When you use update like this, you are using keyword arguments, where you can pass parameters by name instead of by position in the list of arguments. The keyword in such arguments is treated as a name; the only way to have a variable keyword is to use the **kwargs syntax to pass an entire dict as a set of key=value parameters. That means that this would work . . .

update = { Var1: Var2 }
Dict.update(**update)

. . . except that keywords have to be strings, not numbers; trying to use your Var1 with value 1 as a keyword triggers a TypeError.

Because dict.update is designed for updating a dictionary, however, it also accepts an actual dictionary as an argument, without conversion to keyword arguments via **. So this does work:

Dict.update({ Var1: Var2 })

But you don't need update or its keyword arguments in your case; you can just use assignment instead:

Dict[Var1] = Var2

A couple style things: don't name variables with capital letters; that is usually an indication that a value is a constant. Also, don't use "Dict" as a variable name. Such names risk shadowing builtin functions, since functions and variables share the same namespace in Python. For example, if you assign a value to dict, you've just removed your pointer to the dict global function.

The problem that when I try to add more than one K:V to the dictionary, it adds a key but replaces all the previously stored values(in this case it is an array/list) with the newest one.

The technique to solve this problem is that every key, badgeNum, news a list of values rather than a single value.

>>> mydict = {}
>>> mydict.setdefault(1234, []).append('first 1234')
>>> mydict.setdefault(5678, []).append('first 5678')
>>> mydict.setdefault(1234, []).append('second 1234')
>>> mydict.setdefault(5678, []).append('second 5678')
>>> mydict
{1234: ['first 1234', 'second 1234'], 5678: ['first 5678', 'second 5678']}

Another approach is to use collections.defaultdict() which takes a little more effort to set-up but makes it simpler (and a bit faster) to add new values:

>>> from collections import defaultdict
>>> mydict = defaultdict(list)
>>> mydict[1234].append('first 1234')
>>> mydict[5678].append('first 5678')
>>> mydict[1234].append('second 1234')
>>> mydict[5678].append('second 5678')
>>> mydict
defaultdict(<class 'list'>, {1234: ['first 1234', 'second 1234'], 5678: ['first 5678', 'second 5678']})

Python has this feature built-in:

>>> d = {'b': 4}
>>> d.update({'a': 2})
>>> d
{'a': 2, 'b': 4}

Or given you're not allowed to use dict.update:

>>> d = dict(d.items() + {'a': 2}.items())   # doesn't work in python 3

Well you could directly substract from the value by just referencing the key. Which in my opinion is simpler.

>>> books = {}
>>> books['book'] = 3       
>>> books['book'] -= 1   
>>> books   
{'book': 2}   

In your case:

book_shop[ch1] -= 1

1. You can update many keys on the same statement.

my_dict.update(other_dict)

In this case you don't have to know how many keys are in the other_dict. You'll just be sure that all of them will be updated on my_dict.

2. You can use any iterable of key/value pairs with dict.update

As per the documentation you can use another dictionary, kwargs, list of tuples, or even generators that yield tuples of len 2.

3. You can use the update method as an argument for functions that expect a function argument.

Example:

def update_value(key, value, update_function):
    update_function([(key, value)])

update_value("k", 3, update_on_the_db)  # suppose you have a update_on_the_db function
update_value("k", 3, my_dict.update)  # this will update on the dict