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508 lines
24 KiB
Markdown
508 lines
24 KiB
Markdown
# Theano 实例：Logistic 回归
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In [1]:
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```py
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%matplotlib inline
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import numpy as np
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import matplotlib.pyplot as plt
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import theano
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import theano.tensor as T
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```
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```py
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Using gpu device 0: GeForce GTX 850M
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```
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## sigmoid 函数
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一个 `logistic` 曲线由 `sigmoid` 函数给出： $$s(x) = \frac{1}{1+e^{-x}}$$
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我们来定义一个 `elementwise` 的 sigmoid 函数：
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In [2]:
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```py
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x = T.matrix('x')
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s = 1 / (1 + T.exp(-x))
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sigmoid = theano.function([x], s, allow_input_downcast=True)
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```
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这里 `allow_input_downcast=True` 的作用是允许输入 `downcast` 成定义的输入类型：
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In [3]:
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```py
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sigmoid([[ 0, 1],
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[-1,-2]])
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```
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Out[3]:
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```py
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array([[ 0.5 , 0.7310586 ],
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[ 0.26894143, 0.11920293]], dtype=float32)
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```
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其图像如下所示：
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In [4]:
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```py
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X = np.linspace(-6, 6, 100)
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X = X[np.newaxis,:]
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plt.figure(figsize=(12,5))
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plt.plot(X.flatten(), sigmoid(X).flatten(), linewidth=2)
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# 美化图像的操作
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#=========================
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plt.grid('on')
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plt.yticks([0,0.5,1])
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ax = plt.gca()
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ax.spines['right'].set_color('none')
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ax.spines['top'].set_color('none')
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ax.yaxis.set_ticks_position('left')
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ax.spines['left'].set_position(('data', 0))
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plt.legend([r'$s(x)=\frac{1}{1+e^{-x}}$'], loc=0, fontsize=20)
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#=========================
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plt.show()
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```
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## sigmoid 函数与 tanh 函数的关系
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`sigmoid` 函数与 `tanh` 之间有如下的转化关系： $$s(x)=\frac{1}{1+e^{-x}}=\frac{1+\tanh(x/2)}{2}$$
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In [5]:
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```py
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s2 = (1 + T.tanh(x / 2)) / 2
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sigmoid2 = theano.function([x], s2)
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sigmoid2([[ 0, 1],
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[-1,-2]])
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```
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Out[5]:
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```py
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array([[ 0.5 , 0.7310586 ],
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[ 0.26894143, 0.11920291]], dtype=float32)
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```
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## logistic 回归
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简单的二元逻辑回归问题可以这样描述：我们要对数据点 $x = (x_1, ..., x_n)$ 进行 0-1 分类，参数为 $w = (w_1, ..., w_n), b$，我们的假设函数如下：
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$$ \begin{align} h_{w,b}(x) & = P(Y=1|X=x) \\ & = sigmoid(z) \\ & =\frac{1}{1 + e^{-z}}\\ \end{align} $$
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其中
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$$ \begin{align} z & = x_1w_1 + ... + x_nw_n + b\\ & = w^T x + b\\ \end{align} $$
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对于一个数据点 $(x, y), y\in \{0,1\}$ 来说，我们的目标是希望 $h_{w,b}(x)$ 的值尽量接近于 $y$。
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由于数值在 0-1 之间，我们用交叉熵来衡量 $h_{w,b}(x)$ 和 $y$ 的差异：
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$$- y \log(h_{w,b}(x)) - (1-y) \log(1-h_{w,b}(x))$$
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对于一组数据，我们定义损失函数为所有差异的均值，然后通过梯度下降法来优化损失函数，得到最优的参数 $w, b$。
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## 实例
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生成随机数据：
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In [6]:
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```py
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rng = np.random
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# 数据大小和规模
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N = 400
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feats = 784
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# D = (X, Y)
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D = (rng.randn(N, feats), rng.randint(size=N, low=0, high=2))
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```
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定义 `theano` 变量：
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In [7]:
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```py
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x = T.matrix('x')
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y = T.vector('y')
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# 要更新的变量：
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w = theano.shared(rng.randn(feats), name='w')
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b = theano.shared(0., name='b')
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```
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定义模型：
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In [8]:
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```py
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h = 1 / (1 + T.exp(-T.dot(x, w) - b))
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```
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当 $h > 0.5$ 时，认为该类的标签为 1：
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In [9]:
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```py
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prediction = h > 0.5
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```
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损失函数和梯度：
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In [10]:
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```py
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cost = - T.mean(y * T.log(h) + (1 - y) * T.log(1 - h)) + 0.01 * T.sum(w ** 2) # 正则项，防止过拟合
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gw, gb = T.grad(cost, [w, b])
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```
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编译训练和预测函数：
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In [11]:
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```py
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train = theano.function(inputs=[x, y],
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outputs=cost,
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updates=[[w, w - 0.1 * gw], [b, b - 0.1 * gb]],
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allow_input_downcast=True)
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predict = theano.function(inputs=[x],
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outputs=prediction,
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allow_input_downcast=True)
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```
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In [12]:
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```py
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for i in xrange(10001):
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err = train(D[0], D[1])
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if i % 1000 == 0:
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print 'iter %5d, error %f' % (i, err)
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```
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```py
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iter 0, error 19.295896
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iter 1000, error 0.210341
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iter 2000, error 0.126124
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iter 3000, error 0.124872
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iter 4000, error 0.124846
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iter 5000, error 0.124845
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iter 6000, error 0.124845
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iter 7000, error 0.124845
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iter 8000, error 0.124845
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iter 9000, error 0.124845
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iter 10000, error 0.124845
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```
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查看结果：
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In [13]:
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```py
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print D[1]
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```
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```py
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[0 0 0 1 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 0 1 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0
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1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 1 1 0 0 0 0 0 1 0 0
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0 1 0 1 0 0 1 1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 0 0 1 0 0 1 0 1 1 1 1 0 1 0
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0 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1
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1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 1 1 1
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1 0 1 0 0 1 0 0 1 1 1 1 0 1 0 1 0 1 1 1 1 0 1 0 0 1 1 1 0 0 0 1 0 0 0 1 0
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1 0 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1
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0 1 0 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1 0 1 1 0 0 1 1 1 1 1 1
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0 0 1 1 0 0 0 1 0 1 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0
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0 1 0 0 0 0 0 1 1 0 1 1 0 1 1 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 0 1 0 0 1
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0 1 0 1 0 0 1 0 0 1 1 1 0 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 0]
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```
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In [14]:
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```py
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print predict(D[0])
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```
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```py
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[0 0 0 1 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 0 1 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0
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1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 1 1 0 0 0 0 0 1 0 0
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0 1 0 1 0 0 1 1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 0 0 1 0 0 1 0 1 1 1 1 0 1 0
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0 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1
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1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 1 1 1
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1 0 1 0 0 1 0 0 1 1 1 1 0 1 0 1 0 1 1 1 1 0 1 0 0 1 1 1 0 0 0 1 0 0 0 1 0
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1 0 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1
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0 1 0 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1 0 1 1 0 0 1 1 1 1 1 1
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0 0 1 1 0 0 0 1 0 1 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0
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0 1 0 0 0 0 0 1 1 0 1 1 0 1 1 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 0 1 0 0 1
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0 1 0 1 0 0 1 0 0 1 1 1 0 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 0]
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``` |