Appending multiple values to dictionary appends "None" - python-3.x

Question

I try to add a list to a dictionary key, but when I append a value, it returns the value None. I have also tried collections.defaultdict(list) without success.
Code:
text = "ABBBAACCCCAABBCCCCAABCBCBCABCCCA"
chain = dict()
for i in range (0, text.__len__()-1):
key = text[i : i+1]
next_word = text[i +1 : i +2]
if key not in chain.keys():
chain.setdefault(key)
else:
chain.setdefault(key, [].append(next_word))
print(key, next_word, chain[key], chain)
Output:
A B None {'A': None}
B B None {'B': None, 'A': None}
B B None {'B': None, 'A': None}
B A None {'B': None, 'A': None}
…

Answer 1

[].append() returns None. You'd want to append to the return value of dict.setdefault() instead. You also don't need to do a key containment test when using dict.setdefault(); setdefault() already makes that test for you.
Next, don't call object.__len__(). Use len(object) instead. I'd also use {} instead of dict(); the latter has to look up a name and make a function call, the {} literal is compiled to a single bytecode to create a dictionary.
This works:
for i in range(len(text) - 1):
key = text[i:i + 1]
next_word = text[i + 1:i + 2]
chain.setdefault(key, []).append(next_word)
You could also use zip() to pair up the letters:
for key, next_word in zip(text, text[1:]):
chain.setdefault(key, []).append(next_word)
Demo:
>>> text = "ABBBAACCCCAABBCCCCAABCBCBCABCCCA"
>>> chain = {}
>>> for key, next_word in zip(text, text[1:]):
... chain.setdefault(key, []).append(next_word)
...
>>> chain
{'A': ['B', 'A', 'C', 'A', 'B', 'A', 'B', 'B'], 'B': ['B', 'B', 'A', 'B', 'C', 'C', 'C', 'C', 'C'], 'C': ['C', 'C', 'C', 'A', 'C', 'C', 'C', 'A', 'B', 'B', 'A', 'C', 'C', 'A']}
>>> from pprint import pprint
>>> pprint(chain)
{'A': ['B', 'A', 'C', 'A', 'B', 'A', 'B', 'B'],
'B': ['B', 'B', 'A', 'B', 'C', 'C', 'C', 'C', 'C'],
'C': ['C', 'C', 'C', 'A', 'C', 'C', 'C', 'A', 'B', 'B', 'A', 'C', 'C', 'A']}