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Example implementation of Huffman coding in Python
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# Example Huffman coding implementation | |
# Distributions are represented as dictionaries of { 'symbol': probability } | |
# Codes are dictionaries too: { 'symbol': 'codeword' } | |
def huffman(p): | |
'''Return a Huffman code for an ensemble with distribution p.''' | |
assert(sum(p.values()) == 1.0) # Ensure probabilities sum to 1 | |
# Base case of only two symbols, assign 0 or 1 arbitrarily | |
if(len(p) == 2): | |
return dict(zip(p.keys(), ['0', '1'])) | |
# Create a new distribution by merging lowest prob. pair | |
p_prime = p.copy() | |
a1, a2 = lowest_prob_pair(p) | |
p1, p2 = p_prime.pop(a1), p_prime.pop(a2) | |
p_prime[a1 + a2] = p1 + p2 | |
# Recurse and construct code on new distribution | |
c = huffman(p_prime) | |
ca1a2 = c.pop(a1 + a2) | |
c[a1], c[a2] = ca1a2 + '0', ca1a2 + '1' | |
return c | |
def lowest_prob_pair(p): | |
'''Return pair of symbols from distribution p with lowest probabilities.''' | |
assert(len(p) >= 2) # Ensure there are at least 2 symbols in the dist. | |
sorted_p = sorted(p.items(), key=lambda (i,pi): pi) | |
return sorted_p[0][0], sorted_p[1][0] | |
# Example execution | |
ex1 = { 'a': 0.5, 'b': 0.25, 'c': 0.25 } | |
huffman(ex1) # => {'a': '0', 'c': '10', 'b': '11'} | |
ex2 = { 'a': 0.25, 'b': 0.25, 'c': 0.2, 'd': 0.15, 'e': 0.15 } | |
huffman(ex2) # => {'a': '01', 'c': '00', 'b': '10', 'e': '110', 'd': '111'} | |
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