Discussion: pros/cons of array-backed and linked structures¶
Between the array-backed and linked list we have:
- $O(1)$ indexing (array-backed)
- $O(1)$ appending (array-backed & linked)
- $O(1)$ insertion/deletion without indexing (linked)
- $O(\log N)$ binary search, when sorted (array-backed)
Python's other built-in DS: the dict¶
import timeit
def lin_search(lst, x):
for i in range(len(lst)):
if lst[i] == x:
return i
raise ValueError(x)
def bin_search(lst, x):
# assume that lst is sorted!!!
low = 0
hi = len(lst)
mid = (low + hi) // 2
while lst[mid] != x and low <= hi:
if lst[mid] < x:
low = mid + 1
else:
hi = mid - 1
mid = (low + hi) // 2
if lst[mid] == x:
return mid
else:
raise ValueError(x)
def time_lin_search(size):
return timeit.timeit('lin_search(lst, random.randrange({}))'.format(size), # interpolate size into randrange
'import random ; from __main__ import lin_search ;'
'lst = [x for x in range({})]'.format(size), # interpolate size into list range
number=100)
def time_bin_search(size):
return timeit.timeit('bin_search(lst, random.randrange({}))'.format(size), # interpolate size into randrange
'import random ; from __main__ import bin_search ;'
'lst = [x for x in range({})]'.format(size), # interpolate size into list range
number=100)
def time_dict(size):
return timeit.timeit('dct[random.randrange({})]'.format(size),
'import random ; '
'dct = {{x: x for x in range({})}}'.format(size),
number=100)
lin_search_timings = [time_lin_search(n)
for n in range(10, 10000, 100)]
bin_search_timings = [time_bin_search(n)
for n in range(10, 10000, 100)]
dict_timings = [time_dict(n)
for n in range(10, 10000, 100)]
%matplotlib inline
import matplotlib.pyplot as plt
#plt.plot(lin_search_timings, 'ro')
plt.plot(bin_search_timings, 'gs')
plt.plot(dict_timings, 'b^')
plt.show()
A naive lookup DS¶
class LookupDS:
def __init__(self):
self.data = []
def __setitem__(self, key, value):
pass
def __getitem__(self, key):
pass
def __contains__(self, key):
pass
class LookupDS:
def __init__(self):
self.data = []
def __setitem__(self, key, value):
for i in range(len(self.data)):
if self.data[i][0] == key:
self.data[i][1] = value
return
else:
self.data.append([key, value])
def __getitem__(self, key):
for k, v in self.data:
if k == key:
return v
else:
raise KeyError
def __contains__(self, key):
try:
_ = self[key] #calls __getitem__; if getting something return True; if KeyError returns, return False
return True
except:
return False
d = LookupDS()
d['hello'] = 'hola'
d['goodbye'] = 'adios'
d['hello']
d['goodbye']
d['hello'] = 'bonjour'
d['hello']
d.data
Direct lookups via Hashing¶
Hashes (a.k.a. hash codes or hash values) are simply numerical values computed for objects.
hash('hello') #the value could be different on your machine
[hash(s) for s in ['different', 'objects', 'have', 'very', 'different', 'hashes']]
hash('aa'), hash('ab')
5093 % 100
5093 // 100
for i in range(1,20):
print(i, '% 6 => ', i%6)
-4 % 3
-4 // 3
hash('hello') % 85
Hashtables¶
class Hashtable:
def __init__(self, n_buckets=1000):
self.buckets = [None] * n_buckets
def __setitem__(self, key, val):
pass
def __getitem__(self, key):
pass
def __contains__(self, key):
try:
_ = self[key]
return True
except:
return False
class Hashtable:
def __init__(self, n_buckets=1000):
self.buckets = [None] * n_buckets
def __setitem__(self, key, val):
bidx = hash(key) % len(self.buckets)
self.buckets[bidx] = val
def __getitem__(self, key):
bidx = hash(key) % len(self.buckets)
if self.buckets[bidx]:
return self.buckets[bidx]
else:
raise KeyError
def __contains__(self, key):
try:
_ = self[key]
return True
except:
return False
ht = Hashtable()
ht['hello'] = 'hola'
ht['goodbye'] = 'adios'
ht['hello']
ht['goodbye']
ht.buckets
#any problems?
ht = Hashtable(2)
ht['hello'] = 'hola'
ht['byebye'] = 'adios'
ht['hello']
ht['byebye']
ht['eat'] = 'comer'
ht['hello']
ht['eat']
ht['byebye']
class Hashtable:
def __init__(self, n_buckets=1000):
self.buckets = [None] * n_buckets
def __setitem__(self, key, val):
bidx = hash(key) % len(self.buckets)
self.buckets[bidx] = [key, val]
def __getitem__(self, key):
bidx = hash(key) % len(self.buckets)
if self.buckets[bidx] and self.buckets[bidx][0] == key:
return self.buckets[bidx][1]
else:
raise KeyError
def __contains__(self, key):
try:
_ = self[key]
return True
except:
return False
ht = Hashtable(2)
ht['hello'] = 'hola'
ht['byebye'] = 'adios'
ht['hello']
ht['byebye']
ht['eat']
On Collisions¶
The "Birthday Problem"¶
Problem statement: Given $N$ people at a party, how likely is it that at least two people will have the same birthday?
