{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "### 一.参数的梯度下降求解\n", "\n", "#### 损失函数推导\n", "CRF的参数学习同HMM一样,采用的极大似然估计的方式,其对数似然函数为: \n", "\n", "$$\n", "L(w)=L_{\\tilde{P}}(P_w)\\\\\n", "=log\\prod_{x,y}P_w(y\\mid x)^{\\tilde{P}(x,y)}\\\\\n", "=\\sum_{x,y}\\tilde{P}(x,y)logP_w(y\\mid x)\\\\\n", "=\\sum_{j=1}^NlogP_w(y_j\\mid x_j)(假设样本量为N)\n", "$$ \n", "\n", "将CRF的势函数: \n", "\n", "$$\n", "P_w(y\\mid x)=\\frac{exp(\\sum_{k=1}^Kw_kf_k(y,x))}{Z_w(x)}\n", "$$ \n", "\n", "带入上面的对数似然函数可得: \n", "\n", "$$\n", "L(w)=\\sum_{j=1}^N\\sum_{k=1}^Kw_kf_k(y_j,x_j)-\\sum_{j=1}^NlogZ_w(x_j)\n", "$$ \n", "\n", "最终,我们的问题等价于: \n", "\n", "$$\n", "w_k^*=arg\\min_{w_k}\\sum_{j=1}^N(logZ_w(x_j)-\\sum_{k=1}^Kw_kf_k(y_j,x_j))\n", "$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "#### 梯度下降\n", "\n", "最简单的就是梯度下降,可推导出其梯度函数为: \n", "\n", "$$\n", "g(w^t)=\\sum_{j=1}^N(P_{w^t}(y_j\\mid x_j)-1)F(y_j,x_j))\n", "$$ \n", "\n", "这里,$F(y_j,x_j)=[f_1(y_j,x_j),f_2(y_j,x_j),...,f_K(y_j,x_j)]^T$,所以梯度更新公式也就出来了: \n", "\n", "$$\n", "w^{t+1}=w^t-\\eta g(w^t)\n", "$$ \n", "\n", "这里,$\\eta$为学习率,用梯度更新公式来理解一下训练过程,每次迭代会将特征函数输出为1的权重增加,而输出为0的保持不变,则看起来也蛮合理的~~~" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### 二.代码实现" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "import os\n", "os.chdir('../')\n", "from ml_models.pgm import CRFFeatureFunction\n", "import numpy as np\n", "\n", "\"\"\"\n", "线性链CRF的实现,封装到ml_models.pgm\n", "\"\"\"\n", "\n", "\n", "class CRF(object):\n", " def __init__(self, epochs=10, lr=1e-3, tol=1e-5, output_status_num=None, input_status_num=None, unigram_rulers=None,\n", " bigram_rulers=None):\n", " \"\"\"\n", " :param epochs: 迭代次数\n", " :param lr: 学习率\n", " :param tol:梯度更新的阈值\n", " :param output_status_num:标签状态数\n", " :param input_status_num:输入状态数\n", " :param unigram_rulers: 状态特征规则\n", " :param bigram_rulers: 状态转移规则\n", " \"\"\"\n", " self.epochs = epochs\n", " self.lr = lr\n", " self.tol = tol\n", " # 为输入序列和标签状态序列添加一个头尾id\n", " self.output_status_num = output_status_num + 2\n", " self.input_status_num = input_status_num + 2\n", " self.input_status_head_tail = [input_status_num, input_status_num + 1]\n", " self.output_status_head_tail = [output_status_num, output_status_num + 1]\n", " # 特征函数\n", " self.FF = CRFFeatureFunction(unigram_rulers, bigram_rulers)\n", " # 模型参数\n", " self.w = None\n", "\n", " def fit(self, x, y):\n", " \"\"\"\n", " :param x: [[...],[...],...,[...]]\n", " :param y: [[...],[...],...,[...]]\n", " :return\n", " \"\"\"\n", " # 为 x,y加头尾\n", " x = [[self.input_status_head_tail[0]] + xi + [self.input_status_head_tail[1]] for xi in x]\n", " y = [[self.output_status_head_tail[0]] + yi + [self.output_status_head_tail[1]] for yi in y]\n", " self.FF.fit(x, y)\n", " self.w = np.ones(len(self.FF.feature_funcs)) * 1e-5\n", " for _ in range(0, self.epochs):\n", " # 偷个懒,用随机梯度下降\n", " for i in range(0, len(x)):\n", " xi = x[i]\n", " yi = y[i]\n", " \"\"\"\n", " 1.求F(yi \\mid xi)以及P_w(yi \\mid xi)\n", " \"\"\"\n", " F_y_x = []\n", " Z_x = np.ones(shape=(self.output_status_num, 1)).T\n", " for j in range(1, len(xi)):\n", " F_y_x.append(self.FF.map(yi[j - 1], yi[j], xi, j))\n", " # 构建M矩阵\n", " M = np.zeros(shape=(self.output_status_num, self.output_status_num))\n", " for k in range(0, self.output_status_num):\n", " for t in range(0, self.output_status_num):\n", " M[k, t] = np.exp(np.dot(self.w, self.FF.map(k, t, xi, j)))\n", " # 前向算法求 Z(x)\n", " Z_x = Z_x.dot(M)\n", " F_y_x = np.sum(F_y_x, axis=0)\n", " Z_x = np.sum(Z_x)\n", " # 求P_w(yi \\mid xi)\n", " P_w = np.exp(np.dot(self.w, F_y_x)) / Z_x\n", " \"\"\"\n", " 2.