Do regular expressions from the re module support word boundaries (\b)?

Question

You should be using raw strings in your code

>>> x = 'one two three'
>>> y = re.search(r"\btwo\b", x)
>>> y
<_sre.SRE_Match object at 0x100418a58>
>>> 

Also, why don't you try

word = 'two'
re.compile(r'\b%s\b' % word, re.I)

Output:

>>> word = 'two'
>>> k = re.compile(r'\b%s\b' % word, re.I)
>>> x = 'one two three'
>>> y = k.search( x)
>>> y
<_sre.SRE_Match object at 0x100418850>

This will work: re.search(r"\btwo\b", x)

When you write "\b" in Python, it is a single character: "\x08". Either escape the backslash like this:

"\\b"

or write a raw string like this:

r"\b"

Just to explicitly explain why re.search("\btwo\b", x) doesn't work, it's because \b in a Python string is shorthand for a backspace character.

print("foo\bbar")
fobar

So the pattern "\btwo\b" is looking for a backspace, followed by two, followed by another backspace, which the string you're searching in (x = 'one two three') doesn't have.

To allow re.search (or compile) to interpret the sequence \b as a word boundary, either escape the backslashes ("\\btwo\\b") or use a raw string to create your pattern (r"\btwo\b").

Python documentation

https://docs.python.org/2/library/re.html#regular-expression-syntax

\b

Matches the empty string, but only at the beginning or end of a word. A word is defined as a sequence of alphanumeric or underscore characters, so the end of a word is indicated by whitespace or a non-alphanumeric, non-underscore character. Note that formally, \b is defined as the boundary between a \w and a \W character (or vice versa), or between \w and the beginning/end of the string, so the precise set of characters deemed to be alphanumeric depends on the values of the UNICODE and LOCALE flags. For example, r'\bfoo\b' matches 'foo', 'foo.', '(foo)', 'bar foo baz' but not 'foobar' or 'foo3'. Inside a character range, \b represents the backspace character, for compatibility with Python’s string literals.