条件随机场的预测问题是给定条件随机场$P(Y\mid X)$和输入序列(观测序列)$x$,求条件概率最大的输出序列(标记序列)$y^*$的过程,即:
$$ y^*=arg\max_yP_w(y\mid x)\\ =arg\max_y\frac{exp(w^TF(y,x))}{Z_w(x)}\\ =arg\max_yexp(w^TF(y,x))\\ =arg\max_yw^TF(y,x) $$可以发现,少了求$Z_w(x)$,计算量少了很多,上面的式子中:
$$ w=(w_1,w_2,...,w_K)^T\\ F(y,x)=(f_1(y,x),f_2(y,x),...,f_K(y,x))^T\\ f_k(y,x)=\sum_{i=1}^nf_k(y_{i-1},y_i,x,i),k=1,2,...,K $$接下来将上面的$w^TF(y,x)$进行改写:
$$ y^*=arg\max_y\sum_{i=1}^nw^TF_i(y_{i-1},y_i,x) $$其中:
$$ F_i(y_{i-1},y_i,x)=(f_1(y_{i-1},y_i,x,i)),f_2(y_{i-1},y_i,x,i)),...,f_K(y_{i-1},y_i,x,i)))^T $$这个问题的求解,其实与之前的HMM问题一样(链接>>>),同样是求DAG图中的最优路径问题,同样的,我们用符号$\delta_i(l)$表示时刻$i$标签状态为$l$的所有路径中的最优值,$\psi_i(l)$记录$i$的前一步的标签状态,下面对算法流程做一个完整的梳理
输入:模型特征向量$F(y,x)$和权值向量$w$,观测序列$x=(x_1,x_2,...,x_n)$;
输出:最优路径$y^*=(y_1^*,y_2^*,...,y_n^*)$
$$ \delta_1(j)=w^TF_1(y_0=start,y_1=j,x),j=1,2,...,m $$(1)初始化
$$ \delta_i(l)=\max_{1\leq j\leq m}(\delta_{i-1}(j)+w^TF_i(y_{i-1}=j,y_i=l,x)),l=1,2,...,m\\ \psi_i(l)=arg\max_{1\leq j\leq m}(\delta_{i-1}(j)+w^TF_i(y_{i-1}=j,y_i=l,x)),l=1,2,...,m $$(2)对$i=2,3,...,n$
$$ y_n^*=arg\max_{1\leq j \leq m}\delta_n(j) $$(3)终止
$$ y_i^*=\psi_{i+1}(y_{i+1}^*),i=n-1,n-2,...,1 $$(4)回溯路径
求得最后路径:$y^*=(y_1^*,y_2^*,...,y_n^*)$
在上一小节的内容上追加...
import os
os.chdir('../')
from ml_models.pgm import CRFFeatureFunction
import numpy as np
"""
线性链CRF的实现,封装到ml_models.pgm
"""
class CRF(object):
def __init__(self, epochs=10, lr=1e-3, tol=1e-5, output_status_num=None, input_status_num=None, unigram_rulers=None,
bigram_rulers=None):
"""
:param epochs: 迭代次数
:param lr: 学习率
:param tol:梯度更新的阈值
:param output_status_num:标签状态数
:param input_status_num:输入状态数
:param unigram_rulers: 状态特征规则
:param bigram_rulers: 状态转移规则
"""
self.epochs = epochs
self.lr = lr
self.tol = tol
# 为输入序列和标签状态序列添加一个头尾id
self.output_status_num = output_status_num + 2
self.input_status_num = input_status_num + 2
self.input_status_head_tail = [input_status_num, input_status_num + 1]
self.output_status_head_tail = [output_status_num, output_status_num + 1]
# 特征函数
self.FF = CRFFeatureFunction(unigram_rulers, bigram_rulers)
# 模型参数
self.w = None
def fit(self, x, y):
"""
:param x: [[...],[...],...,[...]]
:param y: [[...],[...],...,[...]]
:return
"""
# 为 x,y加头尾
x = [[self.input_status_head_tail[0]] + xi + [self.input_status_head_tail[1]] for xi in x]
y = [[self.output_status_head_tail[0]] + yi + [self.output_status_head_tail[1]] for yi in y]
self.FF.fit(x, y)
self.w = np.ones(len(self.FF.feature_funcs)) * 1e-5
for _ in range(0, self.epochs):
# 偷个懒,用随机梯度下降
for i in range(0, len(x)):
xi = x[i]
yi = y[i]
"""
1.求F(yi \mid xi)以及P_w(yi \mid xi)
"""
F_y_x = []
Z_x = np.ones(shape=(self.output_status_num, 1)).T
for j in range(1, len(xi)):
F_y_x.append(self.FF.map(yi[j - 1], yi[j], xi, j))
# 构建M矩阵
M = np.zeros(shape=(self.output_status_num, self.output_status_num))
for k in range(0, self.output_status_num):
for t in range(0, self.output_status_num):
M[k, t] = np.exp(np.dot(self.w, self.FF.map(k, t, xi, j)))
# 前向算法求 Z(x)
Z_x = Z_x.dot(M)
F_y_x = np.sum(F_y_x, axis=0)
Z_x = np.sum(Z_x)
# 求P_w(yi \mid xi)
P_w = np.exp(np.dot(self.w, F_y_x)) / Z_x
"""
2.求梯度,并更新
"""
dw = (P_w - 1) * F_y_x
self.w = self.w - self.lr * dw
if (np.sqrt(np.dot(dw, dw) / len(dw))) < self.tol:
break
def predict(self, x):
"""
维特比求解最优的y
:param x:[...]
:return:
"""
# 为x加头尾
x = [self.input_status_head_tail[0]] + x + [self.input_status_head_tail[1]]
# 初始化
delta = np.asarray([np.dot(self.w, self.FF.map(self.output_status_head_tail[0], j, x, 1)) for j in
range(0, self.output_status_num)])
psi = [[0] * self.output_status_num]
# 递推
for visible_index in range(2, len(x) - 1):
new_delta = np.zeros_like(delta)
new_psi = []
# 当前节点
for i in range(0, self.output_status_num):
best_pre_index_i = -1
best_pre_index_value_i = 0
delta_i = 0
# 上一轮节点
for j in range(0, self.output_status_num):
delta_i_j = delta[j] + np.dot(self.w, self.FF.map(j, i, x, visible_index))
if delta_i_j > delta_i:
delta_i = delta_i_j
best_pre_index_value_i_j = delta[j] + np.dot(self.w, self.FF.map(j, i, x, visible_index))
if best_pre_index_value_i_j > best_pre_index_value_i:
best_pre_index_value_i = best_pre_index_value_i_j
best_pre_index_i = j
new_delta[i] = delta_i
new_psi.append(best_pre_index_i)
delta = new_delta
psi.append(new_psi)
# 回溯
best_hidden_status = [np.argmax(delta)]
for psi_index in range(len(x) - 3, 0, -1):
next_status = psi[psi_index][best_hidden_status[-1]]
best_hidden_status.append(next_status)
best_hidden_status.reverse()
return best_hidden_status
# 测试一下
x = [
[1, 2, 3, 0, 1, 3, 4],
[1, 2, 3],
[0, 2, 4, 2],
[4, 3, 2, 1],
[3, 1, 1, 1, 1],
[2, 1, 3, 2, 1, 3, 4]
]
y = x
crf = CRF(output_status_num=5, input_status_num=5)
crf.fit(x, y)
crf.predict(x[0])
[1, 2, 3, 0, 1, 3, 4]
预测结果一样就对了...