Python3 正则表达式 findall

Question

Here is my issue. Given below list:

a = ['COP' , '\t\t\t', 'Basis', 'Notl', 'dv01', '6m', '9m', '1y',
     '18m', '2y', '3y', "15.6", 'mm', '4.6', '4y', '5y', '10', 'mm',
     '4.6', '6y', '7y', '8y', '9y', '10y', '20y', 'TOTAL', '\t\t9.2' ]

I'm trying to get some outputs like this one. The most important note is the rows After the first number ended on "y" or "m" will come a number only if it is there in the list Example : ('3y', '15.6', '')

SAMPLE OUTPUT ( forget about the structure that is a tuple, jsut want teh values)

('6m', '', '')
('9m', '', '')
('1y', '', '')
('18m', '', '')
('2y', '', '')
('3y', '15.6', '')
('4y', '', '')
('5y', '10', '')
('6y', '', '')
('7y', '', '')
('8y', '', '')
('9y', '', '')
('10y', '', '')
('20y', '', '')

I used the following regex that should have returned :

1. all numbers followed by "y" or "m" => (\\b\\d+[ym]\\b)
2. and then any number (integer or not) if it appears (meaning zero or more times)=> (\\b[0-9]+. [0-9] \\b)

Here is what I did, using Python3 regex and re.findall(), but still got no result

rule2 = re.compile(r"(\b\d+[ym]\b)(\b[0-9]+.*[0-9]*\b)+")
a_str = " ".join(a)
OUT2 = re.findall(rule2, a_str)
print(OUT2)
# OUT2 >>[]

Why I'm not getting the correct result?

Answer 1

You cannot use word boundary twice. Since data is separated by non-letter/digits use \\W+ instead.

Then, escape the dot, and make it optional, or you're not going to match 10 . Don't use .* as it will match too much (regex greediness)

that yields more or less what you're looking for (note that matching strict numbers, integers or floats, is trickier than that, so this isn't perfect):

rule2 = re.compile(r"\b(\d+[ym])\W+([0-9]+\.?[0-9]*)\b")
a_str = " ".join(a)
OUT2 = re.findall(rule2, a_str)
print(OUT2)

[('3y', '15.6'), ('5y', '10')]