搭建一个简单的问答系统（v2.0）

客官°小女子只卖身不卖艺 2022-03-21 02:52 164阅读 0赞

之前刚接触机器学习的时候，写过一篇[《基于sklearn库，搭建一个简单的问答系统》][sklearn]。此篇文章是在上篇的逻辑上，对一些函数进行了优化，并对检索方式进行了一些优化，再各个环节上时间复杂度都提高了很多。  
下面讲解一下具体的代码：

##### 第一部分： 读取文件，并把内容分别写到两个list里（一个list对应问题集，另一个list对应答案集） #####

import json
    def read_corpus():
        """ 读取给定的语料库，并把问题列表和答案列表分别写入到 qlist, alist 里面。 在此过程中，不用对字符换做任何的处理（这部分需要在 Part 2.3里处理） qlist = ["问题1"， “问题2”， “问题3” ....] alist = ["答案1", "答案2", "答案3" ....] 务必要让每一个问题和答案对应起来（下标位置一致） """
        qlist = []
        alist = []
        with open('data/train-v2.0.json') as f:
            data_array = json.load(f)['data']
            for data in data_array:
                paragraphs = data['paragraphs']
                for paragraph in paragraphs:
                    qas = paragraph['qas']
                    for qa in qas:
                        if 'plausible_answers' in qa:
                            qlist.append(qa['question'])
                            alist.append(qa['plausible_answers'][0]['text'])
                        else:
                            qlist.append(qa['question'])
                            alist.append(qa['answers'][0]['text'])
                            
        assert len(qlist) == len(alist)  # 确保长度一样
        return qlist, alist

##### 理解数据（可视化分析/统计信息） #####

对数据的理解是任何AI工作的第一步，需要充分对手上的数据有个更直观的理解。

# TODO: 统计一下在qlist 总共出现了多少个单词？ 总共出现了多少个不同的单词？
    # 这里需要做简单的分词，对于英文我们根据空格来分词即可，其他过滤暂不考虑（只需分词）
    from collections import Counter
    qlist , alist = read_corpus()
    # 分词
    def cut(input_list):
        list_new = []
        for q in input_list:
            list_new.append(q.replace('?','').split(' '))
        return list_new
    qlist_new = [q for l in cut(qlist) for q in l]
    dif_word_total = len(qlist_new)
    word_dict = Counter(qlist_new)
    word_total = len(dict(word_dict))
    print ("一共出现了 %d 个单词"%dif_word_total)
    print ("共有 %d 个不同的单词"%word_total)

我们使用matplotlib来绘制每个词出现的频率：

# TODO: 统计一下qlist中每个单词出现的频率，并把这些频率排一下序，然后画成plot. 比如总共出现了总共7个不同单词，而且每个单词出现的频率为 4, 5,10,2, 1, 1,1
    # 把频率排序之后就可以得到(从大到小) 10, 5, 4, 2, 1, 1, 1. 然后把这7个数plot即可（从大到小）
    # 需要使用matplotlib里的plot函数。y轴是词频
    import matplotlib.pyplot as plt
    y = []
    for i in word_dict:
        y.append(word_dict[i])
    
    plt.plot(sorted(y,reverse=True)[50:])
    plt.show()

词频分布如下：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2x0MzI2MDMwNDM0_size_16_color_FFFFFF_t_70]

##### 文本预处理 #####

此部分需要尝试做文本的处理。在这里我们面对的是英文文本，所以任何对英文适合的技术都可以考虑进来。

# TODO: 对于qlist, alist做文本预处理操作。 可以考虑以下几种操作：
    # 1. 停用词过滤 （去网上搜一下 "english stop words list"，会出现很多包含停用词库的网页） 
    # 2. 转换成lower_case： 这是一个基本的操作 
    # 3. 去掉一些无用的符号： 比如连续的感叹号！！！， 或者一些奇怪的单词。
    # 4. 去掉出现频率很低的词：比如出现次数少于10,20....
    # 5. 对于数字的处理： 分词完只有有些单词可能就是数字比如44，415，把所有这些数字都看成是一个单词，这个新的单词我们可以定义为 "#number"
    # 6. stemming（利用porter stemming): 因为是英文，所以stemming也是可以做的工作
    # 7. 其他（如果有的话）
    # # 请注意，不一定要按照上面的顺序来处理，具体处理的顺序思考一下，然后选择一个合理的顺序
    import string
    from nltk.stem.porter import PorterStemmer
    
