Question

A newbie in regex here, I will appreciate any help.

cstring = "[0,90,(+45,45)3,0/]S"
regex = re.compile(r'^(\[)(\S+)(\/?)(\][ST]$)')
match = regex.search(cstring)
for s in match.groups():
    print s

The results are “[” “0,90,(+45,45)3,0/” “]S” but I would like to get “/” in a separate (and optional – the only one from the above) group. I’ve tried to replace \S with any combinations like [0-9(),+-] – (these are the only characters expected in the second group) but no avail.

Answer 1

Regex quantity specifiers + and * are greedy, you can add a ? onto the end of them (+? and *?) to turn them into their non greedy forms.

Greedy means that the operator will try consume everything it can before checking the next token.

so for the case of

\S+\/?

The S will try consume everything it can before the / is checked, and as the / is optional nothing needs to be done for it.

Once we make it into its non greedy form

\S+?\/?

The S will consume as little as possible before trying the / meaning that the / gets ‘first dibs’ on any tokens, and once it fails to grab them these tokens will be tried against the \S+?

I found success using the following:

regex = re.compile(r'^(\[)(\S+?)(\/?)(\][ST]$)')

For more information you can see the python re docs search for greedy.

As a side note if you pass the re.VERBOSE flag into re.compile then it will ignore whitespace within your string meaning you can structure it as

regex = re.compile(r'^ (\[) (\S+?) (\/?) (\][ST]$) ', re.VERBOSE)

which I found quite helpful when learning regex.

Also you have the start of string token outside a group ‘^ ([‘ but the end of string token within a group ‘(][ST]$)’, this shouldn’t make a difference except to readability.