a2fa88f561
Instead of calling `filter(..)`.
72 lines
2.1 KiB
Python
72 lines
2.1 KiB
Python
def compute_hash(x):
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"""
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Compute a unique fingerprint for the string input, `x`, as an integer using
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the following equation:
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x[0] * P^0 + x[1] * P^1 + ... x[n-1] * P^(n-1) % M
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P and M are constants where P represents the next available prime number
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that's GTE the number of unique characters you'll be hashing. In the case of
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all lowercase characters, of which there are 26, the next available prime
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number is 31.
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"""
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p = 31
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m = int(10e9) + 9 # large prime number
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power = 0
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result = 0
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for c in x:
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result += ord(c) * p**power
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power += 1
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return result % m
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class HashTable(object):
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def __init__(self, size):
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"""
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Create a hash table with `size` buckets.
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"""
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buckets = []
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for _ in range(size):
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buckets.append([])
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self.xs = buckets
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self.compute_hash = lambda k: compute_hash(k) % size
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def __repr__(self):
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result = []
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for bucket in self.xs:
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for entry in bucket:
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result.append(entry)
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return "HashTable({})".format(",".join(str(x) for x in result))
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def get(self, key):
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"""
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Attempt to retrieve value stored under `key`.
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"""
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h = self.compute_hash(key)
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for k, v in self.xs[h]:
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if k == key:
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return v
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return None
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def put(self, key, val):
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"""
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Set `key` to `val`; update value at `key` if it already exists.
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"""
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h = self.compute_hash(key)
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for i in range(len(self.xs[h])):
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# Update entry if the key exists...
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if self.xs[h][i][0] == key:
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self.xs[h][i] = (key, val)
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return None
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# ...create a new entry otherwise
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self.xs[h].append((key, val))
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def delete(self, key):
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"""
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Remove entry `key` from the hash table.
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"""
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h = self.compute_hash(key)
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for i in range(len(self.xs[h])):
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k, v = self.xs[h][i]
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if k == key:
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self.xs[h].remove((k, v))
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return
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