# Explanation of "Birthday Paradox"
# What is the probability that two people in the room have the same birthday?
# P(n) = 1 – (365! / (365n * (365-n)!)
# because: any 2 persons do NOT have the same bDay is
# 364/365 * 363/365 * 362/365 … 365-(n-1)/365
# = [365! / (365-n)!] / 365n
# Observation: P(n) > 0.5 for n = 23
def birthday_p(n_people):
p_inv = 1
for n in range(365, 365-n_people, -1):
p_inv *= n / 365
return 1 - p_inv
birthday_p(2)
birthday_p(10)
birthday_p(50)
%matplotlib inline
import matplotlib.pyplot as plt
n_people = range(1, 80)
plt.plot(n_people, [birthday_p(n) for n in n_people])
plt.show()
General collision statistics¶
Repeat the birthday problem, but with a given number of values and "buckets" that are allotted to hold them. How likely is it that two or more values will map to the same bucket?
def collision_p(n_values, n_buckets):
p_inv = 1
for n in range(n_buckets, n_buckets-n_values, -1):
p_inv *= n / n_buckets
return 1 - p_inv
collision_p(23, 365) # same as birthday problem, for 23 people, 365 days
collision_p(10, 100) # number of values = 10, number of buckets = 100
collision_p(100, 1000) # number of values = 100, number of buckets = 1,000
collision_p(100, 10000) # number of values = 100, number of buckets = 10,000
# keeping number of values fixed at 100, but vary number of buckets: visualize probability of collision
%matplotlib inline
import matplotlib.pyplot as plt
n_buckets = range(100, 100001, 1000)
plt.plot(n_buckets, [collision_p(100, nb) for nb in n_buckets])
plt.show()
def avg_num_collisions(n, b):
"""Returns the expected number of collisions for n values uniformly distributed
over a hashtable of b buckets. Based on (fairly) elementary probability theory.
(Pay attention in MATH 474!)"""
return n - b + b * (1 - 1/b)**n
avg_num_collisions(28, 365) #number of values = 28, number of buckets = 365
avg_num_collisions(1000, 1000)
avg_num_collisions(1000, 10000)
Dealing with Collisions¶
To deal with collisions in a hashtable, we simply create a "chain" of key/value pairs for each bucket where collisions occur. The chain needs to be a data structure that supports quick insertion — natural choice: the linked list!
class Hashtable:
class Node:
def __init__(self, key, val, next=None):
self.key = key
self.val = val
self.next = next
def __init__(self, n_buckets=1000):
self.buckets = [None] * n_buckets
def __setitem__(self, key, val):
bucket_idx = hash(key) % len(self.buckets)
pass
def __getitem__(self, key):
bucket_idx = hash(key) % len(self.buckets)
pass
def __contains__(self, key):
try:
_ = self[key]
return True
except:
return False
class Hashtable:
class Node:
def __init__(self, key, val, next=None):
self.key = key
self.val = val
self.next = next
def __init__(self, n_buckets=1000):
self.buckets = [None] * n_buckets
def __setitem__(self, key, val):
bucket_idx = hash(key) % len(self.buckets)
# code logic
# get the node at the bucket_idx [] location,
# while not none, walk the list looking for the key,
# if found, set the value
# else insert a new node at start of the list
if not self.buckets[bucket_idx]:
# chain is empty
self.buckets[bucket_idx] = Hashtable.Node(key, val)
else:
n = self.buckets[bucket_idx]
while n:
if n.key == key: #found the key in an existing node
n.val = val
break
n = n.next
else: #prepend the node
self.buckets[bucket_idx] = Hashtable.Node(key, val,
next=self.buckets[bucket_idx])
def __getitem__(self, key):
bucket_idx = hash(key) % len(self.buckets)
#code logic
# get the node at the bucket_idx [] location
# while not none, walk the list looking for the key
# if found, return the value
#else raise KeyError
n = self.buckets[bucket_idx]
while n:
if n.key == key:
return n.val
n = n.next
raise KeyError
def __contains__(self, key):
try:
_ = self[key]
return True
except:
return False
ht = Hashtable(2)
for k, v in (('a', 'apple'), ('b', 'banana'), ('c', 'cat')):
ht[k] = v
for k in 'abc':
print(k, '=>', ht[k])
