Match zero or more occurrence of string Regex

Question

I am trying to match some pattern, reference this online tool for easy test

pattern = r"(^/\w+)\s*?(\w+)"
string_1 = "/path_one path_two"
string_2 = "/path_one_only"

While string_1 matches as expected by returning both parts, for string_2 it cuts the last character of part one. But I would like the pattern to always return both parts, rather return None/empty string if path two does not exist

Answer 1

Use this pattern:

pattern = r"(^/\w+)\s*(\w*)"

Now the second string will match entirely in the first capture group.

The reason the final character was being clipped in the second string can be seen here:

(^/\w+)  - matches 'path_one_onl'
\s*?     - matches nothing (there are no spaces)
(\w+)    - matches 'y'

In other words, the second capture group was imposing that at least one character be matched there.

Answer 2

Change your pattern string to

1. either

    pattern = r"(^/\\w+)\\s*(\\w+)?" 

2. or

    pattern = r"(^/\\w+)\\s*(\\w*)" 

You do not need \\s*? , just \\s* is fine.

Answer 3

您还可以通过在其周围添加一个非捕获组来使整个第二部分（包括空格）成为可选的：

r"(^/\w+)(?:\s+(\w+))?"