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By default, the count is set to zero, which means the re. sub() method will replace all pattern occurrences in the target string.
If you want to replace a string that matches a regular expression (regex) instead of perfect match, use the sub() of the re module. In re. sub() , specify a regex pattern in the first argument, a new string in the second, and a string to be processed in the third.
Put a capture group around the part that you want to preserve, and then include a reference to that capture group within your replacement text. @Amber: I infer from your answer that unlike str. replace(), we can't use variables a) in raw strings; or b) as an argument to re. sub; or c) both.
There is a difference between the use of both functions. Both return the first match of a substring found in the string, but re. match() searches only from the beginning of the string and return match object if found. But if a match of substring is found somewhere in the middle of the string, it returns none.
Simply use \1 instead of $1:
In [1]: import re
In [2]: method = 'images/:id/huge'
In [3]: re.sub(r'(:[a-z]+)', r'<span>\1</span>', method)
Out[3]: 'images/<span>:id</span>/huge'
Also note the use of raw strings (r'...') for regular expressions. It is not mandatory but removes the need to escape backslashes, arguably making the code slightly more readable.
Use \1 instead of $1.
\number Matches the contents of the group of the same number.
http://docs.python.org/library/re.html#regular-expression-syntax
A backreference to the whole match value is \g<0>, see re.sub documentation:
The backreference
\g<0>substitutes in the entire substring matched by the RE.
See the Python demo:
import re
method = 'images/:id/huge'
print(re.sub(r':[a-z]+', r'<span>\g<0></span>', method))
# => images/<span>:id</span>/huge
For the replacement portion, Python uses \1 the way sed and vi do, not $1 the way Perl, Java, and Javascript (amongst others) do. Furthermore, because \1 interpolates in regular strings as the character U+0001, you need to use a raw string or \escape it.
Python 3.2 (r32:88445, Jul 27 2011, 13:41:33)
[GCC 4.0.1 (Apple Inc. build 5465)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> method = 'images/:id/huge'
>>> import re
>>> re.sub(':([a-z]+)', r'<span>\1</span>', method)
'images/<span>id</span>/huge'
>>>
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