You can do this:

s = dict([ (k,r) for k,r in mydict.iteritems() if r['x'] > 92 and r['x'] < 95 and r['y'] > 70 and r['y'] < 75 ])

This takes a dict as you specified and returns a 'filtered' dict.

Use a dict comprehension (Python 2.7 and later):

{key: value for key, value in zip(keys, values)}

Alternatively, use the dict constructor:

pairs = [('a', 1), ('b', 2)]
dict(pairs)                          # → {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs)  # → {'a': 11, 'b': 12}

Given separate lists of keys and values, use the dict constructor with zip:

keys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values))              # → {'a': 1, 'b': 2}

Using defaultdict:

from collections import defaultdict
D = defaultdict(list)
[D[v].append(i) for i, v in enumerate(A)]

Using setdefault:

D = {}
[D.setdefault(v, []).append(i) for i, v in enumerate(A)]

I can't figure any mean to use a dictionnary comprehension without sorting the data:

from itertools import groupby
from operator import itemgetter
{v: ids for v, ids in groupby(enumerate(sorted(A)), itemgetter(1))}

Performances:

from collections import defaultdict
from itertools import groupby
from operator import itemgetter
from random import randint

A = tuple(randint(0, 100) for _ in range(1000))

def one():
    D = defaultdict(list)
    [D[v].append(i) for i, v in enumerate(A)]

def two():
    D = {}
    [D.setdefault(v, []).append(i) for i, v in enumerate(A)]

def three():
    {v: ids for v, ids in groupby(enumerate(sorted(A)), itemgetter(1))}


from timeit import timeit

for func in (one, two, three):
    print(func.__name__ + ':', timeit(func, number=1000))

Results (as always, the simplest win):

one: 0.25547646999984863
two: 0.3754340969971963
three: 0.5032370890003222