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How to create a list so that when I append a variable the first element gets removed from list after a certain threshold

Let's say I want to create a list. The list need to have a MAX length of 5. The list would operate as such:
list = []
list.append(1)
list = [1]
list.append(2)
list = [1,2]
..
list.append(5)
list = [1,2,3,4,5]
But, when I append another number the first element is removed:
list.append(6)
list = [2,3,4,5,6]
This is super basic and I can't figure this one out.
I don't want to use classes - can this be done with basic functions such as slices?

You could use a collections.deque for this:
>>> import collections
>>> data = collections.deque(maxlen=5)
>>> data.append(1)
>>> data.append(2)
>>> data.append(3)
>>> data.append(4)
>>> data.append(5)
>>> data
deque([1, 2, 3, 4, 5], maxlen=5)
>>> data.append(6)
>>> data
deque([2, 3, 4, 5, 6], maxlen=5)
>>>
Note, this can pretty much work like a list object, however, it does have different performance characteristics, perhaps most importantly, no random access (basically, a linked list of arrays, but it won't have constant-time access like an array-list, which is what a list object is).
However, it is particularly meant for these types of operations, a "double-ended queue", it allows for efficient removal and addition at both ends. A list can only efficiently append and pop, but adding / removing from the beginning will be inefficient.

If you need it to be a list, you can use a slightly different approach to add the items (in-place):
L = []
L[:] = L[-4:] + [1] # [1]
L[:] = L[-4:] + [2] # [1, 2]
L[:] = L[-4:] + [3] # [1, 2, 3]
L[:] = L[-4:] + [4] # [1, 2, 3, 4]
L[:] = L[-4:] + [5] # [1, 2, 3, 4, 5]
L[:] = L[-4:] + [6] # [2, 3, 4, 5, 6]
Which you could place in a function to make it easier to use:
def appendMax(L,maxLen,value): # maxLen > 1
L[:] = L[1-maxLen:] + [value]
L = []
appendMax(L,5,1) # [1]
appendMax(L,5,2) # [1, 2]
appendMax(L,5,3) # [1, 2, 3]
appendMax(L,5,4) # [1, 2, 3, 4]
appendMax(L,5,5) # [1, 2, 3, 4, 5]
appendMax(L,5,6) # [2, 3, 4, 5, 6]

Use deque() from collections. It will hold certain amount of variables, after that it will delete the first item while appending.
If you don't want to use libraries or classes, you can simply create a list with N none values inside it, then loop inside the list to replace the values inside it.

Writing an implementation of Selection Sort in Python and my code is throwing "min() arg is an empty set" on a populated list

As an exercise, I am writing an implementation of selection sort (as well as insertion sort and quick sort); for me, the best way to gain a deeper understanding of a function is implementing it myself. I have the following code:
def selectionSort(L):
S = []
i = 0
a = len(L)
while i <= a:
S.append(min(L)) #Find min(L) and add it to set S
L.remove(min(L)) #Remove that same element min(L) from L
i += 1
return S #Return the sorted list
L = [int(x) for x in input('Input a list to be sorted: ').split()]
print(selectionSort(L))
The idea here is to have the user input a list of integers to be sorted and run the selectionSort() function on the list. For the life of me, I cannot figure out why min(L) is throwing the error min() arg is an empty sequence. I have written other code which takes a list as input in the same manner and it works fine.

You are looping once too many:
i = 0
a = len(L)
while i <= a:
#
i += 1
Here i goes from 0 through to len(a) inclusive, so you loop len(a) + 1 times. But there are only len(a) items in the list.
Your options are to pick one of the following
start i at 1, not at 0
stop at i < a (dropping the =) to not include len(a)
Use a for i in range(len(a)) loop; this produces values from 0 through to len(a) - 1.
Simply test if L is empty with
while L:
and remove a and i from your code altogether.
The latter option leads to less code and is clearer:
def selectionSort(L):
S = []
while L:
pick = min(L)
S.append(pick)
L.remove(pick)
return S #Return the sorted list
Note that I store the result of min(L) first to avoid scanning the list twice.

Your issue should already be fixed by #AChampion's comment, I just want to add some clarifications about what you're trying to do. The following code should work for you, but it still has some other issues:
def selectionSort(L):
S = []
i = 0
a = len(L)
while i < a: # Or even better: for i in range(len(L)) and without i += 1
S.append(min(L)) #Find min(L) and add it to set S
L.remove(min(L)) #Remove that same element min(L) from L
i += 1
return S
But, look at the following outputs:
>>> my_list = [3, 5, 1, 9, 2, 7, 4, 7, 4]
>>> selectionSort(my_list)
[1, 2, 3, 4, 4, 5, 7, 7, 9]
>>>
>>> my_list
[]
You see, the input my_list is empty now, because it is altered inside your function. To fix this issue, you can do: (because you may still need to use your original list)
def my_sort(my_list):
temp = my_list[:]
sorted_list = []
for i in range(len(temp)):
sorted_list.append(min(temp))
temp.remove(min(temp))
return sorted_list
Now, inside our function, we use a copy of our original list temp = my_list[:], this is equivalent to temp = copy.copy(my_list):
>>> import copy
>>>
>>> my_list = [3, 5, 1, 9, 2, 7, 4, 7, 4]
>>> id(my_list)
36716416
>>> l1 = my_list
>>> id(l1)
36716416 # In this case the id of l1 is the same as my_list's id!
>>> l2 = my_list[:]
>>> id(l2)
36738832 # This is NOT the same as my_list's id
>>> l3 = copy.copy(my_list)
>>> id(l3)
36766424 # This is NOT the same as my_list's id
Output of the new function:
>>> my_list = [3, 5, 1, 9, 2, 7, 4, 7, 4]
>>> my_sort(my_list)
[1, 2, 3, 4, 4, 5, 7, 7, 9]
>>>
>>> my_list
[3, 5, 1, 9, 2, 7, 4, 7, 4] # Our original list is still alive!
You may need:
As per your goal (implementing your own function for learning purpose), you may also need to use your own min function instead of the built-in one, the following is an example that you can try:
def my_min(my_list):
min = my_list[0]
for i in range(1, len(my_list)):
if min > my_list[i]:
min = my_list[i]
return min
Output:
>>> l1 = [3, 7, 2, 5]
>>> my_min(l1)
2

