Huffman coding python实现

python实现的Huffman coding，给26个英文字母编码，inspired by Dave. 他只给出了Huffman tree的构建，并将walk_tree留给了提问者自己完成。我将walk_tree实现了一下并输出结果，做个记录，也顺便分享给有需要的同学。

import queue

class HuffmanNode(object):
    def __init__(self, left=None, right=None, root=None):
        self.left = left
        self.right = right
    def children(self):
        return((self.left, self.right))

freq = [
    (8.167, 'a'), (1.492, 'b'), (2.782, 'c'), (4.253, 'd'),
    (12.702, 'e'),(2.228, 'f'), (2.015, 'g'), (6.094, 'h'),
    (6.966, 'i'), (0.153, 'j'), (0.747, 'k'), (4.025, 'l'),
    (2.406, 'm'), (6.749, 'n'), (7.507, 'o'), (1.929, 'p'), 
    (0.095, 'q'), (5.987, 'r'), (6.327, 's'), (9.056, 't'), 
    (2.758, 'u'), (1.037, 'v'), (2.365, 'w'), (0.150, 'x'),
    (1.974, 'y'), (0.074, 'z') ]

def create_tree(frequencies):
    p = queue.PriorityQueue()
    for value in frequencies:    # 1. Create a leaf node for each symbol
        p.put(value)             #    and add it to the priority queue
    while p.qsize() > 1:         # 2. While there is more than one node
        l, r = p.get(), p.get()  # 2a. remove two highest nodes
        node = HuffmanNode(l, r) # 2b. create internal node with children
        p.put((l[0]+r[0], node)) # 2c. add new node to queue      
    return p.get()               # 3. tree is complete - return root node

node = create_tree(freq)

#=================== 以上是Dave提供的思路  ==================
# 树里的每一个节点是一个tuple，node[0]是频率，node[1]是HuffmanNode 或者 character
# 下面是我的实现
def walk_tree(node, prefix="", code={}):
    """
    node 是一个tuple(freq, HuffmanNode|character)
    """
    if isinstance(node[1], HuffmanNode): # node[1]是一个HuffmanNode
        code1 = walk_tree(node[1].left, '0', code.copy()) # 这里如果直接传入code会出错
        code2 = walk_tree(node[1].right, '1', code.copy())
        if len(code1) > 0:
            for k, v in code1.items():
                code[k] = prefix + v
        if len(code2) > 0:
            for k, v in code2.items():
                code[k] = prefix + v
    else: # node[1]是一个字符
        code[node[1]] = prefix
    return(code) 

code = walk_tree(node) 
# 输出每个字母的编码
for i in sorted(freq, reverse=True):
    try:
        print(i[1], '{:6.2f}'.format(i[0]), code[i[1]])
    except Exception as e:
        print(e)
        continue

结果如下：

e  12.70 100
t   9.06 000
a   8.17 1110
o   7.51 1101
i   6.97 1011
n   6.75 1010
s   6.33 0111
h   6.09 0110
r   5.99 0101
d   4.25 11111
l   4.03 11110
c   2.78 01001
u   2.76 01000
m   2.41 00111
w   2.37 00110
f   2.23 00100
g   2.02 110011
y   1.97 110010
p   1.93 110001
b   1.49 110000
v   1.04 001010
k   0.75 0010111
j   0.15 001011011
x   0.15 001011010
q   0.10 001011001
z   0.07 001011000