If you only need to check keys that are starting with "seller_account", you don't need regex, just use startswith()

my_dict={'account_0':123445,'seller_account':454545,'seller_account_0':454676, 'seller_account_number':3433343}

for key, value in my_dict.iteritems():   # iter on both keys and values
        if key.startswith('seller_account'):
                print key, value

or in a one_liner way :

result = [(key, value) for key, value in my_dict.iteritems() if key.startswith("seller_account")]

NB: for a python 3.X use, replace iteritems() by items() and don't forget to add () for print.

I wouldn't try to make a function with all of these responsibilities. It will make testing extremely difficult. Instead I would use the simplest list comprehensions possible for each case:

import re

dic = (
    {"name": "Kan",    "number": "2ABC345", "year": 2000},
    {"name": "Jhon",   "number": "2TTC345", "year": 2001},
    {"name": "Louise", "number": "2ABC366", "year": 2001},
    {"name": "Kevin",  "number": "2ABY000", "year": 2002},
)


# So let's say I want the year 2000
year_2000 = [d for d in dic if d["year"] == 2000]

# or the first letter in name be a k
name_k = [d for d in dic if d["name"].startswith("K")]

# or to see if the number starts with 2 and has 7 numbers
starts_2_digits_7 = [d for d in dic if re.match(r"^2\d{6}$", d["number"])]

This should match all the keys in your example as well as your exception case:

new_dict = {k:dict1[k] for k in dict1 if re.match('[^\d\s]+\d{1,2}$', k)}

Using a new example dict with your exception in it:

>>> dict1 = {"key1": 2345, "key2": 356, "key3": 773, "key44": 88, "key333": 12, "key3X": 13, "hello13": 435, "hello4325": 345, "3hi33":3}
>>> new_dict = {k:dict1[k] for k in dict1 if re.match('[^\d\s]+\d{1,2}$', k)}
>>> print(new_dict)
{'hello13': 435, 'key44': 88, 'key3': 773, 'key2': 356, 'key1': 2345}

IIUC, I might do do something like

allowed = [msg for msg in collected 
           if not any( dm.search(msg['service']) 
                       for dm in denied_metrics) ]

For example:

>>> pprint.pprint(collected)
[{'denied': False, 'metric': 1.0, 'service': 'ab'},
 {'denied': False, 'metric': 1.0, 'service': 'bc'},
 {'denied': False, 'metric': 1.0, 'service': 'ca'},
 {'denied': False, 'metric': 1.0, 'service': 'cb'},
 {'denied': False, 'metric': 1.0, 'service': 'bc'}]
>>> denied_metrics = [re.compile("a"), re.compile("c$")]
>>> allowed = [msg for msg in collected 
               if not any(dm.search(msg['service'])
               for dm in denied_metrics)]
>>> allowed
[{'metric': 1.0, 'service': 'cb', 'denied': False}]

Whether you want search or match depends upon your regexes, of course. [BTW, wouldn't 'denied_services' be a better name?]

Constructing a new dict:

dict_you_want = {key: old_dict[key] for key in your_keys}

Uses dictionary comprehension.

If you use a version which lacks them (ie Python 2.6 and earlier), make it dict((key, old_dict[key]) for ...). It's the same, though uglier.

Note that this, unlike jnnnnn's version, has stable performance (depends only on number of your_keys) for old_dicts of any size. Both in terms of speed and memory. Since this is a generator expression, it processes one item at a time, and it doesn't looks through all items of old_dict.

Removing everything in-place:

unwanted = set(old_dict) - set(your_keys)
for unwanted_key in unwanted: del your_dict[unwanted_key]

You do not have to pass d.items() as an iterable to filter. Just pass the dict object (same as dict.keys()) and filter the content based on dict[key] with lambda expression as:

>>> number_key = {35135135: 5, 60103513: 3, 10981179: 2, 18637724 : 4}

>>> filter(lambda x: number_key[x]<=3, number_key)
[60103513, 10981179]

You say that key is "city" and value is "Philadelphi", if we replace these in the query you make you basically write:

table.objects.all().filter(city= r'.*?Philadelphi*?$')

If the problem is not yet visible, then when one doesn't specify a lookup it is automatically assumed to be an exact lookup, you instead want to either use the regex or iregex lookup. Also you should escape your value when you put it in a regex:

import re

key = '%s__iregex' % key
value = r'.*?%s.*?$' % re.escape(value)
dict = {key: value}

table.objects.all().filter(**dict)

Also as suggested in the comments your regex doesn't seem to be different from a contains lookup so unless you have some different regex you should use:

key = '%s__contains' % key
dict = {key: value}

table.objects.all().filter(**dict)

How about a dict comprehension:

filtered_dict = {k: v for k, v in d.iteritems() if filter_string in k}

One you see it, it should be self-explanatory, as it reads like English pretty well.

This syntax requires Python 2.7 or greater.

In Python 3, there is only dict.items(), not iteritems() so you would use:

filtered_dict = {k: v for k, v in d.items() if filter_string in k}