D:", D[1]
print "prediction on D:", predict(D[0])

但-T.dot（x，w）产品因此错误而失败：

TypeError :(＆＃39; theano函数的错误输入参数，名称为＆＃34; train＆＃34; at index 0（0-based）＆＃39;，＆＃39;错误的维数：预期2 ，有3个形状（5,10,2）。＆＃39;）

x具有形状（5,2,10）和W（1,2,10）。我希望点积具有形状（5,2）。

我的问题是：反正有没有做这个内在的产品？您认为有更好的方法来实现多标签逻辑回归吗？

谢谢！

----编辑-----

所以这是我想用numpy做的实现。

x = rng.randn(examples,labels,features)
w = rng.randn (labels,features)
dot = numpy.zeros((examples,labels))
for example in range(examples):
    for label in range(labels):
        dot[example,label] = x[example,label,:].dot(w[label,:])
print dot

输出：

[[-1.70321498  2.51088139]
 [-5.73608956  0.1066286 ]
 [ 2.31334531  3.31892284]
 [ 1.56301872 -0.56150922]
 [-1.98815855 -2.98866706]]

但我不知道如何使用Theano象征性地做到这一点。

Answer 1

经过几个小时的战斗后，这似乎产生了正确的结果：

我有一个错误，输入为rng.randn（示例，功能，标签）而不是rng.randn（示例，功能）。这意味着，除了有更多标签外，输入应该是相同的大小。

以正确的方式计算点积的方法是使用theano.scan方法，如： results，updates = theano.scan（lambda label：T.dot（x，w [label，：]） - b [label]，sequences = T.arange（labels））

感谢大家的帮助！

import numpy as np
import theano
import theano.tensor as T
rng = np.random

examples = 5
features = 10
labels = 2
D = (rng.randn(examples,features), rng.randint(size=(labels, examples), low=0, high=2))
training_steps = 10000

# Declare Theano symbolic variables
x = T.matrix("x")
y = T.matrix("y")
w = theano.shared(rng.randn(labels ,features), name="w")
b = theano.shared(np.zeros(labels), name="b")
print "Initial model:"
print w.get_value(), b.get_value()

results, updates = theano.scan(lambda label: T.dot(x, w[label,:]) - b[label], sequences=T.arange(labels))

# Construct Theano expression graph
p_1 = 1 / (1 + T.exp(- results))   # Probability that target = 1
prediction = p_1 > .5                         # The prediction thresholded
xent = -y * T.log(p_1) - (1-y) * T.log(1-p_1) # Cross-entropy loss function
cost = xent.mean() + 0.01 * (w ** 2).sum()# The cost to minimize
gw, gb = T.grad(cost, [w, b])             # Compute the gradient of the cost
                                          # (we shall return to this in a
                                          # following section of this tutorial)

# Compile
train = theano.function(
          inputs=[x,y],
          outputs=[prediction, xent],
          updates=((w, w - 0.1 * gw), (b, b - 0.1 * gb)),
          name='train')
predict = theano.function(inputs=[x], outputs=prediction , name='predict')

# Train
for i in range(training_steps):
    pred, err = train(D[0], D[1])

print "Final model:"
print w.get_value(), b.get_value()
print "target values for D:", D[1]
print "prediction on D:", predict(D[0])

Theano内在产品3d矩阵

1 个答案: