softmax回归

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softmax回归常用于多分类问题，其输出可直接看成对类别的预测概率

假设对k类标签（[1, 2, ..., k]）进行分类，那么经过softmax回归计算后，输出一个k维向量，向量中每个值都代表对一个类别的预测概率

下面先以单个输入数据为例，进行评分函数、损失函数的计算和求导，然后扩展到多个输入数据同步计算

对数函数操作

对数求和

$$
log_{a}^{x}+log_{a}^{y} = log_{a}^{xy}
$$

对数求差

$$
log_{a}^{x}-log_{a}^{y} = log_{a}^{\frac{x}{y}}
$$

指数乘法

$$
e^{x}\cdot e^{y} = e^{x+y}
$$

求导公式

若函数$u(x),v(x)均可导$，那么

$$
\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{u^{\prime}(x) v(x)-v^{\prime}(x) u(x)}{v^{2}(x)}
$$

单个输入数据进行softmax回归计算

评分函数

假设使用softmax回归分类数据$x$，共$k$个标签，首先进行线性回归操作

$$
z_{\theta}(x)=\theta^T\cdot x
=\begin{bmatrix}
\theta_{1}^T\
\theta_{2}^T\
…\
\theta_{k}^T
\end{bmatrix}\cdot x
=\begin{bmatrix}
\theta_{1}^T\cdot x\
\theta_{2}^T\cdot x\
…\
\theta_{k}^T\cdot x
\end{bmatrix}
$$

其中输入数据$x$大小为$(n+1)\times 1$，$\theta$大小为$(n+1)\times k$，$n$表示权重数量，$m$表示训练数据个数，$k$表示类别标签数量

输出结果$z$大小为$k\times 1$，然后对计算结果进行归一化操作，使得输出值能够表示类别概率，如下所示

$$
h_{\theta}\left(x\right)=\left[ \begin{array}{c}{p\left(y=1 | x ; \theta\right)} \ {p\left(y=2 | x ; \theta\right)} \ {\vdots} \ {p\left(y=k | x ; \theta\right)}\end{array}\right]
=\frac{1}{\sum_{j=1}^{k} e^{\theta_{j}^{T} x}} \left[ \begin{array}{c}{e^{\theta_{1}^{T} x}} \ {e^{\theta_{2}^{T} x}} \ {\vdots} \ {e^{\theta_{k}^{T} x}}\end{array}\right]
$$

其中$\theta_{1}、\theta_{2},…,\theta_{k}$的大小为$(n+1)\times 1$，输出结果是一个$k\times 1$大小向量，每列表示$k$类标签的预测概率

所以对于输入数据$x$而言，其属于标签$j$的概率是

$$
p\left(y=j | x; \theta\right)=\frac{e^{\theta_{j}^{T} x}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x}}
$$

损失函数

利用交叉熵损失（cross entropy loss）作为softmax回归的损失函数，用于计算训练数据对应的真正标签的损失值

$$
J(\theta)
= (-1)\cdot \sum_{j=1}^{k} 1\left{y=j\right} \ln p\left(y=j | x; \theta\right)
= (-1)\cdot \sum_{j=1}^{k} 1\left{y=j\right} \ln \frac{e^{\theta_{j}^{T} x}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x}}
$$

其中函数$1{\cdot}$是一个示性函数（indicator function），其取值规则为

1
1{a true statement} = 1, and 1{a false statement} = 0

也就是示性函数输入为True时，输出为1；否则，输出为0

对权重向量$\theta_{s}$进行求导：

$$
\frac{\varphi J(\theta)}{\varphi \theta_{s}}
=(-1)\cdot \frac{\varphi }{\varphi \theta_{s}}
\left[ \
\sum_{j=1,j\neq s}^{k} 1\left{y=j \right} \ln p\left(y=j | x; \theta\right)
+1\left{y=s \right} \ln p\left(y=s | x; \theta\right) \
\right]
$$

$$
=(-1)\cdot \sum_{j=1,j\neq s}^{k} 1\left{y=j \right} \frac{1}{p\left(y=j | x; \theta\right)}\frac{\varphi p\left(y=j | x; \theta\right)}{\varphi \theta_{s}}
+(-1)\cdot 1\left{y=s \right} \frac{1}{p\left(y=s | x; \theta\right)}\frac{\varphi p\left(y=s | x; \theta\right)}{\varphi \theta_{s}}
$$

分为两种情况

  • 当计算结果正好由$\theta_{s}$计算得到，此时线性运算为$z=\theta_{s}^{T} x$，计算结果为$p\left(y=s | x; \theta\right)=\frac{e^{\theta_{s}^{T} x}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x}}$，求导如下

$$
\frac{\varphi p\left(y=s | x; \theta\right)}{\varphi \theta_{s}}
=\frac{u^{\prime}(x) v(x)-v^{\prime}(x) u(x)}{v^{2}(x)}
$$

其中

$$
u(x) = e^{\theta_{s}^{T} x}, v(x)=\sum_{l=1}^{k} e^{\theta_{l}^{T} x}
$$

所以

$$
\frac{\varphi u(x)}{\varphi \theta_s} = e^{\theta_{s}^{T} x}\cdot x=u(x)\cdot x,
\frac{\varphi v(x)}{\varphi \theta_s} = e^{\theta_{s}^{T} x}\cdot x=u(x)\cdot x \
\frac{\varphi p\left(y=s | x; \theta\right)}{\varphi \theta_{s}} = p\left(y=s | x; \theta\right)\cdot x-p\left(y=s | x; \theta\right)^2\cdot x
$$

  • 当计算结果不是由$\theta_{s}$计算得到，此时线性运算为$z=\theta_{j}^{T} x, j\neq s$，计算结果为$p\left(y=j | x; \theta\right)=\frac{e^{\theta_{j}^{T} x}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x}}$

$$
\frac{\varphi p\left(y=j | x; \theta\right)}{\varphi \theta_{s}}
=\frac{u^{\prime}(x) v(x)-v^{\prime}(x) u(x)}{v^{2}(x)}
$$

其中

$$
u(x) = e^{\theta_{j}^{T} x}, v(x)=\sum_{l=1}^{k} e^{\theta_{l}^{T} x}
$$

所以

$$
\frac{\varphi u(x)}{\varphi \theta_s} = e^{\theta_{j}^{T} x}\cdot x=0,
\frac{\varphi v(x)}{\varphi \theta_s} = e^{\theta_{s}^{T} x}\cdot x \
\frac{\varphi p\left(y=s | x; \theta\right)}{\varphi \theta_{s}} = -p\left(y=s | x; \theta\right)p\left(y=j | x; \theta\right)\cdot x
$$

综合上述两种情况可知，求导结果为

$$
\frac{\varphi J(\theta)}{\varphi \theta_{s}}
=(-1)\cdot \sum_{j=1,j\neq s}^{k} 1\left{y=j \right} \frac{1}{p\left(y=j | x; \theta\right)}\frac{\varphi p\left(y=j | x; \theta\right)}{\varphi \theta_{s}}
+(-1)\cdot 1\left{y=s \right} \frac{1}{p\left(y=s | x; \theta\right)}\frac{\varphi p\left(y=s | x; \theta\right)}{\varphi \theta_{s}} \
=(-1)\cdot \sum_{j=1,j\neq s}^{k} 1\left{y=j \right} \frac{1}{p\left(y=j | x; \theta\right)})\cdot (-1)\cdot p\left(y=s | x; \theta\right)p\left(y=j | x; \theta\right)\cdot x + (-1)\cdot 1\left{y=s \right} \frac{1}{p\left(y=s | x; \theta\right)}\left[p\left(y=s | x; \theta\right)\cdot x-p\left(y=s | x; \theta\right)^2\cdot x\right] \
=(-1)\cdot \sum_{j=1,j\neq s}^{k} 1\left{y=j \right}\cdot (-1)\cdot p\left(y=s | x; \theta\right)\cdot x + (-1)\cdot 1\left{y=s \right} \left[x-p\left(y=s | x; \theta\right)\cdot x\right] \
=(-1)\cdot 1\left{y=s \right} x - (-1)\cdot \sum_{j=1}^{k} 1\left{y=j \right} p\left(y=s | x; \theta\right)\cdot x
$$

因为$\sum_{j=1}^{k} 1\left{y=j \right}=1$，所以最终结果为

$$
\frac{\varphi J(\theta)}{\varphi \theta_{s}}
=(-1)\cdot \left[ 1\left{y=s \right} - p\left(y=s | x; \theta\right) \right]\cdot x
$$

批量数据进行softmax回归计算

上面实现了单个数据进行类别概率和损失函数的计算以及求导，进一步推导到批量数据进行操作

评分函数

假设使用softmax回归分类数据$x$，共$k$个标签，首先进行线性回归操作

$$
z_{\theta}(x_{i})=\theta^T\cdot x_{i}
=\begin{bmatrix}
\theta_{1}^T\
\theta_{2}^T\
…\
\theta_{k}^T
\end{bmatrix}\cdot x_{i}
=\begin{bmatrix}
\theta_{1}^T\cdot x_{i}\
\theta_{2}^T\cdot x_{i}\
…\
\theta_{k}^T\cdot x_{i}
\end{bmatrix}
$$

其中输入数据$x$大小为$(n+1)\times m$，$\theta$大小为$(n+1)\times k$，$n$表示权重数量，$m$表示训练数据个数，$k$表示类别标签数量

输出结果$z$大小为$k\times m$，然后对计算结果进行归一化操作，使得输出值能够表示类别概率，如下所示

$$
h_{\theta}\left(x_{i}\right)=\left[ \begin{array}{c}{p\left(y=1 | x_{i} ; \theta\right)} \
{p\left(y=2 | x_{i} ; \theta\right)} \
{\vdots} \
{p\left(y=k | x_{i} ; \theta\right)}\end{array}\right]
=\frac{1}{\sum_{j=1}^{k} e^{\theta_{j}^{T} x}} \left[ \begin{array}{c}{e^{\theta_{1}^{T} x_{i}}} \
{e^{\theta_{2}^{T} x_{i}}} \
{\vdots} \
{e^{\theta_{k}^{T} x_{i}}}\end{array}\right]
$$

其中$\theta_{1}、\theta_{2},…,\theta_{k}$的大小为$(n+1)\times 1$，输出结果是一个$k\times m$大小向量，每列表示$k$类标签的预测概率

所以对于输入数据$x_{i}$而言，其属于标签$j$的概率是

$$
p\left(y_{i}=j | x_{i}; \theta\right)=\frac{e^{\theta_{j}^{T} x_{i}}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x_{i}}}
$$

代价函数

利用交叉熵损失（cross entropy loss）作为softmax回归的代价函数，用于计算训练数据对应的真正标签的损失值

$$
J(\theta)
= (-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m} \sum_{j=1}^{k} 1\left{y_{i}=j\right} \ln p\left(y_{i}=j | x_{i}; \theta\right)
= (-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m} \sum_{j=1}^{k} 1\left{y_{i}=j\right} \ln \frac{e^{\theta_{j}^{T} x_{i}}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x_{i}}}
$$

其中函数$1{\cdot}$是一个示性函数（indicator function），其取值规则为

1
1{a true statement} = 1, and 1{a false statement} = 0

也就是示性函数输入为True时，输出为1；否则，输出为0

对权重向量$\theta_{s}$进行求导：

$$
\frac{\varphi J(\theta)}{\varphi \theta_{s}}
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \frac{\varphi }{\varphi \theta_{s}}
\left[ \sum_{j=1,j\neq s}^{k} 1\left{y_{i}=j \right} \ln p\left(y_{i}=j | x_{i}; \theta\right)+1\left{y_{i}=s \right} \ln p\left(y_{i}=s | x_{i}; \theta\right) \right]
$$

$$
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \sum_{j=1,j\neq s}^{k} 1\left{y_{i}=j \right} \frac{1}{p\left(y_{i}=j | x_{i}; \theta\right)}\frac{\varphi p\left(y_{i}=j | x_{i}; \theta\right)}{\varphi \theta_{s}}
+(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot 1\left{y_{i}=s \right} \frac{1}{p\left(y_{i}=s | x_{i}; \theta\right)}\frac{\varphi p\left(y_{i}=s | x_{i}; \theta\right)}{\varphi \theta_{s}}
$$

分为两种情况

  • 当计算结果正好由$\theta_{s}$计算得到，此时线性运算为$z=\theta_{s}^{T} x_{i}$，计算结果为$p\left(y_{i}=s | x_{i}; \theta\right)=\frac{e^{\theta_{s}^{T} x_{i}}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x_{i}}}$，求导如下

$$
\frac{\varphi p\left(y_{i}=s | x_{i}; \theta\right)}{\varphi \theta_{s}}
=\frac{u^{\prime}(x) v(x)-v^{\prime}(x) u(x)}{v^{2}(x)}
$$

其中

$$
u(x) = e^{\theta_{s}^{T} x}, v(x)=\sum_{l=1}^{k} e^{\theta_{l}^{T} x}
$$

所以

$$
\frac{\varphi u(x)}{\varphi \theta_s} = e^{\theta_{s}^{T} x}\cdot x=u(x)\cdot x,
\frac{\varphi v(x)}{\varphi \theta_s} = e^{\theta_{s}^{T} x}\cdot x=u(x)\cdot x \
\frac{\varphi p\left(y=s | x_{i}; \theta\right)}{\varphi \theta_{s}} = p\left(y=s | x_{i}; \theta\right)\cdot x_{i}-p\left(y=s | x_{i}; \theta\right)^2\cdot x_{i}
$$

  • 当计算结果不是由$\theta_{s}$计算得到，此时线性运算为$z=\theta_{j}^{T} x_{i}, j\neq s$，计算结果为$p\left(y_{i}=j | x_{i}; \theta\right)=\frac{e^{\theta_{j}^{T} x_{i}}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x_{i}}}$

$$
\frac{\varphi p\left(y_{i}=j | x_{i}; \theta\right)}{\varphi \theta_{s}}
=\frac{u^{\prime}(x) v(x)-v^{\prime}(x) u(x)}{v^{2}(x)}
$$

其中

$$
u(x) = e^{\theta_{j}^{T} x}, v(x)=\sum_{l=1}^{k} e^{\theta_{l}^{T} x}
$$

所以

$$
\frac{\varphi u(x)}{\varphi \theta_s} = e^{\theta_{j}^{T} x}\cdot x=0,
\frac{\varphi v(x)}{\varphi \theta_s} = e^{\theta_{s}^{T} x}\cdot x \
\frac{\varphi p\left(y_{i}=s | x_{i}; \theta\right)}{\varphi \theta_{s}} = -p\left(y_{i}=s | x_{i}; \theta\right)p\left(y_{i}=j | x_{i}; \theta\right)\cdot x_{i}
$$

综合上述两种情况可知，求导结果为

$$
\frac{\varphi J(\theta)}{\varphi \theta_{s}}
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \sum_{j=1,j\neq s}^{k} 1\left{y_{i}=j \right} \frac{1}{p\left(y_{i}=j | x_{i}; \theta\right)}\frac{\varphi p\left(y_{i}=j | x_{i}; \theta\right)}{\varphi \theta_{s}}
+(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot 1\left{y_{i}=s \right} \frac{1}{p\left(y_{i}=s | x_{i}; \theta\right)}\frac{\varphi p\left(y_{i}=s | x_{i}; \theta\right)}{\varphi \theta_{s}} \
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \sum_{j=1,j\neq s}^{k} 1\left{y_{i}=j \right} \frac{1}{p\left(y_{i}=j | x_{i}; \theta\right)})\cdot (-1)\cdot p\left(y_{i}=s | x_{i}; \theta\right)p\left(y_{i}=j | x_{i}; \theta\right)\cdot x_{i} + (-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot 1\left{y_{i}=s \right} \frac{1}{p\left(y_{i}=s | x_{i}; \theta\right)}\left[p\left(y_{i}=s | x_{i}; \theta\right)\cdot x_{i}-p\left(y_{i}=s | x_{i}; \theta\right)^2\cdot x_{i}\right] \
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \sum_{j=1,j\neq s}^{k} 1\left{y_{i}=j \right}\cdot (-1)\cdot p\left(y_{i}=s | x_{i}; \theta\right)\cdot x_{i} + (-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot 1\left{y_{i}=s \right} \left[x_{i}-p\left(y_{i}=s | x_{i}; \theta\right)\cdot x_{i}\right] \
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot 1\left{y_{i}=s \right} x_{i} - (-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \sum_{j=1}^{k} 1\left{y_{i}=j \right} p\left(y_{i}=s | x_{i}; \theta\right)\cdot x_{i}
$$

因为$\sum_{j=1}^{k} 1\left{y_{i}=j \right}=1$，所以最终结果为

$$
\frac{\varphi J(\theta)}{\varphi \theta_{s}}
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \left[ 1\left{y_{i}=s \right} - p\left(y_{i}=s | x_{i}; \theta\right) \right]\cdot x_{i}
$$

梯度下降

权重$W$大小为$(n+1)\times k$，输入数据集大小为$m\times (n+1)$，输出数据集大小为$m\times k$

矩阵求导如下：

$$
\frac{\varphi J(\theta)}{\varphi \theta}
=\frac{1}{m}\cdot \sum_{i=1}^{m}\cdot
\begin{bmatrix}
(-1)\cdot\left[ 1\left{y=1 \right} - p\left(y=1 | x; \theta\right) \right]\cdot x\
(-1)\cdot\left[ 1\left{y=2 \right} - p\left(y=2 | x; \theta\right) \right]\cdot x\
…\
(-1)\cdot\left[ 1\left{y=k \right} - p\left(y=k | x; \theta\right) \right]\cdot x
\end{bmatrix}
=(-1)\cdot \frac{1}{m}\cdot X_{m\times n+1}^T \cdot (I_{m\times k} - Y_{m\times k})
$$

参考：

Softmax regression for Iris classification

Derivative of Softmax loss function

上述计算的是输入单个数据时的评分、损失和求导，所以使用随机梯度下降法进行权重更新，分类

参数冗余和权重衰减

softmax回归存在参数冗余现象，即对参数向量$\theta_{j}$减去向量$\varphi $不改变预测结果。证明如下：

$$
\begin{aligned} p\left(y^{(i)}=j | x^{(i)} ; \theta\right) &=\frac{e^{\left(\theta_{j}-\psi\right)^{T} x^{(i)}}}{\sum_{l=1}^{k} e^{\left(\theta_{l}-\psi\right)^{T} x^{(i)}}} \ &=\frac{e^{\theta_{j}^{T} x^{(i)}} e^{-\psi^{T} x^{(i)}}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x^{(i)}} e^{-\psi^{T} x^{(i)}}} \ &=\frac{e^{\theta_{j}^{T} x^{(i)}}}{\sum_{l=1}^{k} e^{\theta_{t}^{T} x^{(i)}}} \end{aligned}
$$

假设$(\theta_{1},\theta_{2},…,\theta_{k})$能得到$j(\theta)$的极小值点，那么$(\theta_{1}-\varphi,\theta_{2}-\varphi,…,\theta_{k}-\varphi)$同样能得到相同的极小值点

与此同时，因为损失函数是凸函数，局部最小值就是全局最小值，所以会导致权重在参数过大情况下就停止收敛，影响模型泛化能力

在代价函数中加入权重衰减，能够避免过度参数化，得到泛化性能更强的模型

在代价函数中加入L2正则化项，如下所示：

$$
J(\theta)
= (-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m} \sum_{j=1}^{k} 1\left{y_{i}=j\right} \ln p\left(y_{i}=j | x_{i}; \theta\right) + \frac{\lambda}{2} \sum_{i=1}^{k} \sum_{j=0}^{n} \theta_{i j}^{2}
= (-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m} \sum_{j=1}^{k} 1\left{y_{i}=j\right} \ln \frac{e^{\theta_{j}^{T} x_{i}}}{\sum_{l=1}^{k} e^{\theta_{l}^{T} x_{i}}} + \frac{\lambda}{2} \sum_{i=1}^{k} \sum_{j=0}^{n} \theta_{i j}^{2}
$$

求导结果如下：

$$
\frac{\varphi J(\theta)}{\varphi \theta_{s}}
=(-1)\cdot \frac{1}{m}\cdot \sum_{i=1}^{m}\cdot \left[ 1\left{y_{i}=s \right} - p\left(y_{i}=s | x_{i}; \theta\right) \right]\cdot x_{i}+ \lambda \theta_{j}
$$

代价实现如下：

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def compute_loss(scores, indicator, W):
    """
    计算损失值
    :param scores: 大小为(m, n)
    :param indicator: 大小为(m, n)
    :param W: (n, k)
    :return: (m,1)
    """
    return -1 * np.sum(np.log(scores) * indicator, axis=1) + 0.001/2*np.sum(W**2)


def compute_gradient(indicator, scores, x, W):
    """
    计算梯度
    :param indicator: 大小为(m,k)
    :param scores: 大小为(m,k)
    :param x: 大小为(m,n)
    :param W: (n, k)
    :return: (n,k)
    """
    return -1 * x.T.dot((indicator - scores)) + 0.001*W

鸢尾数据集

使用鸢尾（iris）数据集，参考Iris Species

共4个变量：

  • SepalLengthCm - 花萼长度
  • SepalWidthCm - 花萼宽度
  • PetalLengthCm - 花瓣长度
  • PetalWidthCm - 花瓣宽度

以及3个类别：

  • Iris-setosa
  • Iris-versicolor
  • Iris-virginica
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def load_data(shuffle=True, tsize=0.8):
    """
    加载iris数据
    """
    data = pd.read_csv(data_path, header=0, delimiter=',')
    # print(data.columns)

    draw_data(data.values, data.columns)

def draw_data(data, columns):
    data_a = data[:50, 1:5]
    data_b = data[50:100, 1:5]
    data_c = data[100:150, 1:5]

    fig = plt.figure(1)
    plt.scatter(data_a[:, 0], data_a[:, 1], c='b', marker='8')
    plt.scatter(data_b[:, 0], data_b[:, 1], c='r', marker='s')
    plt.scatter(data_c[:, 0], data_c[:, 1], c='y', marker='*')
    plt.xlabel(columns[1])
    plt.ylabel(columns[2])
    plt.show()

    fig = plt.figure(2)
    plt.scatter(data_a[:, 2], data_a[:, 3], c='b', marker='8')
    plt.scatter(data_b[:, 2], data_b[:, 3], c='r', marker='s')
    plt.scatter(data_c[:, 2], data_c[:, 3], c='y', marker='*')
    plt.xlabel(columns[3])
    plt.ylabel(columns[4])
    plt.show()

    # 验证是否有重复数据
    # for i in range(data_b.shape[0]):
    #     res = list(filter(lambda x: x[0] == data_b[i][0] and x[1] == data_b[i][1], data_c[:, :2]))
    #     if len(res) != 0:
    #         res2 = list(filter(lambda x: x[2] == data_b[i][2] and x[3] == data_b[i][3], data_c[:, 2:4]))
    #         if len(res2) != 0:
    #             print(b[i])

numpy实现

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# -*- coding: utf-8 -*-

# @Time    : 19-4-25 上午10:30
# @Author  : zj

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from sklearn import utils
from sklearn.model_selection import train_test_split
import warnings

warnings.filterwarnings('ignore')

data_path = '../data/iris-species/Iris.csv'


def load_data(shuffle=True, tsize=0.8):
    """
    加载iris数据
    """
    data = pd.read_csv(data_path, header=0, delimiter=',')

    if shuffle:
        data = utils.shuffle(data)

