Ruby-like expression substitution in Python

I don’t know much Ruby, and probably won’t learn; all that syntax and magic scare me away. But I have to admit it has some darned useful gadgets. Here’s a python function I hacked up to do something much like Ruby’s expression-substitution, using the same #{ } syntax. It doesn’t allow curly braces inside the #{ }; were I a little less lazy I would put in some escaping.


import re
import sys

def esub(s):
    """
    Perform Ruby-like expression substitution.

    >>> x=3
    >>> y='A'
    >>> esub('abc#{x}def#{3+5}hij#{"".join([y, y])}')
    'abc3def8hijAA'
    """
    restr = r'(?:#{(?P[^{}]*)})|(?:[^#])+|#'
    fr = sys._getframe(1)
    def process(m):
        txt = m.group('exp')
        if txt is not None:
            val = eval(txt, fr.f_globals, fr.f_locals)
            return type(s)(val)
        else:
            return m.group()
    return ''.join(process(m) for m in re.finditer(restr, s))

Tags: Programming, python, ruby

This entry was posted on February 28, 2008 at 9:38 am and is filed under Programming. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Ruby-like expression substitution in Python”

ibtiwuksiwpdg Says:
March 1, 2008 at 10:10 am | Reply
What’s about this:

def m(a):
def n():
return esub(‘#{a}’)
return n()

m(42)

your approach cannot deal with free variables in inner functions, unfortunately.
mrlauer Says:
March 1, 2008 at 12:38 pm | Reply
D’oh! Oh well.

A not-great workaround:

def m(a):
def n():
a
return esub(’#{a}’)
return n()

m(42)

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