Python正则表达式。 匹配一次或两次模式

Question

I have a bunch of lines in a file with either one or two occurences of same pattern (id=):

Linetype1 : ...id=1234...id=4321...value=5678... # "..." means whatever
Linetype2 : ...id=7890...value=8765

I thought I could write such a regex to grep all my ids and associated values:

>>> l="...id=1234...id=4321...value=5678...\n...id=7890...value=8765\n"
>>> ret = re.findall('(id=[0-9]+).*?(id=[0-9]+)*.*?(value=[0-9]+)',l)
[('id=1234', '', 'value=5678'), ('id=7890', '', 'value=8765')]

I can't get the second "id=4321" part. This looks very strange to me since I use the non-greedy .*? between first id=[0-9]+ and second.

Answer 1

The middle of your regex has

(id=[0-9]+)*

The empty string matches this, since it is under the Kleene star * . So the regex engine proceeds through the string as follows:

• find the first id=[0-9]+ group
• expand .*? to the empty string, since it matches
• expand (id=[0-9]+) * to the empty string, since it matches
• expand .*? to the rest of the string

If you replace the middle group's quantifier with + , or just remove it entirely, then it works.