3 ms·

But couldn't that inefficiency be optimized away in the implementation? (Load imports lazily).

marmaduke 9y ago
It can't be done automatically because imported can execute code. Since that's part of the semantics, you have to be lazy by hand.
- deckiedan 9y ago
  Only on the first load ever. When modules are compiled to their byte code, they could get a "clean" flag if they don't execute code, which means they get loaded lazily next time.
  - marmaduke 9y ago
    Ok but figuring out what executes code might require executing code: consider non trivial use of meta classes, where an inconspicuous subclass actually invokes e.g. A registration process like a Django model class.
    - smitherfield 9y ago
      Sure, but, just spitballing here, could a substantial fraction of imported code benefit even with a very conservative heuristic? In any case, given your points, it seems like a future Python version ought to introduce a new version of the import statement with lazy semantics (which, besides eliminating dead code, is also IMO the more correct/explicit behavior when importing symbols).
      - deckiedan 9y ago
        That's kind of what I was hoping. I suspect a vast proportion of normal Python code doesn't use the meta programming and run on import and so on.
      - joshuamorton 9y ago
        If a module imports anything, it cannot be assumed clean, since any of its imports may execute code, and they will not be available at compile time.
      - marmaduke 9y ago
        It'd be pretty easy to do the experiment, since you can override the import logic and do lazy imports, eg. class LazyModule: def __init__(self, name): self.name = name def __getattr__(self, key): mod = {} exec('import %s as mod', mod) self.mod = mod['mod'] return getattr(self.mod, key) class LazyImporter: def get_module(self, name): return LazyModule(name)

It can't be done automatically because imported can execute code. Since that's part of the semantics, you have to be lazy by hand.

deckiedan 9y ago
Only on the first load ever. When modules are compiled to their byte code, they could get a "clean" flag if they don't execute code, which means they get loaded lazily next time.
- marmaduke 9y ago
  Ok but figuring out what executes code might require executing code: consider non trivial use of meta classes, where an inconspicuous subclass actually invokes e.g. A registration process like a Django model class.
  - smitherfield 9y ago
    Sure, but, just spitballing here, could a substantial fraction of imported code benefit even with a very conservative heuristic? In any case, given your points, it seems like a future Python version ought to introduce a new version of the import statement with lazy semantics (which, besides eliminating dead code, is also IMO the more correct/explicit behavior when importing symbols).
    - deckiedan 9y ago
      That's kind of what I was hoping. I suspect a vast proportion of normal Python code doesn't use the meta programming and run on import and so on.
    - joshuamorton 9y ago
      If a module imports anything, it cannot be assumed clean, since any of its imports may execute code, and they will not be available at compile time.
    - marmaduke 9y ago
      It'd be pretty easy to do the experiment, since you can override the import logic and do lazy imports, eg. class LazyModule: def __init__(self, name): self.name = name def __getattr__(self, key): mod = {} exec('import %s as mod', mod) self.mod = mod['mod'] return getattr(self.mod, key) class LazyImporter: def get_module(self, name): return LazyModule(name)