Andrew Savige <ajsav...@yahoo.com.au> wrote: > > I'd like to convert the following Perl code to Python: > > use strict; > { > my %private_hash = ( A=>42, B=>69 ); > sub public_fn { > my $param = shift; > return $private_hash{$param}; > } > } > print public_fn("A"); # good: prints 42 > my $x = $private_hash{"A"}; # error: good, hash not in scope > > The real code is more complex; the above is a simplified example. > > Notice that this code uses Perl's lexical scope to hide the > %private_hash variable, but not the public_fn() function. > > While I could convert this code to the following Python code: > > private_hash = dict( A=42, B=69 ) > def public_fn(param): > return private_hash[param] > print public_fn("A") # good: prints 42 > x = private_hash["A"] # works: oops, hash is in scope > > I'm not happy with that because I'd like to limit the scope of the > private_hash variable so that it is known only inside public_fn. > > Of course, I could hide the hash like so: > > def public_fn(param): > private_hash = dict( A=42, B=69 ) > return private_hash[param] > > yet I'm not happy with that either because of the repeated > initialization the hash each time the function is called. > > What is the Pythonic equivalent of Perl's lexical scope, as > illustrated by the code snippet above?
Either _private_hash = dict( A=42, B=69) def public_fn(param): return _private_hash[param] Or def public_fn(param, _private_hash = dict( A=42, B=69)): return _private_hash[param] Is probably the pythonic equivalents. Note that private_hash starts with an underscore which means it won't be exported from a module by default and it is a convention that it is private and shouldn't be fiddled with. I'd probably go with the latter of the two examples. -- Nick Craig-Wood <n...@craig-wood.com> -- http://www.craig-wood.com/nick -- http://mail.python.org/mailman/listinfo/python-list