You are correct in that your list contains a reference to the original dictionary.

a.append(b.copy()) should do the trick.

Bear in mind that this makes a shallow copy. An alternative is to use copy.deepcopy(b), which makes a deep copy.

list.append returns None, since it is an in-place operation and you are assigning it back to dates_dict[key]. So, the next time when you do dates_dict.get(key, []).append you are actually doing None.append. That is why it is failing. Instead, you can simply do

dates_dict.setdefault(key, []).append(date)

But, we have collections.defaultdict for this purpose only. You can do something like this

from collections import defaultdict
dates_dict = defaultdict(list)
for key, date in cur:
    dates_dict[key].append(date)

This will create a new list object, if the key is not found in the dictionary.

Note: Since the defaultdict will create a new list if the key is not found in the dictionary, this will have unintented side-effects. For example, if you simply want to retrieve a value for the key, which is not there, it will create a new list and return it.

You just have to iterate the list and increment the count against the key if it is already there, otherwise set it to 1.

>>> d = {'a': 1, 'b': 2}
>>> l = ['a', 'b', 'c', 'd', 'e']
>>> for item in l:
...     if item in d:
...         d[item] += 1
...     else:
...         d[item] = 1
>>> d
{'a': 2, 'c': 1, 'b': 3, 'e': 1, 'd': 1}

You can write the same, succinctly, with dict.get, like this

>>> d = {'a': 1, 'b': 2}
>>> l = ['a', 'b', 'c', 'd', 'e']
>>> for item in l:
...     d[item] = d.get(item, 0) + 1
>>> d
{'a': 2, 'c': 1, 'b': 3, 'e': 1, 'd': 1}

dict.get function will look for the key, if it is found it will return the value, otherwise it will return the value you pass in the second parameter. If the item is already a part of the dictionary, then the number against it will be returned and we add 1 to it and store it back in against the same item. If it is not found, we will get 0 (the second parameter) and we add 1 to it and store it against item.

Now, to get the total count, you can just add up all the values in the dictionary with sum function, like this

>>> sum(d.values())
8

The dict.values function will return a view of all the values in the dictionary. In our case it will numbers and we just add all of them with sum function.