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Python True Copy Within a Class

For some reason, I have a list that keeps getting modified despite being explicitly a deep copy. It appears to be unmodified as it goes through the loop, but it suddenly is modified once it exists? I must be missing something that pertains to Python's rules and logic, but I can't figure it out for the life of me.
def all_possible_states(self):
#creates many variations of a board from the available set of moves
to_return = [] #list of possible board states to return
#print(self.availible_moves)
list_to_copy = copy.deepcopy(self.availible_moves)
for item in self.availible_moves:
print('loop')
#append possible board state to list. set of availible moves for that board is one less. Done by removing item from that move list
to_return.append(board(self.player, copy.deepcopy(self.board.copy()), list_to_copy, self.plays, self.score, copy.deepcopy(item)))
#print(item)
print( self.avalible_moves) #shows the total set of moves. This is unmodified whenever it prints
print(list_to_copy)#deep copy of the original list. This is unmodified when it prints
print(to_return[len(to_return) - 1].availible_moves) #List of moves left availible for the board, this slowly shrinks for some reason each loop
print(self.availible_moves) #this is the original list, but it's not basically been cut all the way down for some reason
return to_return

Notice, the local variable list_to_copy is the deepcopy, not self.availible_moves. You are saving a deepcopy of self.availible_moves and it is being stored in the list_to_copy variable you defined. list_to_copy never changes as expected from a deepcopy. I'm not exactly sure what you are trying to do, but if you want, at the end you can reset self.availible_moves to be equal to list_to_copy and then it will be as if it never changed.
EDIT:
Actually, I think you have a spelling mistake, notice that you print(self.avalible_moves) and you are saying its not changing, when really what is changing is self.availible_moves. Notice the extra letter i in the first expression. This definitely is one of your problems.

How the linked list works?

I do not have computer science background. I am trying to learn coding by myself, and I'm doing it, partly, by solving the problems on LeetCode.
Anyway, there are the problems that use Linked Lists. And I already found info that linked list have to be simulated in Phython. My problem is that I really cannot get what is behind linked list. For instance, what kind of problems those are suppose to target?
And in general how linked list function. Any link for such info would be really helpfull.
The recent problem I looked at LeetCode asks to swap every two adjacent nodes and return its head. And LeetCode offers following solution, that I cannot actually figure out how it acutaly works.
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def swapPairs(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
pre = self
pre.next = head
while pre.next and pre.next.next:
a = pre.next
b = a.next
pre.next =b
b.next =a
a.next =b.next
pre = a
return self.next
As I said, I do not understand this solution. I tried to use example list 1->2->3->4 that should return list 2->1->4->3
All I managed is to make only one pass through the loop, and then computer should exit the loop, but then what happens? How are the last two numbers switched? How does this code work at all if list has only 2 elements, to me it seems impossible.
If you could just direct me to the online literature that explains something like this, I would be most grateful.
Thanks.

a linked-list acts almost the same as an array. There are a few main differences though. In a linked-list, the memory used doesn't (and almost never is) contiguous memory. So in an array, if u have 5 items and you look at the memory all 5 items will be right next to each other (for the most part). However each 'item' in a linked list has a pointer that points directly to the next item, removing the need to have contiguous memory. So an array is a 'list' of items that exist contiguously in memory and a linked-list is a 'list' of objects that each hold an item and a pointer to the next item. This is considered a single linked-list as traversal is only possible from one direction. There is also a double linked-list where each node now has a pointer to the next node and another pointer for the previous node allowing traversal from both directions.
https://www.cs.cmu.edu/~adamchik/15-121/lectures/Linked%20Lists/linked%20lists.html
the link will help you get familiar with visualizing how these linked-lists work. I would probably focus on inserting before and after as these should help you understand what your loop is doing.

Linked lists don't "exist" in Python as the language basically has an iterable builtin list object. Under the hood I'm sure this is implemented as a linked list in C code (most common implementation of Python).
The main feature is that a linked list is easily extendible, wheras an array has to be manually resized if you wish to expand it. Again, in Python these details are all abstracted away. So trying to work an example of linked lists in Python is pointless in my opinion, as you won't learn anything.
You should be doing this in C to get an actual understanding of memory allocation and pointers.
That said, given your example, each ListNode contains a value (like an array), but rather than just that, it has a variable 'next' where you store another ListNode object. This object, just like the first, has a value, and a variable that stores another ListNode object.This can continue for as many objects as desired.
The way the code works is that when we say pre.next, this refers to the ListNode object stored there, and the next object after that is pre.next.next. This works because pre.next is a ListNode object, which has a variable next.
Again, read up on linked lists in C. If you plan to work in higher level languages, I would say you don't really need an understanding of linked lists, as these data structures come "free" with most high level languages.

