仅在第一个字母数字字符之前放置特殊字符

时间:2019-06-04 03:13:22

标签: python regex

我有一个正则表达式,可以从数据中提取地址。 正则表达式查找“​​地址|地址|等等。'并选择地址,直到遇到4位数的邮政编码为止。 有时,在地址的开头添加了_,-等特殊字符。我需要代码来忽略第一个字母数字字符之前的特殊字符。

到目前为止,我运行了一个循环以排除所有特殊字符,如下面的代码所示。我需要代码仅删除出现在第一个字母数字字符前面的特殊字符。

输入(来自图像上的OCR):
服务地址—_Unit8-10 LEWIS St,BERRI,SA 5343

possible_addresses = list(re.findall('Address(.* \d{4}|)|address(.*\d{4})|address(.*)', data)) 
address = str(possible_addresses[0])
for k in address.split("\n"): 
    if k != ['(A-Za-Z0-9)']:
        address_2 = re.sub(r"[^a-zA-Z0-9]+", ' ', k)

现在获得:
地址:—_Unit 8-10 LEWIS ST,BERRI SA 5343
地址_2:8单元10 LEWIS ST BERRI SA 5343

2 个答案:

答案 0 :(得分:1)

[\W_]*捕获特殊字符。

import re
data='Service address —_Unit8-10 LEWIS St, BERRI,SA 5343'
possible_addresses = re.search('address[\W_]*(.*?\d{4})', data,re.I)
address = possible_addresses[1]
print('Address : ' address)
  

地址:SA 5343,BERRI,LEWIS St8-10单位

答案 1 :(得分:0)

我猜想我们希望在此处设计的表达式应该将address到四位数的zip刷掉,不包括诸如_之类的某些已定义字符。然后让我们从带有i标志的简单表达式开始,例如:

(address:)[^_]*|[^_]*\d{4}

Demo

我们不希望有的任何字符都将进入[^_]。例如,如果我们排除!,则表达式将变为:

(address:)[^_!]*|[^_!]*\d{4}

测试

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"(address:)[^_]*|[^_]*\d{4}"

test_str = ("Address: Unit 8 - 10 LEWIS ST, BERRI SA 5343 \n"
    "Address: Got now: __Unit 8 - 10 LEWIS ST, BERRI SA 5343\n"
    "aDDress: Unit 8 10 LEWIS ST BERRI SA 5343\n"
    "address: any special chars here !@#$%^&*( 5343\n"
    "address: any special chars here !@#$%^&*( 5343")

matches = re.finditer(regex, test_str, re.MULTILINE | re.IGNORECASE)

for matchNum, match in enumerate(matches, start=1):

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.