使用Regex捕获组解析地址

Question

我正在尝试将这些地址解析成组，并且我有一个正则表达式：

(^.*?(?:Lane|Street|Boulevard|Crescent|Place|Road|Highway|Avenue|Drive|Circuit|Parade|Telopea|Nicklin Way|Terrace|Square|Court|Close|Endeavour Way|Esplanade|East|The Centreway|Mall|Quay|Gateway|Low Way|Point|Rd|Morinda|Way|Ave|St|South Steyne|Broadway|HQ|Expressway|Street|Castlereagh|Meadow Way|Track|Kulkyne Way|Narabang Way|Bank)),(?:)? ?(.*?),? ?([A-Z]{2,3}),? ?(\d{,4})$

它捕获并分组这些地址：

139 McKinnon Road, PINELANDS, NT, 829
108 East Point Road, Fannie Bay, NT, 820
3-11 Hamilton Street, Townsville City, QLD, 4810
40 17 Geranium Street, THE GARDENS, NT, 820
Lot 9 Island Point Road, ST GEORGES BASIN, NSW, 2540
316 Sturt Street and 511 Flinders Street, Townsville City, QLD, 4810

但没有用以下格式捕获地址：

1, 3, 5 Demeter Street & 12 Hermes Avenue ROUSE HILL NSW 2155
31 Stephen Street SOUTH TOOWOOMBA QLD 4350

我想把这些地址分成几个不同的类别，例如：

street_address = 1, 3, 5 Demeter Street & 12 Hermes Avenue
subrub = ROUSE HILL
state = QLD
postcode = 4350

如何使用上述表达式捕获两个地址？这是我的Regex码

Answer 1

您可以使用特定的regex分别匹配四个组中的每个组，使用以下方法：

• Group 1，包含名为<street_address>的地址

.*(?:Lane|Street|Boulevard|Crescent|Place|Road|Highway|Avenue|Drive|Circuit|Parade|Telopea|Nicklin Way|Terrace|Square|Court|Close|Endeavour Way|Esplanade|East|The Centreway|Mall|Quay|Gateway|Low Way|Point|Rd|Morinda|Way|Ave|St|South Steyne|Broadway|HQ|Expressway|Street|Castlereagh|Meadow Way|Track|Kulkyne Way|Narabang Way|Bank)

• 群2，包含子摩擦，称为<subrub>

[A-Za-z ]+

• Group 3，包含状态，称为<state>

[A-Z]+

• Group 4，包含邮政编码，名为：

\d+

您的最终regex就是使用可选逗号和强制空格,?连接这些regex。

(?P<street_address>.*(?:Lane|Street|Boulevard|Crescent|Place|Road|Highway|Avenue|Drive|Circuit|Parade|Telopea|Nicklin Way|Terrace|Square|Court|Close|Endeavour Way|Esplanade|East|The Centreway|Mall|Quay|Gateway|Low Way|Point|Rd|Morinda|Way|Ave|St|South Steyne|Broadway|HQ|Expressway|Street|Castlereagh|Meadow Way|Track|Kulkyne Way|Narabang Way|Bank)),? (?P<subrub>[A-Za-z ]+),? (?P<state>[A-Z]+),? (?P<postcode>\d+)

检查演示这里。

注意:在Python代码中，您将能够按相应的名称提取每个组。