How is the for-loop able to use a variable that isn't defined yet

I'm new to coding and I'm a little confused. How/why can a for loop use a variable that isn't defined yet?
For example:
demond = {'green':'grass', 'red':'fire', 'yellow':'sun'}
for i in demond:
print(i)
Output:
green
yellow
red

In python, you don't need to declare variables. In C/C++/JAVA etc. you will have to declare them first and then use them.
Variables are nothing but reserved memory locations to store values.Based on the data type of a variable, the interpreter allocates memory and decides what can be stored in the reserved memory.Python variables do not need explicit declaration to reserve memory space. The declaration happens automatically when you assign a value to a variable.

There are two things that you need to keep in mind:
because Python is a weakly-typed language, you do not need to explicitly declare any variable to a certain object type. This is something you already know, and why you can assign things without having to state what type they will be.
For loop constructs do a lot of things in the background that you don't explicitly see. This means that although it doesnt LOOK like anything is being defined, it is.
With that in mind, I dont want to really explain how for loops work, because there are already answers available for that but the main point is that a for loop in python is the same as the following pseudo code.
#set up your iterable
demond = SOME_ITERABLE_OBJECT (this can be a list, string, dict, etc)
#this
for i in demond:
do_something(i)
#is the same as this
i = demond[0] # the first item in demond
do_something(i)
i = demond[1] # the second item in demond
do_something(i)
i = demond[2]
...
...
..
i = demond[n] # the last item in demond
do_something(i)
Now your follow up question may be this: what makes it so that, in your code, for i in demond sets i to equal to it's keys? Well that is just part of the design of python, specifically how dicts work. What the for loop is ACTUALLY doing is calling an iterables next() function until the iterable generator is done. Each iterable can have a different result from a for loop (see the first link).
NOTE:
In my code example, I am setting i = demond[some_index]. This looks like a list index grab but it is really meant to just show that is iterating through the list in some sort of order. IT IS PSUEDO CODE. Just keep that in mind.

Modifying variables in Python function is affecting variables with different names outside the function

I have a nested dictionary containing a bunch of data on a number of different objects (where I mean object in the non-programming sense of the word). The format of the dictionary is allData[i][someDataType], where i is a number designation of the object that I have data on, and someDataType is a specific data array associated with the object in question.
Now, I have a function that I have defined that requires a particular data array for a calculation to be performed for each object. The data array is called cleanFDF. So I feed this to my function, along with a bunch of other things it requires to work. I call it like this:
rm.analyze4complexity(allData[i]['cleanFDF'], other data, other data, other data)
Inside the function itself, I straight away re-assign the cleanFDF data to another variable name, namely clFDF. I.e. The end result is:
clFDF = allData[i]['cleanFDF']
I then have to zero out all of the data that lies below a certain threshold, as such:
clFDF[ clFDF < threshold ] = 0
OK - the function works as it is supposed to. But now when I try to plot the original cleanFDF data back in the main script, the entries that got zeroed out in clFDF are also zeroed out in allData[i]['cleanFDF']. WTF? Obviously something is happening here that I do not understand.
To make matters even weirder (from my point of view), I've tried to do a bodgy kludge to get around this by 'saving' the array to another variable before calling the function. I.e. I do
saveFDF = allData[i]['cleanFDF']
then run the function, then update the cleanFDF entry with the 'saved' data:
allData[i].update( {'cleanFDF':saveFDF} )
but somehow, simply by performing clFDF[ clFDF < threshold ] = 0 within the function modifies clFDF, saveFDF and allData[i]['cleanFDF'] in the main friggin' script, zeroing out all the entires at the same array indexes! It is like they are all associated global variables somehow, but I've made no such declarations anywhere...
I am a hopeless Python newbie, so no doubt I'm not understanding something about how it works. Any help would be greatly appreciated!

You are passing the value at allData[i]['cleanFDF'] by reference (decent explanation at https://stackoverflow.com/a/430958/337678). Any changes made to it will be made to the object it refers to, which is still the same object as the original, just assigned to a different variable.
Making a deep copy of the data will likely fix your issue (Python has a deepcopy library that should do the trick ;)).