    # 示性函数
    pd_indicator = pd.get_dummies(data['Species'])
    indicator = np.array(
        [pd_indicator['Iris-setosa'], pd_indicator['Iris-versicolor'], pd_indicator['Iris-virginica']]).T

    species_dict = {
        'Iris-setosa': 0,
        'Iris-versicolor': 1,
        'Iris-virginica': 2
    }
    data['Species'] = data['Species'].map(species_dict)

    data_x = np.array(
        [data['SepalLengthCm'], data['SepalWidthCm'], data['PetalLengthCm'], data['PetalWidthCm']]).T
    data_y = data['Species']

    x_train, x_test, y_train, y_test = train_test_split(data_x, data_y, train_size=tsize, test_size=(1 - tsize),
                                                        shuffle=False)

    y_train = np.atleast_2d(y_train).T
    y_test = np.atleast_2d(y_test).T

    y_train_indicator = np.atleast_2d(indicator[:y_train.shape[0]])
    y_test_indicator = indicator[y_train.shape[0]:]

    return x_train, x_test, y_train, y_test, y_train_indicator, y_test_indicator


def linear(x, w):
    """
    线性操作
    :param x: 大小为(m,n+1)
    :param w: 大小为(n+1,k)
    :return: 大小为(m,k)
    """
    return x.dot(w)


def softmax(x):
    """
    softmax归一化计算
    :param x: 大小为(m, k)
    :return: 大小为(m, k)
    """
    x -= np.atleast_2d(np.max(x, axis=1)).T
    exps = np.exp(x)
    return exps / np.atleast_2d(np.sum(exps, axis=1)).T


def compute_scores(X, W):
    """
    计算精度
    :param X: 大小为(m,n+1)
    :param W: 大小为(n+1,k)
    :return: (m,k)
    """
    return softmax(linear(X, W))


def compute_loss(scores, indicator, W, la=2e-4):
    """
    计算损失值
    :param scores: 大小为(m, k)
    :param indicator: 大小为(m, k)
    :param W: (n+1, k)
    :return: (1)
    """
    cost = -1 / scores.shape[0] * np.sum(np.log(scores) * indicator)
    reg = la / 2 * np.sum(W ** 2)
    return cost + reg


def compute_gradient(scores, indicator, x, W, la=2e-4):
    """
    计算梯度
    :param scores: 大小为(m,k)
    :param indicator: 大小为(m,k)
    :param x: 大小为(m,n+1)
    :param W: (n+1, k)
    :return: (n+1,k)
    """
    return -1 / scores.shape[0] * x.T.dot((indicator - scores)) + la * W


def compute_accuracy(scores, Y):
    """
    计算精度
    :param scores: (m,k)
    :param Y: (m,1)
    """
    res = np.dstack((np.argmax(scores, axis=1), Y.squeeze())).squeeze()

    return len(list(filter(lambda x: x[0] == x[1], res[:]))) / len(res)


def draw(res_list, title=None, xlabel=None):
    if title is not None:
        plt.title(title)
    if xlabel is not None:
        plt.xlabel(xlabel)
    plt.plot(res_list)
    plt.show()


def compute_gradient_descent(batch_size=8, epoches=2000, alpha=2e-4):
    x_train, x_test, y_train, y_test, y_train_indicator, y_test_indicator = load_data()

    m, n = x_train.shape[:2]
    k = y_train_indicator.shape[1]
    # 初始化权重(n+1,k)
    W = 0.01 * np.random.normal(loc=0.0, scale=1.0, size=(n + 1, k))
    x_train = np.insert(x_train, 0, np.ones(m), axis=1)
    x_test = np.insert(x_test, 0, np.ones(x_test.shape[0]), axis=1)

    loss_list = []
    accuracy_list = []
    bestW = None
    bestA = 0
    range_list = np.arange(0, x_train.shape[0] - batch_size, step=batch_size)
    for i in range(epoches):
        for j in range_list:
            data = x_train[j:j + batch_size]
            labels = y_train_indicator[j:j + batch_size]

