如何在 Python 中使用 str.replace() 函数将地址替换为只有数字和一些字母？

Question

我正在尝试匹配参考索引 (coClean) 上的左侧和紧缩地址（来自单独的表），这是我在 #Python #JupyterNotebook 中使用以下公式创建的

import pandas as pd
df1=pd.read_csv("/content/Addmatchdf1.csv")
df2=pd.read_csv("/content/Addmatchdf2.csv")

import re
def cleanAddress(series):
return series.str.lower().str.replace(r"[a-z\s\,]","")
df1["coClean"]=cleanAddress(df1["Address"])

df1["coClean"]=cleanAddress(df1["Address"]) 
df = pd.merge(df1, df2, 
                      on =['coClean'], 
                      how ='inner') 

这会产生一个 coClean 作为参考索引。

Address_x	coClean	Address_y
7 Pindara Bvd LANGWARRIN VIC 3910	73910	7 Pindara Blv, Langwarrin, VIC 3910
2a Manor St BACCHUS MARSH VIC 3340	23340	2a Manor Street, BACCHUS MARSH, VIC 3340
38 Sommersby Rd POINT COOK VIC 3030	383030	38 Sommersby Road, Point Cook, VIC 3030
17 Moira Avenue, Carnegie, Vic 3163	173163	17 Moira Avenue, Carnegie, Vic 3163
17 Moira Avenue, Carnegie, Vic 3163	173163	17 Newman Avenue, Carnegie, VIC 3163
17 Moira Avenue, Carnegie, Vic 3163	173163	17 Maroona Rd, Carnegie VIC 3163

显然，我面临的问题是，相同邮政编码下的某些地址具有相同的门牌号。但是由于参考索引相同，连接变得困难。

我怎样才能修改这个函数，使参考索引只包含

a. the house numbers
b. first four letters
c. postcode

所以 '23340'(2a manor street bacchus marsh vic 3340) 的新参考变成 '2aman3340'？因此返回一个列表，如下所示：

coClean
7pind3910
2aman3340
38somm3030
17moir3163
17newm3163
17maroo3163

我试图修改函数以包含所有字母和数字

def cleanAddress(series):
return series.str.lower().str.replace(r"[^a-z\d]","")

但是包含所有字母并不能解决问题，因为不同的表包含街道作为 st。和道路作为路。因此，更好的策略是使用带有一些首字母的门牌号和邮政编码。

感谢您的友好建议。

更新： 我换了

def cleanAddress(series):
return series.str.lower().str.replace(r"[a-z\s\,]","")
df1["coClean"]=cleanAddress(df1["Address"])

与

def cleanAddress(series):
    coclen=""
    number_of_letters=0
    if series:
        for i in range(len(series)):
            if series[i].isnumeric():
                coclen+=series[i]
            elif series[i].isalpha():
                number_of_letters+=1
                coclen+=series[i]
                if number_of_letters==4:
                    break
        for i in range(i,len(series)):
            if series[i].isnumeric():
                coclen+=series[i]
    return coclen

这会在我执行时返回错误

cleanAddress(df1["Address"])

The full error is as follows:
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-17-b653a19f5638> in <module>()
----> 1 df1["coClean"]=cleanAddress(df1["Address"])

1 frames
/usr/local/lib/python3.7/dist-packages/pandas/core/generic.py in __nonzero__(self)
   1328     def __nonzero__(self):
   1329         raise ValueError(
-> 1330             f"The truth value of a {type(self).__name__} is ambiguous. "
   1331             "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
   1332         )

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Answer 1

import pandas as pd
df1 = pd.DataFrame({"Address_x":["7 Pindara Bvd LANGWARRIN VIC 3910","2a Manor St BACCHUS MARSH VIC 3340","38 Sommersby Rd POINT COOK VIC 3030","17 Moira Avenue, Carnegie, Vic 3163"],"Address_y":["7 Pindara Blv, Langwarrin, VIC 3910","2a Manor Street, BACCHUS MARSH, VIC 3340","38 Sommersby Road, Point Cook, VIC 3030","17 Moira Avenue, Carnegie, Vic 3163"]})
def cleanAddress(series):
    cocleans=[]
    for address in series:
        number_of_letters=0
        coclean=""
        for i in range(len(address)):
            if address[i].isnumeric():
                coclean+=address[i]
            elif address[i].isalpha():
                number_of_letters+=1
                coclean+=address[i]
                if number_of_letters==4:
                    break
        for i in range(i,len(address)):
            if address[i].isnumeric():
                coclean+=address[i]
        cocleans.append(coclean.lower())
    return cocleans
df1["coClean"]=cleanAddress(df1["Address_x"])