Conditional based on whether or not a module is being used
in [Python]
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From: MRAB on 5 Mar 2010 14:57 Pete Emerson wrote: > In a module, how do I create a conditional that will do something > based on whether or not another module has been loaded? > > Suppose I have the following: > > import foo > import foobar > > print foo() > print foobar() > > ########### foo.py > def foo: > return 'foo' > > ########### foobar.py > def foobar: > if foo.has_been_loaded(): # This is not right! > return foo() + 'bar' # This might need to be foo.foo() ? > else: > return 'bar' > > If someone is using foo module, I want to take advantage of its > features and use it in foobar, otherwise, I want to do something else. > In other words, I don't want to create a dependency of foobar on foo. > > My failed search for solving this makes me wonder if I'm approaching > this all wrong. > Look for its name in sys.modules, for example: 'foo' in sys.modules
From: Martin P. Hellwig on 5 Mar 2010 15:06 On 03/05/10 19:24, Pete Emerson wrote: > In a module, how do I create a conditional that will do something > based on whether or not another module has been loaded? <cut>> > If someone is using foo module, I want to take advantage of its > features and use it in foobar, otherwise, I want to do something else. > In other words, I don't want to create a dependency of foobar on foo. > > My failed search for solving this makes me wonder if I'm approaching > this all wrong. > > Thanks in advance, > Pete Hmm how about the module is available, just not imported yet, I would assume that you still would like to use the module then. Perhaps playing around with the imp module might get you what you mean instead of what you say? -- mph
From: Chris Rebert on 5 Mar 2010 15:17 On 3/5/10, Pete Emerson <pemerson(a)gmail.com> wrote: > In a module, how do I create a conditional that will do something > based on whether or not another module has been loaded? > > Suppose I have the following: > > import foo > import foobar > > print foo() > print foobar() > > ########### foo.py > def foo: > return 'foo' > > ########### foobar.py > def foobar: > if foo.has_been_loaded(): # This is not right! > return foo() + 'bar' # This might need to be foo.foo() ? > else: > return 'bar' > > If someone is using foo module, I want to take advantage of its > features and use it in foobar, otherwise, I want to do something else. > In other words, I don't want to create a dependency of foobar on foo. > > My failed search for solving this makes me wonder if I'm approaching > this all wrong. Just try importing foo, and then catch the exception if it's not installed. #foobar.py try: import foo except ImportError: FOO_PRESENT = False else: FOO_PRESENT = True if FOO_PRESENT: def foobar(): return foo.foo() + 'bar' else: def foobar(): return 'bar' You could alternately do the `if FOO_PRESENT` check inside the function body rather than defining separate versions of the function. Cheers, Chris -- http://blog.rebertia.com
From: Pete Emerson on 5 Mar 2010 15:25 On Fri, Mar 5, 2010 at 12:17 PM, Chris Rebert <clp2(a)rebertia.com> wrote: > On 3/5/10, Pete Emerson <pemerson(a)gmail.com> wrote: >> In a module, how do I create a conditional that will do something >> based on whether or not another module has been loaded? >> >> Suppose I have the following: >> >> import foo >> import foobar >> >> print foo() >> print foobar() >> >> ########### foo.py >> def foo: >> return 'foo' >> >> ########### foobar.py >> def foobar: >> if foo.has_been_loaded(): # This is not right! >> return foo() + 'bar' # This might need to be foo.foo() ? >> else: >> return 'bar' >> >> If someone is using foo module, I want to take advantage of its >> features and use it in foobar, otherwise, I want to do something else. >> In other words, I don't want to create a dependency of foobar on foo. >> >> My failed search for solving this makes me wonder if I'm approaching >> this all wrong. > > Just try importing foo, and then catch the exception if it's not installed. > > #foobar.py > try: > import foo > except ImportError: > FOO_PRESENT = False > else: > FOO_PRESENT = True > > if FOO_PRESENT: > def foobar(): > return foo.foo() + 'bar' > else: > def foobar(): > return 'bar' > > > You could alternately do the `if FOO_PRESENT` check inside the > function body rather than defining separate versions of the function. > > Cheers, > Chris > -- > http://blog.rebertia.com > Except I want to use the module only if the main program is using it too, not just if it's available for use. I think that I found a way in my follow-up post to my own message, but not sure it's the best way or conventional. Pete
From: Pete Emerson on 5 Mar 2010 15:31
On Mar 5, 12:06 pm, "Martin P. Hellwig" <martin.hell...(a)dcuktec.org> wrote: > On 03/05/10 19:24, Pete Emerson wrote: > > > In a module, how do I create a conditional that will do something > > based on whether or not another module has been loaded? > <cut>> > > If someone is using foo module, I want to take advantage of its > > features and use it in foobar, otherwise, I want to do something else. > > In other words, I don't want to create a dependency of foobar on foo. > > > My failed search for solving this makes me wonder if I'm approaching > > this all wrong. > > > Thanks in advance, > > Pete > > Hmm how about the module is available, just not imported yet, I would > assume that you still would like to use the module then. > Perhaps playing around with the imp module might get you what you mean > instead of what you say? > > -- > mph I can certainly see why one might want to use it if it's available but not yet imported. In that case I could do a try / exception block. But in this case, I actually don't want to use the module unless the main program is doing it too. But you've got me thinking, I need to make sure that's really the desired behavior. Pete |