Determining a replacement dictionary from scratch

[Dan]

Hello,

I'd like to be able to take a formatted string and determine the
replacement dictionary necessary to do string interpolation with it.
For example:

str = 'his name was %(name)s and i saw him %(years)s ago.'
createdict( str ) {'name':'', 'years':''}

Notice how it would automatically fill in default values based on
type. I figure since python does this automatically maybe there is a
clever way to solve the problem. Otherwise I will just have to parse
the string myself. Any clever solutions to this?

Thanks
-dan

[Dave Benjamin]

In article <4b**************************@posting.google.com >, Dan wrote:

I'd like to be able to take a formatted string and determine the
replacement dictionary necessary to do string interpolation with it.
For example:

str = 'his name was %(name)s and i saw him %(years)s ago.'
createdict( str ) {'name':'', 'years':''}

Notice how it would automatically fill in default values based on
type. I figure since python does this automatically maybe there is a
clever way to solve the problem. Otherwise I will just have to parse
the string myself. Any clever solutions to this?

Here's a solution that uses a regular expression. It only handles strings
and doesn't cache the regular expression; I'll leave this up to you. =)

import re
def createdict(fmt):
keys = re.findall('%\(([A-Za-z]+)\)s', fmt)
return dict(zip(keys, [''] * len(keys)))

The regular expression works like this:

%\( - match a percent character and opening parenthesis
([A-Za-z]+) - match a sequence of one or more alpha characters as a GROUP
\)s - match a closing parenthesis and 's' character

For more details on "re", see the documentation:
http://python.org/doc/current/lib/module-re.html

HTH,
Dave

--
..:[ dave benjamin (ramenboy) -:- www.ramenfest.com -:- www.3dex.com ]:.
: d r i n k i n g l i f e o u t o f t h e c o n t a i n e r :

[Christophe Delord]

On 18 Jan 2004 11:16:31 -0800, Dan wrote:

Hello,

I'd like to be able to take a formatted string and determine the
replacement dictionary necessary to do string interpolation with it.
For example:

str = 'his name was %(name)s and i saw him %(years)s ago.'
createdict( str ) {'name':'', 'years':''}

Notice how it would automatically fill in default values based on
type. I figure since python does this automatically maybe there is a
clever way to solve the problem. Otherwise I will just have to parse
the string myself. Any clever solutions to this?

Thanks
-dan

You can use the % operator to look for such constructions. You just
need to define the __getitem__ method of a class to store the
variable names. What about this function:

def createdict(format):
class grab_variables:
def __init__(self):
self.variables = {}
def __getitem__(self, item):
self.variables[item] = ''
g = grab_variables()
format%g
return g.variables

print createdict('his name was %(name)s and i saw him %(years)s ago.')

{'name': '', 'years': ''}
Best regards,
Christophe.

[Jeff Epler]

On Sun, Jan 18, 2004 at 07:34:07PM -0000, Dave Benjamin wrote:

Here's a solution that uses a regular expression. It only handles strings
and doesn't cache the regular expression; I'll leave this up to you. =)

Of course, since it uses regular expressions it's also wrong...

createdict("%%(foo)s")

{'foo': ''}

This version might be more correct:

pat = re.compile("%\([^)]*\)s|%%")

def createdict(fmt):
keys = [x[2:-2] for x in pat.findall(fmt) if x != '%%']
return dict(zip(keys, [''] * len(keys)))

Here's another way, one I like better:
class D:
def __init__(self): self.l = []
def __getitem__(self, item): self.l.append(item)

def createdict(fmt):
d = D()
fmt % d
return dict(zip(d.l, [''] * len(d.l)))
... I like it better because it uses the exact same logic to determine
the needed names as it will when the string is actually formatted.

Jeff

[Dan]

Brilliant, thank you all for your help.

-Dan