How do I reverse a list using while loop?

Input list: [1, 2, 3, 4, 5]
Output: [5, 4, 3, 2, 1]
I know how to do it with for loop, but my assignment is to do it with while loop; which I have no idea to do. Here is the code I have so far:
def while_version(items):
a = 0
b = len(items)
r_list = []
while (a!=b):
items[a:a] = r_list[(-a)-1]
a+=1
return items

I would say to make the while loop act like a for loop.
firstList = [1,2,3]
secondList=[]
counter = len(firstList)-1
while counter >= 0:
secondList.append(firstList[counter])
counter -= 1

The simplest way would be:
def while_version(items):
new_list = []
while items: # i.e. until it's an empty list
new_list.append(items.pop(-1))
return new_list
This will reverse the list:
>>> l1 = [1, 2, 3]
>>> l2 = while_version(l)
>>> l2
[3, 2, 1]
Note, however, that it also empties the original list:
>>> l1
[]
To avoid this, call e.g. l2 = while_version(l1[:]).

The trivial answer
Given
a = [1, 2, 3, 4, 5]
then
a[::-1]
returns
[5, 4, 3, 2, 1]
In your code:
You use r_list[(-a)+1], buy you have never assigned r_list any value (just "r_list = []")
I think your are confusing "items" with "r_list". So I think you want to return "r_list" instead of "items" (the input parameter)
The assignment should be "r_list[a] = items[-a-1]", but that doesn't work. You should use "r_list.append(items[-a-1])"
The return should be "return r_list"
"while (a!=b)" should be "while (a < b)" for readeability
Hope this helps

How to delete a fixed number of items from end of list in python

If you have a list
myList = [1,2,3,4,5,6,7,8,9,0]
Is there a pythonic way to delete a defined number of items from the end of a list.
EG (pseudocode):
removeFromend(myList, 3)
print myList
>>>[1,2,3,4,5,6,7]

You can use list slicing, which I think is the most pythonic way of doing it:
end_trimm = 3
myList = myList[:-end_trimm]

If you want to mutate the list, setting a slice to an empty list is equivalent to deleting those indices.
myList = [1,2,3,4,5,6,7,8,9,0]
myList[-3:] = []
myList
Out[16]: [1, 2, 3, 4, 5, 6, 7]
This works in cases where you can't simply rebind myList to a new list, e.g. you pass your list to a function and want that function to mutate your list.

deling the slice is the direct approach
>>> myList = [1,2,3,4,5,6,7,8,9,0]
>>> del myList[-3:]
>>> myList
[1, 2, 3, 4, 5, 6, 7]
>>>
The -3 means the slice starts 3 from the end, So the general form is del myList[-n:]

How to append multiple values to a list in Python

I am trying to figure out how to append multiple values to a list in Python. I know there are few methods to do so, such as manually input the values, or put the append operation in a for loop, or the append and extend functions.
However, I wonder if there is a more neat way to do so? Maybe a certain package or function?

You can use the sequence method list.extend to extend the list by multiple values from any kind of iterable, being it another list or any other thing that provides a sequence of values.
>>> lst = [1, 2]
>>> lst.append(3)
>>> lst.append(4)
>>> lst
[1, 2, 3, 4]
>>> lst.extend([5, 6, 7])
>>> lst.extend((8, 9, 10))
>>> lst
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> lst.extend(range(11, 14))
>>> lst
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
So you can use list.append() to append a single value, and list.extend() to append multiple values.

Other than the append function, if by "multiple values" you mean another list, you can simply concatenate them like so.
>>> a = [1,2,3]
>>> b = [4,5,6]
>>> a + b
[1, 2, 3, 4, 5, 6]

If you take a look at the official docs, you'll see right below append, extend. That's what your looking for.
There's also itertools.chain if you are more interested in efficient iteration than ending up with a fully populated data structure.

if the number of items was saved in a variable say n. you can use list comprehension and plus sign for list expansion.
lst = ['A', 'B']
n = 1
new_lst = lst + ['flag'+str(x) for x in range(n)]
print(my_lst)
>>> ['A','B','flag0','flag1']

One way you can work around this type of problem is -
Here we are inserting a list to the existing list by creating a variable new_values.
Note that we are inserting the values in the second index, i.e. a[2]
a = [1, 2, 7, 8]
new_values = [3, 4, 5, 6]
a.insert(2, new_values)
print(a)
But here insert() method will append the values as a list.
So here goes another way of doing the same thing, but this time, we'll actually insert the values in between the items.
a = [1, 2, 7, 8]
a[2:2] = [3,4,5,6]
print(a)