For example String is '10S#9D7T*'.
My desire result is 3 arrays. [('10S#'), ('9D'), ('7T*')]
There are 3 condition.
'#' or '*'. But it is not essential.That is my code.
rex = re.compile(r'\d?\d\w?[\*\#]')
str = '10S#9D7T'
print(rex.findall(str))
Acutal Result -> ['10S#']
There is only one array.
please fixed my regex pattern.
You could change \d?\d to \d{1,2} and as the following character is always there it should not be optional so you could omit the question mark. The * and # do not have to be escaped in the character class and add a quesion mark to that to make it optional.
You might use:
\d{1,2}\w[*#]?
That will match:
\d{1,2} Match 1-2 digits\w Match a word character[*#]? Optional character class to match one of * or #
import re str = "10S#9D7T*" rex = re.compile(r'\d{1,2}[A-Z][*#]?') print(rex.findall(str))
Result
['10S#', '9D', '7T*']
You can get three results by making the [*#] class optional. (Also, note that those characters don't need to be escaped.)
str = '10S#9D7T*'
rex = re.compile(r'\d?\d\w?[*#]?')
print(rex.findall(str))
Result: ['10S#', '9D', '7T*']
As for your second rule, "And follow one character is located always," you probably want to make the \w non-optional by removing the ? directly after it.
r'\d?\d\w[*#]?'
*, and[*#]does not have a?after so you enforce one of these after the match. Just add a*in your string and a?after the[*#]. Note you do not need to escape the*in the brackets.