Everything is a reference in Python.
def function(y):
y.append('yes')
return y
example = list()
function(example)
print(example)
it would return ['yes'] even though i am not directly changing the variable 'example'.
See Why does list.append evaluate to false?, Python append() vs. + operator on lists, why do these give different results?, Python lists append return value.

how to release used memory immediately in python list?

In many cases, you are sure you definitely won't use the list again, so you want the memory to be released right now.
a = [11,22,34,567,9999]
del a
I'm not sure if the above really releases the memory. You can use:
del a[:]
that actually removes all the elements in list a.
Is that the best way to release the memory?
def realse_list(a):
del a[:]
del a
I have the same question about tuples and sets.

def release_list(a):
del a[:]
del a
Do not ever do this. Python automatically frees all objects that are not referenced any more, so a simple del a ensures that the list's memory will be released if the list isn't referenced anywhere else. If that's the case, then the individual list items will also be released (and any objects referenced only from them, and so on and so on), unless some of the individual items were also still referenced.
That means the only time when del a[:]; del a will release more than del a on its own is when the list is referenced somewhere else. This is precisely when you shouldn't be emptying out the list: someone else is still using it!!!
Basically, you shouldn't be thinking about managing pieces of memory. Instead, think about managing references to objects. In 99% of all Python code, Python cleans up everything you don't need pretty soon after the last time you needed it, and there's no problem. Every time a function finishes all the local variables in that function "die", and if they were pointing to objects that are not referenced anywhere else they'll be deleted, and that will cascade to everything contained within those objects.
The only time you need to think about it is when you have a large object (say a huge list), you do something with it, and then you begin a long-running (or memory intensive) sub-computation, where the large object isn't needed for the sub-computation. Because you have a reference to it, the large object won't be released until the sub-computation finishes and then you return. In that sort of case (and only that sort of case), you can explicitly del your reference to the large object before you begin the sub-computation, so that the large object can be freed earlier (if no-one else is using it; if a caller passed the object in to you and the caller does still need it after you return, you'll be very glad that it doesn't get released).

Python uses Reference Count to manage its resource.
import sys
class foo:
pass
b = foo()
a = [b, 1]
sys.getrefcount(b) # gives 3
sys.getrefcount(a) # gives 2
a = None # delete the list
sys.getrefcount(b) # gives 2
In the above example, b's reference count will be incremented when you put it into a list, and as you can see, when you delete the list, the reference count of b get decremented too. So in your code
def release_list(a):
del a[:]
del a
was redundant.
In summary, all you need to do is assigning the list into a None object or use del keyword to remove the list from the attributes dictionary. (a.k.a, to unbind the name from the actual object). For example,
a = None # or
del a
When the reference count of an object goes to zero, python will free the memory for you. To make sure the object gets deleted, you have to make sure no other places reference the object by name, or by container.
sys.getrefcount(b) # gives 2
If sys.getrefcount gives you 2, that means you are the only one who had the reference of the object and when you do
b = None
it will get freed from the memory.

As #monkut notes, you probably shouldn't worry too much about memory management in most situations. If you do have a giant list that you're sure you're done with now and it won't go out of the current function's scope for a while, though:
del a simply removes your name a for that chunk of memory. If some other function or structure or whatever has a reference to it still, it won't be deleted; if this code has the only reference to that list under the name a and you're using CPython, the reference counter will immediately free that memory. Other implementations (PyPy, Jython, IronPython) might not kill it right away because they have different garbage collectors.
Because of this, the del a statement in your realse_list function doesn't actually do anything, because the caller still has a reference!
del a[:] will, as you note, remove the elements from the list and thus probably most of its memory usage.
You can do the_set.clear() for similar behavior with sets.
All you can do with a tuple, because they're immutable, is del the_tuple and hope nobody else has a reference to it -- but you probably shouldn't have enormous tuples!

If your worried about memory management and performance for data types why not use something like a linked double queue.
First its memory footprint is scattered though out the memory so you won't have to allocate a large chunk of continuous memory right off the bat.
Second you will see faster access times for enqueueing and dequeueing because unlike in a standard list when you remove lets say a middle element there is no need for sliding the rest of the list over in the index which takes time in large lists.
I should also note if you are using just integers I would suggest looking into a binary heap as you will see O(log^2n) access times compared to mostly O(N) with lists.

If you need to release list's memory, keeping the list's name, you can simply write a=[]