escapes in regular expressions

[James Thiele]

I was helping a guy at work with regular expressions and found
something I didn't expect:

re.match('\d', '7').group() '7' re.match('\\d', '7').group() '7'

It's not clear to me why these are the same. Could someone please
explain?

[Dennis Benzinger]

James Thiele schrieb:

I was helping a guy at work with regular expressions and found
something I didn't expect:

re.match('\d', '7').group()
'7'
'\d' is not recognized as a escape sequence by Python and therefore it
is left unchanged <http://docs.python.org/ref/strings.html> in the
string which is passed to re.match.
re.match('\\d', '7').group()

'7'
[...]

This is the correct version. The first backslash escapes the second one
and this version will work even if a future version of Python recognizes
\d as an escape sequence.
Dennis

[Paul McGuire]

"James Thiele" <ja****************@gmail.com> wrote in message
news:11*********************@u72g2000cwu.googlegro ups.com...

I was helping a guy at work with regular expressions and found
something I didn't expect:

re.match('\d', '7').group() '7' re.match('\\d', '7').group() '7'
It's not clear to me why these are the same. Could someone please
explain?

This is not a feature of regexp's at all, but of Python strings. If the
backslash precedes a character that is not normally interpreted, then it is
treated like just a backslash. Look at this sample from the Python command
line:

s = "\d"
s '\\d' s = "\t"
s '\t'
This is one reason why Python programmers who use regexp's use the "raw"
notation to create strings (this is often misnomered as a "raw string", but
the resulting string is an ordinary string in every respect - what is "raw"
about it is the disabling of escape behavior of any backslashes that are not
the last character in the string). It is painful enough to litter your
regexp with backslashes, just because you have the misfortune of having to
match a '.', '+', '?', '*', or brackets or parentheses in your expression,
without having to double up the backslashes for escaping purposes. Consider
these sample statements:
"\d" == "\\d" True "\t" == "\\t" False r"\t" == "\\t" True

So your question is really a string question - you just happened to trip
over it while defining a regexp.

-- Paul

[Heiko Wundram]

Am Sonntag 21 Mai 2006 19:49 schrieb James Thiele:

re.match('\d', '7').group() print '\d' \d
re.match('\\d', '7').group() print '\\d'

\d

'\d' evaluates to \d, because d is not a valid escape sequence. '\n' evaluates
to newline, because n is a valid escape sequence. '\\' evaluates to \,
because \ is a valid escape sequence.

--- Heiko.