正则表达式：赛后采取一切措施

Question

I have many strings in a dataframe, an example being:

adrianos ristorante 2930 beverly glen circle los angeles 310475 9807 italian

I want to take all words/characters after the phone number. I have the regex for getting the phone number and space after( ([0-9]{6}\\s[0-9]{4})\\s ). What I want is all characters after this. In this case it's italian but it could be something like asian fusion or indian and thai .

Answer 1

In Pandas, you may use

[0-9]{6}\s[0-9]{4}\s+(.+)
                    ^^^^^

The code will look like

df['col'].str.extract('[0-9]{6}\s[0-9]{4}\s+(.+)')

Note you should wrap the part you need to extract with a capturing group, (...) . See Pandas reference :

pat : string

Regular expression pattern with capturing groups

Answer 2

You can do (in Notepad++):

Find What: (. )([0-9]{6}\\s[0-9]{4})\\s(. ) Replace with: $3

$3 gives the 3rd argument which is everything after your matching phone number.

Answer 3

Using this regular expression: /([0-9]{6}\\s[0-9]{4})\\s(.*$)/m

You will have anything after the phone number, until the end of the line (you can omit the m flag if you want to get everything until the end of the string), in the second capture group.

Answer 4

You can use positive lookbehind:

(?<=[0-9]{6}\s[0-9]{4}\s).*

Regex101 .