Question on importing and function defs

[castironpi at gmail.com]

On Mar 2, 11:44 am, Steve Holden <st... at holdenweb.com> wrote:
> TC wrote:
> > On Mar 2, 11:37 am, Gary Herron <gher... at islandtraining.com> wrote:
> >> TC wrote:
> >>> I have a problem.  Here's a simplified version of what I'm doing:
> >>> I have functions a() and b() in a module called 'mod'.  b() calls a().
> >>> So now, I have this program:
> >>> from mod import *
> >>> def a():
> >>>     blahblah
> >>> b()
> >>> The problem being, b() is calling the a() that's in mod, not the new
> >>> a() that I want to replace it.  (Both a()'s have identical function
> >>> headers, in case that matters.)  How can I fix this?
> >>> Thanks for any help.
> >> Since b calls mod.a, you could replace mod.a with your new a.  Like
> >> this:  (Warning, this could be considered bad style because it will
> >> confuse anyone who examines the mod module in an attempt to understand
> >> you code.)
>
> >>   import mod
>
> >>   def replacement_a():
> >>     ...
>
> >>   mod.a = replacement_a
>
> >>   ...
>
> >> Or another option.  Define b to take, as a parameter, the "a" function
> >> to call.
>
> >> In mod:
>
> >>   def a():
> >>    ...
>
> >>   def b(fn=a):  # to set the default a to call
> >>     ...
>
> >> And you main program:
>
> >>   from mod import *
>
> >>   def my_a():
> >>     ...
>
> >>   b(my_a)
>
> >> Hope that helps
>
> >> Gary Herron
>
> > Thanks for the tips, but no luck.  This is for a homework assignment,
> > so there are a couple of requirements, namely that I can't touch
> > 'mod', and I have to do 'from mod import *' as opposed to 'import
> > mod'.
>
> > So the first method you suggested won't work as written, since the mod
> > namespace doesn't exist.  I tried a = replacement_a, but b() is still
> > calling mod's version of a() for some reason.  And because I can't
> > touch mod, I can't use your second suggestion.
>
> > In case I somehow oversimplified, here's the actual relevant code, in
> > 'mod' (actually called 'search').  The first fn is what I've been
> > calling a(), the second is b().
>
> > (lots of stuff...)
>
> > def compare_searchers(problems, header,
> > searchers=[breadth_first_tree_search,
> >                       breadth_first_graph_search,
> > depth_first_graph_search,
> >                       iterative_deepening_search,
> > depth_limited_search,
> >                       astar_search]):
> >     def do(searcher, problem):
> >         p = InstrumentedProblem(problem)
> >         searcher(p)
> >         return p
> >     table = [[name(s)] + [do(s, p) for p in problems] for s in
> > searchers]
> >     print_table(table, header)
>
> > def compare_graph_searchers():
> >     compare_searchers(problems=[GraphProblem('A', 'B', romania),
> >                                 GraphProblem('O', 'N', romania),
> >                                 GraphProblem('Q', 'WA', australia)],
> >             header=['Searcher', 'Romania(A,B)', 'Romania(O, N)',
> > 'Australia'])
>
> > That's the end of the 'search' file.  And here's my program, which
> > defines an identical compare_searchers() with an added print
> > statement.  That statement isn't showing up.
>
> > from search import *
>
> > def compare_searchers(problems, header,
> > searchers=[breadth_first_tree_search,
> >                       breadth_first_graph_search,
> > depth_first_graph_search,
> >                       iterative_deepening_search,
> > depth_limited_search,
> >                       astar_search, best_first_graph_search]):
> >     def do(searcher, problem):
> >         p = InstrumentedProblem(problem)
> >         searcher(p)
> >         return p
> >     table = [[name(s)] + [do(s, p) for p in problems] for s in
> > searchers]
> >     print 'test'
> >     print_table(table, header)
>
> > compare_graph_searchers()
>
> Since you've admitted it's for homework, here are a couple of hints.
>
> 1. The b() function is *always* going to try and resolve its references
> in the namespace it was defined in;
>
> 2. The technique you need is most likely known as "monkey patching".
> When you say "I can't touch mod", that may mean "the source of mod must
> remain unchanged", which is subtly different. Google is your friend ...
>
> Good luck with your assignment.
>
> regards
>   Steve
> --
> Steve Holden        +1 571 484 6266   +1 800 494 3119
> Holden Web LLC              http://www.holdenweb.com/- Hide quoted text -
>
> - Show quoted text -

You can use 'settrace' to intervene.  You might be able to delete the
'a'.