正则表达式捕获字符串中的确切数字

Question

I have a column that looks like this:

11/33/4500030050
4100000300/4503134501
4100030300+4503114501
11

The regular expression should capture the following:

4500030050
4100000300/4503134501
4100030300+4503114501
''

Here's my current regular expression:

 col.str.findall(r'[/+ #_;.-]?(?<![0-9])[0-9]{10}(?![0-9])').str.join('').str.lstrip('/+ #_;.-')

This however captures all numbers that have 10 digits. How can I modify so that it can only capture numbers that start with 41 and 45?

Answer 1

You can use

[+ #_;.-]?(?<![0-9])4[15][0-9]{8}(?![0-9])

See the regex demo .

Details :

• [+ #_;.-]? - an optional + , space, # , _ , ; , . or -
• (?<![0-9]) - left digit boundary
• 4 - a 4 digit char
• [15] - 1 or 5
• [0-9]{8} - eight digits
• (?![0-9]) - right digit boundary.