    # 低频词库
    low_frequency_words = []
    for (k,v) in  word_dict.items():
        if v < 2:
            low_frequency_words.append(k)
    p = PorterStemmer()
    
    def text_preprocessing(input_list):
        stop_words = ["a","able","about","across","after","all","almost","also","am","among","an","and","any","are","as","at","be","because","been","but","by","can","cannot","could","dear","did","do","does","either","else","ever","every","for","from","get","got","had","has","have","he","her","hers","him","his","how","however","i","if","in","into","is","it","its","just","least","let","like","likely","may","me","might","most","must","my","neither","no","nor","not","of","off","often","on","only","or","other","our","own","rather","said","say","says","she","should","since","so","some","than","that","the","their","them","then","there","these","they","this","tis","to","too","twas","us","wants","was","we","were","what","when","where","which","while","who","whom","why","will","with","would","yet","you","your","ain't","aren't","can't","could've","couldn't","didn't","doesn't","don't","hasn't","he'd","he'll","he's","how'd","how'll","how's","i'd","i'll","i'm","i've","isn't","it's","might've","mightn't","must've","mustn't","shan't","she'd","she'll","she's","should've","shouldn't","that'll","that's","there's","they'd","they'll","they're","they've","wasn't","we'd","we'll","we're","weren't","what'd","what's","when'd","when'll","when's","where'd","where'll","where's","who'd","who'll","who's","why'd","why'll","why's","won't","would've","wouldn't","you'd","you'll","you're","you've"]
        # 分词
        input_list = cut(input_list)    
        new_list = [] #保存处理完的qlist\alist
        for l in input_list:
            l_list = '' # 保存句子
            for word in l:
                # 转换小写/stemming
                word = p.stem(word)
                
                # 去除所有标点符号
                word = ''.join(c for c in word if c not in string.punctuation)
                
                # 处理数字
                if word.isdigit():
                    word = word.replace(word,'#number')
                
                if word not in stop_words and word not in low_frequency_words:
                    l_list += word + ' '
            new_list.append(l_list)
        return new_list
           
    qlist = text_preprocessing(qlist)   # 更新后的

##### 文本表示 #####

当我们做完关键的预处理过程之后，就需要把每一个文本转换成向量。

# TODO: 把qlist中的每一个问题字符串转换成tf-idf向量, 转换之后的结果存储在X矩阵里。 X的大小是： N* D的矩阵。 这里N是问题的个数（样本个数），
    # D是字典库的大小。 
    from sklearn.feature_extraction.text import TfidfVectorizer
    
    vectorizer = TfidfVectorizer() # 定一个tf-idf的vectorizer
    X = vectorizer.fit_transform(qlist)   # 结果存放在X矩阵
    
    X_list = [i for k in X.toarray() for i in k]
    count_total = len(X_list)
    count_zero = X_list.count(0)
    sparsity = (count_total-count_zero)/count_total
    print (sparsity)  # 打印出稀疏度

##### 对于用户的输入问题，找到相似度最高的TOP5问题，并把5个潜在的答案做返回 #####

from sklearn.metrics.pairwise import cosine_similarity
    alist = np.array(alist)
    def top5results(input_q):
        """ 给定用户输入的问题 input_q, 返回最有可能的TOP 5问题。这里面需要做到以下几点： 1. 对于用户的输入 input_q 首先做一系列的预处理，然后再转换成tf-idf向量（利用上面的vectorizer) 2. 计算跟每个库里的问题之间的相似度 3. 找出相似度最高的top5问题的答案 """
        input_q = text_preprocessing([input_q])
        input_vec = vectorizer.transform(input_q)
        res = cosine_similarity(input_vec,X)[0]
        top_idxs = np.argsort(res)[-5:].tolist()  # top_idxs存放相似度最高的（存在qlist里的）问题的下标
        return alist[top_idxs]  # 返回相似度最高的问题对应的答案，作为TOP5答案

##### 利用倒排表的优化 #####

上面的算法，一个最大的缺点是每一个用户问题都需要跟库里的所有的问题都计算相似度。假设我们库里的问题非常多，这将是效率非常低的方法。 这里面一个方案是通过倒排表的方式，先从库里面找到跟当前的输入类似的问题描述。然后针对于这些candidates问题再做余弦相似度的计算。这样会节省大量的时间。

inverted_idx = { }  # 定一个一个简单的倒排表
    
    # 将关键词添加到倒排表中
    for key,value in word_dict.items():
        if value >100 and value <1000:
            inverted_idx[key]=[]
            
