Python dictionary : TypeError: unhashable type: 'list'

The error you gave is due to the fact that in python, dictionary keys must be immutable types (if key can change, there will be problems), and list is a mutable type.

Your error says that you try to use a list as dictionary key, you'll have to change your list into tuples if you want to put them as keys in your dictionary.

According to the python doc :

The only types of values not acceptable as keys are values containing lists or dictionaries or other mutable types that are compared by value rather than by object identity, the reason being that the efficient implementation of dictionaries requires a key’s hash value to remain constant

This is indeed rather odd.

If aSourceDictionary were a dictionary, I don't believe it is possible for your code to fail in the manner you describe.

This leads to two hypotheses:

1. The code you're actually running is not identical to the code in your question (perhaps an earlier or later version?)

2. aSourceDictionary is in fact not a dictionary, but is some other structure (for example, a list).

As per your description, things don't add up. If aSourceDictionary is a dictionary, then your for loop has to work properly.

>>> source = {'a': [1, 2], 'b': [2, 3]}
>>> target = {}
>>> for key in source:
...   target[key] = []
...   target[key].extend(source[key])
... 
>>> target
{'a': [1, 2], 'b': [2, 3]}
>>> 

It works fine : http://codepad.org/5KgO0b1G, your aSourceDictionary variable may have other datatype than dict

aSourceDictionary = { 'abc' : [1,2,3] , 'ccd' : [4,5] }
aTargetDictionary = {}
for aKey in aSourceDictionary:
        aTargetDictionary[aKey] = []
        aTargetDictionary[aKey].extend(aSourceDictionary[aKey])
print aTargetDictionary

You can also use defaultdict to address this situation. It goes something like this:

from collections import defaultdict

#initialises the dictionary with values as list
aTargetDictionary = defaultdict(list)

for aKey in aSourceDictionary:
    aTargetDictionary[aKey].append(aSourceDictionary[aKey])