python-internals

Can I access a list while it is being sorted in the list.sort() b = ['b', 'e', 'f', 'd', 'c', 'g', 'a'] f = 'check this' def m(i): print i, b, f return None b.sort(key=m) print b this returns b [] check this e [] check this f [] check this d [] check this c [] check this g [] check this a [] check this Note that individual items of list b is sent to function m . But at m the list b is empty, however it can see the variable f , which has same scope as list b . Why does function m print b as [] ? Hyperboreus Looking at the source code (of CPython, maybe different behaviour for other

In contrast to functions, a class' body is executed at definition time: class A(object): print 'hello' Out: hello Why is it the case? Is it related to @classmethod / @staticmethod methods and class attributes? Everything is executed at the module level when Python first imports a module. Function bodies (and generator expression bodies) are the exception here, not the rule. Python executes everything to create the objects contained in a module; like everything in Python, classes are objects, and so are functions. The only reason a class body uses a separate code object is because a class body

So i was playing with list objects and found little strange thing that if list is created with list() it uses more memory, than list comprehension? I'm using Python 3.5.2 In [1]: import sys In [2]: a = list(range(100)) In [3]: sys.getsizeof(a) Out[3]: 1008 In [4]: b = [i for i in range(100)] In [5]: sys.getsizeof(b) Out[5]: 912 In [6]: type(a) == type(b) Out[6]: True In [7]: a == b Out[7]: True In [8]: sys.getsizeof(list(b)) Out[8]: 1008 From the docs : Lists may be constructed in several ways: Using a pair of square brackets to denote the empty list: [] Using square brackets, separating items