    # 在倒排表中添加关键词存在于qlist中的索引
    keywords_index = 0
    for q in cut(qlist):
        for word in q:
            if word in inverted_idx.keys():
                inverted_idx[word].append(keywords_index)
        keywords_index += 1

# TODO: 基于倒排表的优化。在这里，我们可以定义一个类似于hash_map, 比如 inverted_index = {}， 然后存放包含每一个关键词的文档出现在了什么位置，
    # 也就是，通过关键词的搜索首先来判断包含这些关键词的文档（比如出现至少一个），然后对于candidates问题做相似度比较。
    # 
    
    def top5results_invidx(input_q):
        # 从倒排表中取出相关联的索引
        index_list = []
        for c in cut([input_q]):
            for i in c:
                if i in inverted_idx.keys():
                    index_list += inverted_idx[i]
        index_list = list(set(index_list))
        
        input_q = text_preprocessing([input_q])
        input_vec = vectorizer.transform(input_q)
        res = cosine_similarity(input_vec,X[index_list])[0]
        top_idxs = np.argsort(res)[-5:].tolist()  # top_idxs存放相似度最高的（存在qlist里的）问题的下标
     
        return alist[top_idxs]

##### 基于词向量的文本表示 #####

# 建立glove矩阵
    with open('data/glove.6B.100d.txt', 'r') as file1:
        emb = []
        vocab = []
        for line in file1.readlines():
            row = line.strip().replace("?",'').split(' ')
            vocab.append(row[0])
            emb.append(row[1:])
    emb =np.asarray(emb) # 读取每一个单词的嵌入。这个是 D*H的矩阵，这里的D是词典库的大小， H是词向量的大小。 这里面我们给定的每个单词的词向量，那句子向量怎么表达？
          # 其中，最简单的方式 句子向量 = 词向量的平均（出现在问句里的）， 如果给定的词没有出现在词典库里，则忽略掉这个词。

# 获取句子的vec值
    def get_words_vec(words):
        glove_index_list = []
        # 取句子对应单词的索引
        for word in words.split(' '):
            if word in vocab:
                index = vocab.index(word)
                glove_index_list.append(index)
        return emb[glove_index_list].astype(float).sum()/len(words.split(' '))

将句子转换为向量

qlist_matrix = []
    for q in qlist:
        q_vec = get_words_vec(q)
        qlist_matrix.append(q_vec)
    qlist_matrix = np.asarray(qlist_matrix)

给定用户输入的问题 input\_q, 返回最有可能的TOP 5问题。这里面需要做到以下几点：  
1. 利用倒排表来筛选 candidate  
2. 对于用户的输入 input\_q，转换成句子向量  
3. 计算跟每个库里的问题之间的相似度  
4. 找出相似度最高的top5问题的答案

def top5results_emb(input_q):    
        abs_v = abs(qlist_matrix[0])
        # 存储所有返回结果的索引
        res = []
        # 从倒排表中取出相关联的索引
        index_list = []
        for c in cut([input_q]):
            for i in c:
                if i in inverted_idx.keys():
                    index_list += inverted_idx[i]
        index_list = list(set(index_list))
        # input_q的vec值
        input_q_vec = get_words_vec(input_q)
        # 遍历倒排表内所有值，将list中所有vec值与input_q的vec值做对比，绝对值差较小的数的索引存入res中
        for value in qlist_matrix[index_list]:
            if abs(input_q_vec-value) < abs_v:
                abs_v = abs(input_q_vec-value)
                res.append(qlist_matrix[index_list].tolist().index(value))
        
        top_indx = np.argsort(res)[-5:].tolist()
        
        return alist[top_indx]

相关代码及测试数据我已上传至github，[点击这里][Link 1]可直接查看。有问题的同学请在博客下方留言或提Issue！

[sklearn]: https://blog.csdn.net/lt326030434/article/details/82909589
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2x0MzI2MDMwNDM0_size_16_color_FFFFFF_t_70]: /images/20220321/2881824a05f141019667846f8b81c3eb.png
[Link 1]: https://github.com/Maginaaa/QuestionAnsweringSystem-V2.0