识别具有相似地址的ID

时间:2019-06-19 04:17:05

标签: python python-3.x pandas

我在一个csv文件中有一个数据,该文件基本上具有一些ID,它们的对应地址以及1个地址与其他地址的匹配相似率。我想确定地址相似的ID及其匹配百分比

我已经完成了文本匹配,找到了将1个地址与其他每个地址进行比较的地址字符串之间的相似度百分比。

import pandas as pd
from fuzzywuzzy import process, fuzz

pd.set_option('display.width', 1000)
pd.set_option('display.max_columns', 10)

data = pd.read_csv(r"address_details.csv", skiprows=0)
id = data['COD_CUST_ID'].values.tolist()
address = data['ADDRESS'].values.tolist()

dict_list=[]

for i in range(0,len(id)):
    for add in range(0,len(address)):
        score=process.extractBests(address[add], address, limit=len(address), score_cutoff=40)
        #print(type(score))

        for sc in score:
            #print(sc)
            for scr in sc:
                print(scr)

            dict_={}
            dict_.update({"Cust_Id": id[i]})
            dict_.update({"Match Ratio": sc})
            dict_.update({"Search String": address[add]})
        #dict_.update({"Address List": address})

            dict_list.append(dict_)

df=pd.DataFrame(dict_list)


matches = df['Match Ratio'].tolist()
matches = [x[0][0] for x in matches]

found  = []
for s in df['Search String']:
    data_list=[]

    if s in matches:
        index=[i for i, x in enumerate(matches) if x == s]
        Cust_Id = list([df['Cust_Id'][i]] for i in index)
        data_list.append(s)
        data_list.append(Cust_Id)
        found.append(data_list)
print(found)

sd=df.to_csv("match_score.csv",sep=',',index=None)

假设我将此数据帧作为代码输出

Cust_Id Match Ratio Search String
1   [('ABC', 100)]  ABC
2   [('DEF', 100)]  DEF
3   [('DEF', 100)]  XYZ
4   [('ABC', 100)]  PQR
5   [('PQR', 100)]  TUV
6   [('DEF', 100)]  LMN

我想在“匹配比率”列下获取具有类似数据的IDS列表

1 个答案:

答案 0 :(得分:1)

我编写了一个代码,该代码给出了一个包含“搜索字符串”的列表,它对应的是匹配的“ Cust_Id”。

代码是

 import pandas as pd

def duplicates(lst, item):
   return [i for i, x in enumerate(lst) if x == item]

# Creating Data frame
data = {'Cust_Id' : ['1 ','2' , '3','4','5','6'],
        'Match Ratio'  : [[('ABC', 100)],[('DEF', 100)],[('DEF', 100)], [('ABC', 100)],[('PQR', 100)],[('DEF', 100)]],
        'Search' : ['ABC','DEF','XYZ','PQR','TUV','LMN']
        }
df = pd.DataFrame(data)

print(df)
# Creating a list of 1'st value of tuple Match Ratio
matches = df['Match Ratio'].tolist()
matches = [x[0][0] for x in matches]

found  = []
for s in df['Search']:
    data_list = []
    if s in matches:
        index = duplicates(matches,s)
        Cust_Id = list([df['Cust_Id'][i]] for i in index)
        data_list.append(s)
        data_list.append(Cust_Id)
        found.append(data_list)
print(found)

数据帧输出

  Cust_Id   Match Ratio Search
0      1   [(ABC, 100)]    ABC
1       2  [(DEF, 100)]    DEF
2       3  [(DEF, 100)]    XYZ
3       4  [(ABC, 100)]    PQR
4       5  [(PQR, 100)]    TUV
5       6  [(DEF, 100)]    LMN

发现列表输出

[['ABC', [['1 '], ['4']]], ['DEF', [['2'], ['3'], ['6']]], ['PQR', [['5']]]]

希望您能找到想要的东西:)

相关问题