Develop Reference

What's the point of assignment to slice?

I found this line in the pip source:
sys.path[:] = glob.glob(os.path.join(WHEEL_DIR, "*.whl")) + sys.path
As I understand the line above is doing the same as below:
sys.path = glob.glob(os.path.join(WHEEL_DIR, "*.whl")) + sys.path
With one difference: in the first case sys.path still points to the same object in memory while in the second case sys.path points to the new list created from two existing.
Another one thing is that the first case is two times slower than second:
>>> timeit('a[:] = a + [1,2]', setup='a=[]', number=20000)
2.111023200035561
>>> timeit('a = a + [1,2]', setup='a=[]', number=20000)
1.0290934000513516
The reason as I think is that in the case of slice assignment objects from a (references to objects) are copied to a new list and then copied back to the resized a.
So what are the benefits of using a slice assignment?

Assigning to a slice is useful if there are other references to the same list, and you want all references to pick up the changes.
So if you do something like:
bar = [1, 2, 3]
foo = bar
bar[:] = [5, 4, 3, 2, 1]
print(foo)
this will print [5, 4, 3, 2, 1]. If you instead do:
bar = [5, 4, 3, 2, 1]
print(foo)
the output will be [1, 2, 3].

With one difference: in the first case sys.path still points to the same object in memory while in the second case sys.path points to the new list created from two existing.
Right: That’s the whole point, you’re modifying the object behind the name instead of the name. Thus all other names referring to the same object also see the changes.
Another one thing is that the first case is two times slower than second:
Not really. Slice assignment performs a copy. Performing a copy is an O(n) operation while performing a name assignment is O(1). In other words, the bigger the list, the slower the copy; whereas the name assignment always takes the same (short) time.

Your assumptions are very good!
In python a variable is a name that has been set to point to an object in memory, which in essence is what gives python the ability to be a dynamically typed language, i.e. you can have the same variable as a number, then reassign it to a string etc.
as shown here whenever you assign a new value to a variable, you are just pointing a name to a different object in memory
>>> a = 1
>>> id(a)
10968800
>>> a = 1.0
>>> id(a)
140319774806136
>>> a = 'hello world'
>>> id(a)
140319773005552
(in CPython the id refers to its address in memory).
Now for your question sys.path is a list, and a python list is a mutable type, thus meaning that the type itself can change, i.e.
>>> l = []
>>> id(l)
140319772970184
>>> l.append(1)
>>> id(l)
140319772970184
>>> l.append(2)
>>> id(l)
140319772970184
even though I modified the list by adding items, the list still points to the same object, and following the nature of python, a lists elements as well are only pointers to different areas in memory (the elements aren't the objects, the are only like variables to the objects held there) as shown here,
>>> l
[1, 2]
>>> id(l[0])
10968800
>>> l[0] = 3
>>> id(l[0])
10968864
>>> id(l)
140319772970184
After reassigning to l[0] the id of that element has changed. but once again the list hasn't.
Seeing that assigning to an index in the list only changes the places where lists elements where pointing, now you will understand that when I reassign l I don't reassign, I just change where l was pointing
>>> id(l)
140319772970184
>>> l = [4, 5, 6]
>>> id(l)
140319765766728
but if I reassign to all of ls indexes, then l stays the same object only the elements point to different places
>>> id(l)
140319765766728
>>> l[:] = [7, 8, 9]
>>> id(l)
140319765766728
That will also give you understanding on why it is slower, as python is reassigning the elements of the list, and not just pointing the list somewhere else.
One more little point if you are wondering about the part where the line finishes with
sys.path[:] = ... + sys.path
it goes in the same concept, python first creates the object on the right side of the = and then points the name on the left side to the new object, so when python is still creating the new list on the right side, sys.path is in essence the original list, and python takes all of its elements and then reassigns all of the newly created elements to the mappings in the original sys.paths addresses (since we used [:])
now for why pip is using [:] instead of reassigning, I don't really know, but I would believe that it might have a benefit of reusing the same object in memory for sys.path.
python itself also does it for the small integers, for example
>>> id(a)
10968800
>>> id(b)
10968800
>>> id(c)
10968800
a, b and c all point to the same object in memory even though all requested to create an 1 and point to it, since python knows that the small numbers are most probably going to be used a lot in programs (for example in for loops) so they create it and reuse it throughout.
(you might also find it being the case with filehandles that python will recycle instead of creating a new one.)

You are right, slice assignment will not rebind, and slice object is one type of objects in Python. You can use it to set and get.
In [1]: a = [1, 2, 3, 4]
In [2]: a[slice(0, len(a), 2)]
Out[2]: [1, 3]
In [3]: a[slice(0, len(a), 2)] = 6, 6
In [4]: a[slice(0, len(a), 1)] = range(10)
In [5]: a
Out[5]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
In [6]: a[:] = range(4)
In [7]: a
Out[7]: [0, 1, 2, 3]

Python: those variable dont point to the same values. Why?