求梯度,并更新\n", " \"\"\"\n", " dw = (P_w - 1) * F_y_x\n", " self.w = self.w - self.lr * dw\n", " if (np.sqrt(np.dot(dw, dw) / len(dw))) < self.tol:\n", " break" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [], "source": [ "# 随便测试一下\n", "x = [\n", " [1, 2, 3, 0, 1, 3, 4],\n", " [1, 2, 3],\n", " [0, 2, 4, 2],\n", " [4, 3, 2, 1],\n", " [3, 1, 1, 1, 1],\n", " [2, 1, 3, 2, 1, 3, 4]\n", "]\n", "y = x\n", "\n", "crf = CRF(output_status_num=5, input_status_num=5)\n", "crf.fit(x, y)" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "array([1.00000000e-05, 1.00009053e-01, 7.00089277e-02, 7.00091106e-02,\n", " 2.00098984e-02, 4.00097839e-02, 6.00090103e-02, 1.00000000e-05,\n", " 2.00092452e-02, 2.00092452e-02, 1.00099996e-02, 1.00099996e-02,\n", " 3.00099986e-02, 2.00099991e-02, 2.00099991e-02, 2.00092452e-02,\n", " 2.00092452e-02, 2.00092452e-02, 1.00099996e-02, 1.00099996e-02,\n", " 3.00099986e-02, 2.00099991e-02, 2.00099991e-02, 1.00000000e-05,\n", " 2.00092452e-02, 2.00092452e-02, 1.00099996e-02, 1.00099996e-02,\n", " 3.00099986e-02, 2.00099991e-02, 2.00099991e-02, 2.00092452e-02,\n", " 2.00092452e-02, 2.00092452e-02, 1.00099996e-02, 1.00099996e-02,\n", " 3.00099986e-02, 2.00099991e-02, 2.00099991e-02, 2.00092452e-02,\n", " 2.00092452e-02, 2.00092452e-02, 1.00099996e-02, 1.00099996e-02,\n", " 3.00099986e-02, 2.00099991e-02, 2.00099991e-02, 1.00092456e-02,\n", " 1.00092456e-02, 1.00092456e-02, 1.00092456e-02, 1.00092456e-02,\n", " 1.00000000e-05, 1.00098988e-02, 1.00098988e-02, 1.00098988e-02,\n", " 1.00098988e-02, 1.00098988e-02, 1.00098988e-02, 1.00098988e-02,\n", " 1.00098988e-02, 1.00098988e-02, 1.00000000e-05, 1.00098988e-02,\n", " 1.00098988e-02, 1.00098988e-02, 1.00098988e-02, 1.00098988e-02,\n", " 1.00098988e-02, 1.00098988e-02, 1.00098988e-02, 1.00098988e-02,\n", " 1.00098988e-02, 1.00098988e-02, 1.00098988e-02, 1.00098988e-02,\n", " 1.00098988e-02, 1.00000000e-05, 1.00098860e-02, 2.00098855e-02,\n", " 3.00098850e-02, 2.00098668e-02, 1.00098860e-02, 1.00098860e-02,\n", " 2.00098855e-02, 3.00098850e-02, 2.00098668e-02, 1.00000000e-05,\n", " 1.00098860e-02, 2.00098855e-02, 3.00098850e-02, 2.00098668e-02,\n", " 1.00098860e-02, 1.00098860e-02, 2.00098855e-02, 3.00098850e-02,\n", " 2.00098668e-02, 1.00098860e-02, 1.00098860e-02, 2.00098855e-02,\n", " 3.00098850e-02, 2.00098668e-02, 1.00000000e-05, 1.00099808e-02,\n", " 3.00099424e-02, 1.00099808e-02, 1.00099808e-02, 3.00099424e-02,\n", " 1.00000000e-05, 1.00099808e-02, 3.00099424e-02, 1.00099808e-02,\n", " 1.00099808e-02, 3.00099424e-02, 1.00099808e-02, 1.00099808e-02,\n", " 3.00099424e-02, 1.00000000e-05, 1.00099995e-02, 1.00000000e-05,\n", " 1.00099995e-02, 1.00099995e-02])" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "crf.w" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "122" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "len(crf.w)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "可知一共有122个特征函数,接下来还剩最后一个问题,那就是标签预测的问题,见下一小节~~~" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.4" } }, "nbformat": 4, "nbformat_minor": 2 }