Use defaultdict

>>> from collections import defaultdict
>>> d = defaultdict(list) # create the dictionary, then populate it.
>>> d.update({"TIM":[['xx', 'yy'], ['aa', 'bb']], "SAM":[['yy', 'cc']]})
>>> d # see its what you wanted.
defaultdict(<type 'list'>, {'TIM': [['xx', 'yy'], ['aa', 'bb']], 'SAM': [['yy', 'cc']]})
>>> d["SAM"].append(['tt','uu']) # add more items to SAM
>>> d["KIM"].append(['ii','pp']) # create and add to KIM
>>> d # see its what you wanted.
defaultdict(<type 'list'>, {'TIM': [['xx', 'yy'], ['aa', 'bb']], 'KIM': [['ii', 'pp']], 'SAM': [['yy', 'cc'], ['tt', 'uu']]})

If you want the dictionary values to be sets, that is no problem:

>>> from collections import defaultdict
>>> d = defaultdict(set)
>>> d.update({"TIM":set([('xx', 'yy'), ('aa', 'bb')]), "SAM":set([('yy', 'cc')])})
>>> d["SAM"].add(('tt','uu'))
>>> d["KIM"].add(('ii','pp'))
>>> d
defaultdict(<type 'set'>, {'TIM': set([('xx', 'yy'), ('aa', 'bb')]), 'KIM': set([('ii', 'pp')]), 'SAM': set([('tt', 'uu'), ('yy', 'cc')])})

Use defaultdict from the collections module.

>>> from collections import defaultdict
>>> dict1 = {1:'a',2:'b',3:'c'}
>>> dict2 = {1:'hello', 4:'four', 5:'five'}
>>> my_dict = defaultdict(list)
>>> for k in dict1:
...    my_dict[k].append(dict1[k])
...
>>> for k in dict2:
...    my_dict[k].append(dict2[k])
...
>>> my_dict[1]
['a', 'hello']

A benchmark shows your suspicions of its performance impact appear to be correct:

$ python -m timeit -s 'd = {"key": "value"}' 'd["key"] = "value"'
10000000 loops, best of 3: 0.0741 usec per loop
$ python -m timeit -s 'd = {"key": "value"}' 'd.update(key="value")'
1000000 loops, best of 3: 0.294 usec per loop
$ python -m timeit -s 'd = {"key": "value"}' 'd.update({"key": "value"})'
1000000 loops, best of 3: 0.461 usec per loop

That is, it's about six times slower on my machine. However, Python is already not a language you'd use if you need top performance, so I'd just recommend use of whatever is most readable in the situation. For many things, that would be the [] way, though update could be more readable in a situation like this:

configuration.update(
    timeout=60,
    host='example.com',
)

…or something like that.

There are two approaches:

1. Use defaultdict
2. Use your own implementation of defaultdict

Assuming that your file looks like this:

a 1
b 4
a 2
...

Then you can do this:

import collections 
answer = collections.defaultdict(list) 
with open('path/to/file') as infile:
    for line in infile:
        key, value = line.strip().split()
        answer[key].append(value)

If you don't want to use defaultdict, then:

answer = {}
with open('path/to/file') as infile:
    for line in infile:
        key, value = line.strip().split()
        if key not in answer:
            answer[key] = []
        answer[key].append(value)

Hope this helps

The append () function is not the property of a dictionary. It is for a list to add new elements to the list. So that you are recommended to use .add() function to add a new element in the dictionary as follow:

dict['jobs'].add('doctor')