I thought that if you assign a variable to another list, it's not copied, but it points to the same location. That's why deepcopy() is for. This is not true with Python 2.7: it's copied.
>>> a=[1,2,3]
>>> b=a
>>> b=b[1:]+b[:1]
>>> b
[2, 3, 1]
>>> a
[1, 2, 3]
>>>
>>> a=(1,2,3)
>>> b=a
>>> b=b[1:]+b[:1]
>>> a
(1, 2, 3)
>>> b
(2, 3, 1)
>>>
What am I missing?

This line changes what b points to:
b=b[1:]+b[:1]
List or tuple addition creates a new list or tuple, and the assignment operator makes b refer to that new list while leaving a referring to the original list or tuple.
Slicing a list or tuple also creates a new object, so that line creates three new objects - one for each slice, and then one for the sum. b = a + b would be a simpler example to demonstrate that addition creates a new object.
You will sometimes see c = b[:] as a way to shallow copy a list, making use of the fact that slicing creates a new object.

When you do b=b[1:]+b[:1] you first create a new object of two b slices and then assign b to reference that object. The same is for both list and tuple cases

Copying and Appending to an empty list

In a piece of code I need to copy a list and append a new value. I have found the following behaviour:
a=[]
b=[n for n in a].append('val')
print(b)
None
While, the following works as I require:
b=[n for n in a]
b.append('val')
print(b)
['val']
Any insight why it is so?

append modifies the list in place, it doesn't return a new list, that's why you're getting None.
See the documentation:
.. The methods that add, subtract, or rearrange their members in place, and don’t return a specific item, never return the collection instance itself but None.

Method append returns None, because it modifies list in place, that is why b in your first example is None. You could use list concatenation in order to copy a list and append an element to it:
In [238]: a = [1, 2, 3]
In [239]: b = a + [4]
In [240]: b
Out[240]: [1, 2, 3, 4]

That is because b=[n for n in a].append('val') does not return anything. In specific append('val') does not return any value.

How to pass a list element as reference?

I am passing a single element of a list to a function. I want to modify that element, and therefore, the list itself.
def ModList(element):
element = 'TWO'
l = list();
l.append('one')
l.append('two')
l.append('three')
print l
ModList(l[1])
print l
But this method does not modify the list. It's like the element is passed by value. The output is:
['one','two','three']
['one','two','three']
I want that the second element of the list after the function call to be 'TWO':
['one','TWO','three']
Is this possible?

The explanations already here are correct. However, since I have wanted to abuse python in a similar fashion, I will submit this method as a workaround.
Calling a specific element from a list directly returns a copy of the value at that element in the list. Even copying a sublist of a list returns a new reference to an array containing copies of the values. Consider this example:
>>> a = [1, 2, 3, 4]
>>> b = a[2]
>>> b
3
>>> c = a[2:3]
>>> c
[3]
>>> b=5
>>> c[0]=6
>>> a
[1, 2, 3, 4]
Neither b, a value only copy, nor c, a sublist copied from a, is able to change values in a. There is no link, despite their common origin.
However, numpy arrays use a "raw-er" memory allocation and allow views of data to be returned. A view allows data to be represented in a different way while maintaining the association with the original data. A working example is therefore
>>> import numpy as np
>>> a = np.array([1, 2, 3, 4])
>>> a
array([1, 2, 3, 4])
>>> b = a[2]
>>> b
3
>>> b=5
>>> a
array([1, 2, 3, 4])
>>> c = a[2:3]
>>> c
array([3])
>>> c[0]=6
>>> a
array([1, 2, 6, 4])
>>>
While extracting a single element still copies by value only, maintaining an array view of element 2 is referenced to the original element 2 of a (although it is now element 0 of c), and the change made to c's value changes a as well.
Numpy ndarrays have many different types, including a generic object type. This means that you can maintain this "by-reference" behavior for almost any type of data, not only numerical values.

Python doesn't do pass by reference. Just do it explicitly:
l[1] = ModList(l[1])
Also, since this only changes one element, I'd suggest that ModList is a confusing name.

Python is a pass by value language hence you can't change the value by assignment in the function ModList. What you could do instead though is pass the list and index into ModList and then modify the element that way
def ModList(theList, theIndex) :
theList[theIndex] = 'TWO'
ModList(l, 1)

In many cases you can also consider to let the function both modify and return the modified list. This makes the caller code more readable:
def ModList(theList, theIndex) :
theList[theIndex] = 'TWO'
return theList
l = ModList(l, 1)

Python: How do you insert into a list by slicing?

I was instructed to prevent this from happening in a Python program but frankly I have no idea how this is even possible. Can someone give an example of how you can slice a list and insert something into it to make it bigger? Thanks

>>> a = [1,2,3]
>>> a[:0] = [4]
>>> a
[4, 1, 2, 3]
a[:0] is the "slice of list a beginning before any elements and ending before index 0", which is initially an empty slice (since there are no elements in the original list before index 0). If you set it to be a non-empty list, that will expand the original list with those elements. You could also do the same anywhere else in the list by specifying a zero-width slice (or a non-zero width slice, if you want to also replace existing elements):
>>> a[1:1] = [6,7]
>>> a
[4, 6, 7, 1, 2, 3]

To prevent this from happening you can subclass the builtin list and then over-ride these methods for details refer here