使用 Python 对大数据集进行模糊逻辑

Question

我的团队一直坚持在两个大型数据集上运行模糊逻辑算法。第一个（子集）大约有 180K 行，包含我们需要在第二个（超集）中匹配的人员的姓名、地址和电子邮件。超集包含 250 万条记录。两者具有相同的结构，并且数据已经被清理，即解析地址、标准化名称等。

• 联系人ID int，
• 全名 varchar(150),
• 地址 varchar(100),
• 电子邮件 varchar(100)

目标是将子集中的行中的值与超集中的相应值进行匹配，因此输出将组合子集和超集以及每个字段（标记）的相应相似性百分比。

• 联系方式，
• 查找联系人ID,
• 全名，
• 查找全名，
• 全名_相似度，
• 地址，
• 查找地址，
• 地址_相似度，
• 电子邮件，
• 查找电子邮件,
• 电子邮件_相似度

为了首先简化和测试代码，我们连接了字符串，并且我们知道代码可以在非常小的超集上运行；然而，一旦我们增加记录数量，它就会陷入困境。我们尝试了不同的算法，Levenshtein、FuzzyWuzzy 等，但均无济于事。在我看来，问题在于 Python 是逐行执行的；不过，我不确定。我们甚至尝试使用流式传输在 Hadoop 集群上运行它；然而，它尚未取得任何积极成果。

#!/usr/bin/env python
import sys
from fuzzywuzzy import fuzz
import datetime
import time
import Levenshtein

#init for comparison
with open('normalized_set_record_set.csv') as normalized_records_ALL_file:
# with open('delete_this/xab') as normalized_records_ALL_file:
    normalized_records_ALL_dict = {}
    for line in normalized_records_ALL_file:
        key, value = line.strip('\n').split(':', 1)
        normalized_records_ALL_dict[key] = value
        # normalized_records_ALL_dict[contact_id] = concat_record

def score_it_bag(target_contact_id, target_str, ALL_records_dict):
    '''
    INPUT target_str, ALL_records_dict
    OUTPUT sorted list by highest fuzzy match
    '''
    return sorted([(value_str, contact_id_index_str, fuzz.ratio(target_str, value_str)) 
        for contact_id_index_str, value_str in ALL_records_dict.iteritems()], key=lambda x:x[2])[::-1]

def score_it_closest_match_pandas(target_contact_id, target_str, place_holder_delete):
    '''
    INPUT target_str, ALL_records_dict
    OUTPUT closest match
    '''
    # simply drop this index target_contact_id
    df_score = df_ALL.concat_record.apply(lambda x: fuzz.ratio(target_str, x))

    return df_ALL.concat_record[df_score.idxmax()], df_score.max(), df_score.idxmax()

def score_it_closest_match_L(target_contact_id, target_str, ALL_records_dict_input):
    '''
    INPUT target_str, ALL_records_dict
    OUTPUT closest match tuple (best matching str, score, contact_id of best match str)
    '''
    best_score = 100

    #score it
    for comparison_contactid, comparison_record_str in ALL_records_dict_input.iteritems():
        if target_contact_id != comparison_contactid:
            current_score = Levenshtein.distance(target_str, comparison_record_str)


            if current_score < best_score:
                best_score = current_score 
                best_match_id = comparison_contactid 
                best_match_str = comparison_record_str 

    return (best_match_str, best_score, best_match_id)



def score_it_closest_match_fuzz(target_contact_id, target_str, ALL_records_dict_input):
    '''
    INPUT target_str, ALL_records_dict
    OUTPUT closest match tuple (best matching str, score, contact_id of best match str)
    '''
    best_score = 0

    #score it
    for comparison_contactid, comparison_record_str in ALL_records_dict_input.iteritems():
        if target_contact_id != comparison_contactid:
            current_score = fuzz.ratio(target_str, comparison_record_str)

            if current_score > best_score:
                best_score = current_score 
                best_match_id = comparison_contactid 
                best_match_str = comparison_record_str 

    return (best_match_str, best_score, best_match_id)

def score_it_threshold_match(target_contact_id, target_str, ALL_records_dict_input):
    '''
    INPUT target_str, ALL_records_dict
    OUTPUT closest match tuple (best matching str, score, contact_id of best match str)
    '''
    score_threshold = 95

    #score it
    for comparison_contactid, comparison_record_str in ALL_records_dict_input.iteritems():
        if target_contact_id != comparison_contactid:
            current_score = fuzz.ratio(target_str, comparison_record_str)

            if current_score > score_threshold: 
                return (comparison_record_str, current_score, comparison_contactid)

    return (None, None, None)


def score_it_closest_match_threshold_bag(target_contact_id, target_str, ALL_records_dict):
    '''
    INPUT target_str, ALL_records_dict
    OUTPUT closest match
    '''
    threshold_score = 80
    top_matches_list = []
    #score it
    #iterate through dictionary
    for comparison_contactid, comparison_record_str in ALL_records_dict.iteritems():
        if target_contact_id != comparison_contactid:
            current_score = fuzz.ratio(target_str, comparison_record_str)

            if current_score > threshold_score:
                top_matches_list.append((comparison_record_str, current_score, comparison_contactid))


    if len(top_matches_list) > 0:  return top_matches_list

def score_it_closest_match_threshold_bag_print(target_contact_id, target_str, ALL_records_dict):
    '''
    INPUT target_str, ALL_records_dict
    OUTPUT closest match
    '''
    threshold_score = 80


    #iterate through dictionary
    for comparison_contactid, comparison_record_str in ALL_records_dict.iteritems():
        if target_contact_id != comparison_contactid:

            #score it
            current_score = fuzz.ratio(target_str, comparison_record_str)
            if current_score > threshold_score:
                print target_contact_id + ':' + str((target_str,comparison_record_str, current_score, comparison_contactid))


    pass


#stream in all contacts ie large set
for line in sys.stdin:
    # ERROR DIAG TOOL
    ts = time.time()
    st = datetime.datetime.fromtimestamp(ts).strftime('%Y-%m-%d %H:%M:%S')
    print >> sys.stderr, line, st

    contact_id, target_str = line.strip().split(':', 1)

    score_it_closest_match_threshold_bag_print(contact_id, target_str, normalized_records_ALL_dict)
    # output = (target_str, score_it_closest_match_fuzz(contact_id, target_str, normalized_records_ALL_dict))
    # output = (target_str, score_it_closest_match_threshold_bag(contact_id, target_str, normalized_records_ALL_dict))
    # print contact_id + ':' + str(output)

Answer 1

您的方法要求您进行 180,000 * 2,500,000 = 450,000,000,000 次比较。

4500亿是很多了。

为了减少比较次数，您可以首先对具有某些共同特征（例如地址字段的前五个字符或共同标记）的记录进行分组。然后，仅比较具有相同特征的记录。这个想法称为“阻塞”，通常会减少您必须与可管理的事物进行的总比较次数。

您试图解决的一般问题称为“记录链接”。由于您使用的是 python，您可能需要查看提供全面方法的重复数据删除库（我是该库的作者）。