Replacing items in a dictionary of lists in Python

Question

I am new to python and programming and need some help replacing items in a dictionary of lists. I would like to replace None with 'None' in the dictionary below:

dict = {'Chester100': ['Caesar, Augustus', '05/10/2012', '09/09/2012', None],
        'Rochester102': ['Henrich, Norton', '08/18/2014', '12/17/2014', None],
        'Rochester100': ['Caeser, Julius', '08/18/2014', '12/17/2014', None],
        'Rochester101': [None, None, None, '08/18/2012']}

My code is as follows:

new_dict = {}

for i in dict: #This accesses each dictionary key.
    temp = []
    for j in dict[i]: #This iterates through the inner lists
        if j is None:
            temp.append('None')
        else:
            temp.append(j)
        temp2 = {str(i):temp}
        new_dict.update(temp2)

    print(new_dict)

yields

{'Chester100': ['Caesar, Augustus', '05/10/2012', '09/09/2012', 'None'], 
'Rochester102': ['Henrich, Norton', '08/18/2014', '12/17/2014', 'None'], 
'Rochester100': ['Caeser, Julius', '08/18/2014', '12/17/2014', 'None'], 
'Rochester101': ['None', 'None', 'None', '08/18/2012']}

Is there a way to do this in a fewer lines of code or more efficiently using list comprehension or other methods? Should nested for loop (as I have it in my code) be avoided? Thanks.

Using Python 3.4.1

Answer 1

Use a dictionary comprehension:

>>> {k:[e if e is not None else 'None' for e in v] for k,v in di.items()}
{'Rochester102': ['Henrich, Norton', '08/18/2014', '12/17/2014', 'None'], 'Rochester100': ['Caeser, Julius', '08/18/2014', '12/17/2014', 'None'], 'Rochester101': ['None', 'None', 'None', '08/18/2012'], 'Chester100': ['Caesar, Augustus', '05/10/2012', '09/09/2012', 'None']}

And don't name a dict dict since it will mask the built in function by that name.

---

If you have huge dicts or lists, you may want to modify your data in place. If so, this may be the most efficient:

for key, value in di.items():
    for i, e in enumerate(value):
        if e is None: di[key][i]='None'