# do we really get O(1) access?
def prep_ht(size):
ht = Hashtable(size*10)
for x in range(size):
ht[x] = x
return ht
def time_ht(size):
return timeit.timeit('ht[random.randrange({})]'.format(size),
'import random ; from __main__ import prep_ht ;'
'ht = prep_ht({})'.format(size),
number=100)
# Explain timeit()
# test code
# ht[random.randrange({})]'.format(size)
# ht[100] ht[99] ht[10]
# setup code
# import random
# from __main__ import prep_ht
# ht = prep_ht({}).format(size)
# ht = prep_ht(100)
# number executions = 100
# order of execution:
# setup code (once)
# repeat test code 100 times
# return the time measurement
ht_timings = [time_ht(n)
for n in range(10, 10000, 100)]
%matplotlib inline
import matplotlib.pyplot as plt
plt.plot(bin_search_timings, 'ro')
plt.plot(ht_timings, 'gs')
plt.plot(dict_timings, 'b^')
plt.show()
Loose ends¶
Iteration¶
class Hashtable(Hashtable):
def __iter__(self):
pass
class Hashtable(Hashtable):
def __iter__(self):
# walk the bucket list
# walk the linked list
# yield keys as you go
for b in self.buckets:
while b:
yield b.key
b = b.next
ht = Hashtable(10)
for k, v in (('a', 'apple'), ('b', 'banana'), ('c', 'cat')):
ht[k] = v
#test iteration
for k in ht:
print(k, '=>', ht[k])
"Load factor" and Rehashing¶
It doesn't often make sense to start with a large number of buckets, unless we know in advance that the number of keys is going to be vast — also, the user of the hashtable would typically prefer to not be bothered with implementation details (i.e., bucket count) when using the data structure.
Instead: start with a relatively small number of buckets, and if the ratio of keys to the number of buckets (known as the load factor) is above some desired threshold — which we can determine using collision probabilities — we can dynamically increase the number of buckets. This requires, however, that we rehash all keys and potentially move them into new buckets (since the hash(key) % num_buckets mapping will likely be different with more buckets).
Other APIs¶
- FIXED
__setitem__(to update value for existing key) __delitem__keys&values(return iterators for keys and values)setdefault
Runtime analysis & Discussion¶
For a hashtable with $N$ key/value entries:
- Insertion: $O(N)$
- Lookup: $O(N)$
- Deletion: $O(N)$
A few words about the HashTable Lab¶
It is often convenient to be able to iterate over the keys in a hashtable in the order in which they were first inserted. But our implementation makes this impossible because of the unpredictable nature of the hash function.
We can fix this by introducing a separate array-backed list (we call this "entries") to which we append key/value pairs as they are added to the hashtable. To facilitate rapid search, we will continue to use the original buckets + chains structure, but it will merely contain indices of entries in the first list (we call this structure "indices").
The operations will work as follows:
get will use the hashcode of the provided key to locate the appropriate bucket in entries, then iterate over indices in the attached linked-list chain to search for a matching key in entries.
set will conduct the same search as get to see if the provided key is already in entries; if it is, the value will be updated. If the key isn't already in the hashtable, the key/value pair will be appended to entries, and a new node will be inserted into indices (in the correct bucket).
del will conduct the same search as get; if the key is present, its entries slot will simply be set to None (so that it can be skipped during iteration), and the corresponding node deleted from indices
iteration: is easy, just walk through the "entries" (an array-backed list)
Vocabulary list¶
- hashtable
- hashing and hashes
- collision
- hash buckets & chains
- birthday problem
- load factor
- rehashing
Addendum: On Hashability¶
Remember: a given object must always hash to the same value. This is required so that we can always map the object to the same hash bucket.
Hashcodes for collections of objects are usually computed from the hashcodes of its contents, e.g., the hash of a tuple is a function of the hashes of the objects in said tuple:
hash(('two', 'strings'))
This is useful. It allows us to use a tuple, for instance, as a key for a hashtable.
However, if the collection of objects is mutable — i.e., we can alter its contents — this means that we can potentially change its hashcode.`
If we were to use such a collection as a key in a hashtable, and alter the collection after it's been assigned to a particular bucket, this leads to a serious problem: the collection may now be in the wrong bucket (as it was assigned to a bucket based on its original hashcode)!
For this reason, only immutable types are, by default, hashable in Python. So while we can use integers, strings, and tuples as keys in dictionaries, lists (which are mutable) cannot be used. Indeed, Python marks built-in mutable types as "unhashable", e.g.,
hash([1, 2, 3])
That said, Python does support hashing on instances of custom classes (which are mutable). This is because the default hash function implementation does not rely on the contents of instances of custom classes. E.g.,
class Student:
def __init__(self, fname, lname):
self.fname = fname
self.lname = lname
s = Student('John', 'Doe')
hash(s)
s.fname = 'Jane'
hash(s) # same as before mutation
We can change the default behavior by providing our own hash function in __hash__, e.g.,
class Student:
def __init__(self, fname, lname):
self.fname = fname
self.lname = lname
def __hash__(self):
return hash(self.fname) + hash(self.lname)
s = Student('John', 'Doe')
hash(s)
s.fname = 'Jane'
hash(s)
But be careful: instances of this class are no longer suitable for use as keys in hashtables (or dictionaries), if you intend to mutate them after using them as keys!