x modify the contents of the list that is passed to it?x does modify the contents of the list that is passed to it?>>> def x(mylist):
... mylist = [ x for x in mylist if x > 5 ]
...
>>> foo = [1,2,3,4,5,6,7,8,9,10]
>>> x(foo)
>>> print foo
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
The function doesn't modify x because it is reassigning the name mylist to a new value. The alternative to this would be surprising.
a = 1
b = a
b = 2
assert a == 1 # would you want a to be 2 here?
If you want to replace the contents you can do so with a slice assignment
def x(mylist):
mylist[:] = [ x for x in mylist if x > 5 ]
The assignment operator in python doesn't go into any method calls. It is the mechanism for name rebinding. However, you can implement the __setitem__ method (as list does), which is more or less the operator []=
With slice assignment you are calling __setitem__ with a slice argument that says "replace this slice of the list with ..." where ... is the right side of the =
a=1 creates a reference to 1, and then b=a creates another reference to the same 1, so that then executing b=2 would in fact change the value of the object (the 1) referred to by a. Can you please help me see my error?
b = 2 makes b reference a different object with the value 2 rather than modifying the 1 in memory.b = a you are saying "make b point to whatever a points to." When you say b = 2 you are saying "make b point to 2" not "change what b is already pointing to, to be 2"As Ryan Haining explained, x is reassigning the list. It should, however, only modify the list items:
def x(mylist):
for i in reversed(mylist):
if not i > 5:
mylist.remove(i)
Note that you should traverse the list in reversed order.
mylist =makesmylistpoint to a different list.