Is there an overhead when nesting functions in Python?

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终归单人心
终归单人心 2020-12-04 23:28

In Python, if I have a child function within a parent function, is the child function \"initialised\" (created) every time the parent function is called? Is there any overhe

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  • 2020-12-04 23:51

    The code object is pre-compiled so that part has no overhead. The function object gets built on every invocation -- it binds the function name to the code object, records default variables, etc.

    Executive summary: It's not free.

    >>> from dis import dis
    >>> def foo():
            def bar():
                    pass
            return bar
    
    >>> dis(foo)
      2           0 LOAD_CONST               1 (<code object bar at 0x1017e2b30, file "<pyshell#5>", line 2>)
                  3 MAKE_FUNCTION            0
                  6 STORE_FAST               0 (bar)
    
      4           9 LOAD_FAST                0 (bar)
                 12 RETURN_VALUE 
    
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  • 2020-12-04 23:53

    Yes, a new object would be created each time. It's likely not an issue unless you have it in a tight loop. Profiling will tell you if it's a problem.

    In [80]: def foo():
       ....:     def bar():
       ....:         pass
       ....:     return bar
       ....: 
    
    In [81]: id(foo())
    Out[81]: 29654024
    
    In [82]: id(foo())
    Out[82]: 29651384
    
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  • 2020-12-04 23:54

    I was curious about this too, so I decided to figure out how much overhead this incurred. TL;DR, the answer is not much.

    Python 3.5.2 (default, Nov 23 2017, 16:37:01) 
    [GCC 5.4.0 20160609] on linux
    Type "help", "copyright", "credits" or "license" for more information.
    >>> from timeit import timeit
    >>> def subfunc():
    ...     pass
    ... 
    >>> def no_inner():
    ...     return subfunc()
    ... 
    >>> def with_inner():
    ...     def s():
    ...         pass
    ...     return s()
    ... 
    >>> timeit('[no_inner() for _ in range(1000000)]', setup='from __main__     import no_inner', number=1)
    0.22971350199986773
    >>> timeit('[with_inner() for _ in range(1000000)]', setup='from __main__ import with_inner', number=1)
    0.2847519510000893
    

    My instinct was to look at percents (with_inner is 24% slower), but that number is misleading in this case, since we'll never actually just return the value of an inner function from an outer function, especially with functions that don't actually do anything.
    After making that mistake, I decided to compare it to other common things, to see when this does and does not matter:

        >>> def no_inner():
        ...     a = {}
        ...     return subfunc()
        ... 
        >>> timeit('[no_inner() for _ in range(1000000)]', setup='from __main__ import no_inner', number=1)
        0.3099582109998664
    

    Looking at this, we can see that it takes less time than creating an empty dict (the fast way), so if you're doing anything non-trivial, this probably does not matter at all.

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  • 2020-12-04 23:56

    The other answers are great and really answer the question well. I wanted to add that most inner functions can be avoided in python using for loops, generating functions, etc.

    Consider the following Example:

    def foo():
        # I need to execute a function on two sets of arguments:
        argSet1 = (1, 3, 5, 7)
        argSet2 = (2, 4, 6, 8)
    
        # A Function could be executed on each set of args
        def bar(arg1, arg2, arg3, arg4):
            return (arg1 + arg2 + arg3 + arg4)
    
        total = 0
        for argSet in [argSet1, argSet2]:
          total += bar(*argSet)
        print( total )
    
        # Or a loop could be used on the argument sets
        total = 0
        for arg1, arg2, arg3, arg4 in [argSet1, argSet2]:
            total += arg1 + arg2 + arg3 + arg4
        print( total )
    

    This example is a little goofy, but I hope you can see my point nonetheless. Inner functions are often not needed.

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  • 2020-12-05 00:07

    Yes. This enables closures, as well as function factories.

    A closure causes the inner function to remember the state of its environment when called.

    def generate_power(number):
    
        # Define the inner function ...
        def nth_power(power):
            return number ** power
    
        return nth_power
    

    Example

    >>> raise_two = generate_power(2)
    >>> raise_three = generate_power(3)
    
    >>> print(raise_two(3))
    8
    >>> print(raise_three(5))
    243
    """
    
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  • 2020-12-05 00:17

    There is an impact, but in most situations it is so small that you shouldn't worry about it - most non-trivial applications probably already have performance bottlenecks whose impacts are several orders of magnitude larger than this one. Worry instead about the readability and reusability of the code.

    Here some code that compares the performance of redefining a function each time through a loop to reusing a predefined function instead.

    import gc
    from datetime import datetime
    
    class StopWatch:
         def __init__(self, name):
             self.name = name
    
         def __enter__(self):
             gc.collect()
             self.start = datetime.now()
    
         def __exit__(self, type, value, traceback):
             elapsed = datetime.now()-self.start
             print '** Test "%s" took %s **' % (self.name, elapsed)
    
    def foo():
         def bar():
              pass
         return bar
    
    def bar2():
        pass
    
    def foo2():
        return bar2
    
    num_iterations = 1000000
    
    with StopWatch('FunctionDefinedEachTime') as sw:
        result_foo = [foo() for i in range(num_iterations)]
    
    with StopWatch('FunctionDefinedOnce') as sw:
        result_foo2 = [foo2() for i in range(num_iterations)]
    

    When I run this in Python 2.7 on my Macbook Air running OS X Lion I get:

    ** Test "FunctionDefinedEachTime" took 0:00:01.138531 **
    ** Test "FunctionDefinedOnce" took 0:00:00.270347 **
    
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