>>> match = re.search(r'\d', 'ca\d')
>>> type(match)
<type 'NoneType'>
From my understanding 'r' means don't do any special processing with blackslashes and just return the raw string.
Also, why do i get the output below:
>>> match = re.search(r'\a', 'ca\a')
>>> match.group()
'\x07'
Because your input string has no digit. \d means capture a digit.
If you want to capture a literal \d, you should use \\d pattern.
See example here.
This program
import re
p = re.compile(ur'\\d')
test_str = u"ca\d"
print re.search(p, test_str).group(0)
Will output \d.
As for r'', please check this re documentation:
The solution is to use Python’s raw string notation for regular expression patterns; backslashes are not handled in any special way in a string literal prefixed with 'r'. So r"\n" is a two-character string containing '\' and 'n', while "\n" is a one-character string containing a newline. Usually patterns will be expressed in Python code using this raw string notation.
It does not mean it does not process slashes anyhow, this just lets you use a single slash instead of a doubled one. The slash is meaningful before d in a regular expression.
And as for \a, there is no such a regex metacharacter, so \ is treated as a literal.
And in addition to stribizhev's comment, probably the 'r' (raw string indicator) is making you confused. That is used to avoid escaping. Escaping is a form of allowing writing in the code special (unprintable) characters like: TAB - ASCII 9 - "\t" CR - ASCII 13 - "\r" (Unix Enter)
But there's no special char that has the code "\d", so placing an r in front of it makes no difference, so the string will still be "\d" (2 chars) that in regex, matches over a digit.