            # 计算分类概率
            scores = np.atleast_2d(compute_scores(data, W))
            # 更新梯度
            tempW = W - alpha * compute_gradient(scores, labels, data, W)
            W = tempW

            if j == range_list[-1]:
                loss = compute_loss(scores, labels, W)
                loss_list.append(loss)

                accuracy = compute_accuracy(compute_scores(x_train, W), y_train)
                accuracy_list.append(accuracy)
                if accuracy >= bestA:
                    bestA = accuracy
                    bestW = W.copy()
                break

    draw(loss_list, title='损失值')
    draw(accuracy_list, title='训练精度')

    print(bestA)
    print(compute_accuracy(compute_scores(x_test, bestW), y_test))


if __name__ == '__main__':
    compute_gradient_descent(batch_size=8, epoches=100000)

训练10万次的最好训练结果以及对应的测试结果：

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# 测试集精度
0.9916666666666667
# 验证集精度
0.9666666666666667

指数计算 - 数值稳定性考虑

在softmax回归中，需要利用指数函数$e^x$对线性操作的结果进行归一化，这有可能会造成数值溢出，常用的做法是对分数上下同乘以一个常数$C$

$$
\frac{e^{f_{i_{i}}}}{\sum_{j} e^{f_{j}}}=\frac{C e^{f_{y_{i}}}}{C \sum_{j} e^{f_{j}}}=\frac{e^{f_{i_{i}}+\log C}}{\sum_{j} e^{f_{j}+\log C}}
$$

这个操作不改变结果，如果取值$C$为线性操作结果最大值负数$\log C=-\max {j} f{j}$，就能够将向量$f$的取值范围降低，最大值为$0$，避免数值不稳定

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def softmax(x):
    """
    softmax归一化计算
    :param x: 大小为(m, k)
    :return: 大小为(m, k)
    """
    x -= np.atleast_2d(np.max(x, axis=1)).T
    exps = np.exp(x)
    return exps / np.atleast_2d(np.sum(exps, axis=1)).T

softmax回归和logistic回归

softmax回归是logistic回归在多分类任务上的扩展，将$k=2$时，softmax回归模型可转换成logistic回归模型

$$
h_{\theta}(x)=\frac{1}{e^{\theta_{1}^{T} x}+e^{\theta_{2}^{T} x^{(i)}}} \left[ \begin{array}{c}{e^{\theta_{1}^{T} x}} \ {e^{\theta_{2}^{T} x}}\end{array}\right]
=\frac{1}{e^{\vec{0}^{T} x}+e^{(\theta_{2}-\theta_{1})^{T} x^{(i)}}} \left[ \begin{array}{c}{e^{\vec{0}^{T} x}} \ {e^{(\theta_{2}-\theta_{1})^{T} x}}\end{array}\right] \
=\frac{1}{1+e^{(\theta_{2}-\theta_{1})^{T} x^{(i)}}} \left[ \begin{array}{c}{1} \ {e^{(\theta_{2}-\theta_{1})^{T} x}}\end{array}\right]
= \left[ \begin{array}{c}{\frac{1}{1+e^{(\theta_{2}-\theta_{1})^{T} x^{(i)}}}} \ {\frac{e^{(\theta_{2}-\theta_{1})^{T} x}}{1+e^{(\theta_{2}-\theta_{1})^{T} x^{(i)}}}}\end{array}\right]
=\left[ \begin{array}{c}{\frac{1}{1+e^{(\theta_{2}-\theta_{1})^{T} x^{(i)}}}} \ {1- \frac{1}{1+e^{(\theta_{2}-\theta_{1})^{T} x^{(i)}}}}\end{array}\right]
$$

针对多分类任务，可以选择softmax回归模型进行多分类，也可以选择logistic回归模型进行若干个二分类

区别在于选择的类别是否互斥，如果类别互斥，使用softmax回归分类更为合适；如果类别不互斥，使用logistic回归分类更为合适

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