Python list slice syntax used for no obvious reason
I occasionally see the list slice syntax used in Python code like this:
newList = oldList[:]
Surely this is just the same as:
newList = oldList
Or am I missing something?
Like NXC said, Python variable names actually point to an object, and not a specific spot in memory.
newList = oldList would create two different variables that point to the same object, therefore, changing oldList would also change newList.
However, when you do newList = oldList[:], it "slices" the list, and creates a new list. The default values for [:] are 0 and the end of the list, so it copies everything. Therefore, it creates a new list with all the data contained in the first one, but both can be altered without changing the other.
[:] Shallow copies the list, making a copy of the list structure containing references to the original list members. This means that operations on the copy do not affect the structure of the original. However, if you do something to the list members, both lists still refer to them, so the updates will show up if the members are accessed through the original.
A Deep Copy would make copies of all the list members as well.
The code snippet below shows a shallow copy in action.
# ================================================================ # === ShallowCopy.py ============================================= # ================================================================ # class Foo: def __init__(self, data): self._data = data aa = Foo ('aaa') bb = Foo ('bbb') # The initial list has two elements containing 'aaa' and 'bbb' OldList = [aa,bb] print OldList._data # The shallow copy makes a new list pointing to the old elements NewList = OldList[:] print NewList._data # Updating one of the elements through the new list sees the # change reflected when you access that element through the # old list. NewList._data = 'xxx' print OldList._data # Updating the new list to point to something new is not reflected # in the old list. NewList = Foo ('ccc') print NewList._data print OldList._data
Running it in a python shell gives the following transcript. We can see the list being made with copies of the old objects. One of the objects can have its state updated by reference through the old list, and the updates can be seen when the object is accessed through the old list. Finally, changing a reference in the new list can be seen to not reflect in the old list, as the new list is now referring to a different object.
>>> # ================================================================ ... # === ShallowCopy.py ============================================= ... # ================================================================ ... # ... class Foo: ... def __init__(self, data): ... self._data = data ... >>> aa = Foo ('aaa') >>> bb = Foo ('bbb') >>> >>> # The initial list has two elements containing 'aaa' and 'bbb' ... OldList = [aa,bb] >>> print OldList._data aaa >>> >>> # The shallow copy makes a new list pointing to the old elements ... NewList = OldList[:] >>> print NewList._data aaa >>> >>> # Updating one of the elements through the new list sees the ... # change reflected when you access that element through the ... # old list. ... NewList._data = 'xxx' >>> print OldList._data xxx >>> >>> # Updating the new list to point to something new is not reflected ... # in the old list. ... NewList = Foo ('ccc') >>> print NewList._data ccc >>> print OldList._data xxx
As it has already been answered, I'll simply add a simple demonstration:
>>> a = [1, 2, 3, 4] >>> b = a >>> c = a[:] >>> b = 10 >>> c = 20 >>> a [1, 2, 10, 4] >>> b [1, 2, 10, 4] >>> c [1, 2, 3, 20]
Never think that 'a = b' in Python means 'copy b to a'. If there are variables on both sides, you can't really know that. Instead, think of it as 'give b the additional name a'.
If b is an immutable object (like a number, tuple or a string), then yes, the effect is that you get a copy. But that's because when you deal with immutables (which maybe should have been called read only, unchangeable or WORM) you always get a copy, by definition.
If b is a mutable, you always have to do something extra to be sure you have a true copy. Always. With lists, it's as simple as a slice: a = b[:].
Mutability is also the reason that this:
def myfunction(mylist=): pass
... doesn't quite do what you think it does.
If you're from a C-background: what's left of the '=' is a pointer, always. All variables are pointers, always. If you put variables in a list: a = [b, c], you've put pointers to the values pointed to by b and c in a list pointed to by a. If you then set a = d, the pointer in position 0 is now pointing to whatever d points to.
See also the copy-module: http://docs.python.org/library/copy.html
Shallow Copy: (copies chunks of memory from one location to another)
a = ['one','two','three'] b = a[:] b = 2 print id(a), a #Output: 1077248300 ['one', 'two', 'three'] print id(b), b #Output: 1077248908 ['one', 2, 'three']
Deep Copy: (Copies object reference)
a = ['one','two','three'] b = a b = 2 print id(a), a #Output: 1077248300 ['one', 2, 'three'] print id(b), b #Output: 1077248300 ['one', 2, 'three']