Pure python implementation of a dictionary of fixed length

DictionaryInPython.py

__author__ = 'svalleru'
"""If 'YOU' can't solve it, you can't ask the interviewee to solve it ;)"""
class Dictionary(object):
    """Pure python implementation of a dictionary of fixed length"""

    def __init__(self, size):
        self.size = size
        self.data = [[] for _ in xrange(size)]

    #hash method
    def _hash(self, key):
        return hash(key)%self.size

    #insert & update
    def __setitem__(self, key, value):
        #no hash collision
        if len(self.data[self._hash(key)]) == 0:
            self.data[self._hash(key)].append([key, value])
        #hash collision
        else:
            #update the value for matching key
            for i, kv in enumerate(self.data[self._hash(key)]):
                if kv[0] == key:
                    self.data[self._hash(key)][i][1] = value
                    return
            #no matching key found in the bucket; append
            else:
                self.data[self._hash(key)].append([key, value])
    #retrieve            
    def __getitem__(self, key):
            for i, kv in enumerate(self.data[self._hash(key)]):
                if kv[0] == key:
                    return self.data[self._hash(key)][i][1]
            #no matching key found in the bucket
            else:
                return None

    #delete
    def __delitem__(self, key):
        #only one kv in the bucket
        if len(self.data[self._hash(key)]) == 1:
            self.data[self._hash(key)].pop(0)
        #bucket of len > 1
        else:
            for i, kv in enumerate(self.data[self._hash(key)]):
                if kv[0] == key:
                    self.data[self._hash(key)].pop(i)
        
d = Dictionary(10)
d['k1'] = 'v1'
print d['k1']
d['k1'] = 'v1.1'
print d['k1']
d['k2'] = 'v2'
d[9] = 11
print d[9]
d[9] = 12
print d[9]
del d[9]
print d[9]
print d['k2']
del d['k2']
print d['k2']

# Output
# v1
# v1.1
# 11
# 12
# None
# v2
# None