如何为霍夫曼编码和解码创建树？

Question

对于我的任务，我要对霍夫曼树进行编码和解码。我在创建树时遇到问题，而且卡住了。

不介意打印语句 - 它们只是供我测试并查看函数运行时的输出。

对于第一个for循环，我从我在主块中用于测试的文本文件中获取了所有值和索引。

在第二个for循环中，我将所有内容插入到优先级队列中。

我对于接下来要去哪里感到困惑 - 我正在尝试制作节点，但我对如何进展感到困惑。有人可以告诉我，如果我这样做了吗？

def _create_code(self, frequencies):
    '''(HuffmanCoder, sequence(int)) -> NoneType
    iterate over index into the sequence keeping it 256 elements long, '''
    #fix docstring
    p = PriorityQueue()
    print frequencies

    index = 0 
    for value in frequencies:
        if value != 0:
            print value #priority
            print index #elm
            print '-----------'       
        index = index + 1


    for i in range(len(frequencies)):
        if frequencies[i] != 0:
            p.insert(i, frequencies[i])  
            print i,frequencies[i]
            if p.is_empty():
                a = p.get_min()
                b = p.get_min()
                n1 = self.HuffmanNode(None, None, a)
                n2 = self.HuffmanNode(None, None, b)
                print a, b, n1, n2
    while not p.is_empty():
        p.get_min()

我手动插入前两个来启动我的树，这是正确的吗？

我该如何继续？我知道它的想法，只是代码方面我很困难。

顺便说一下，这是使用python。我试着看维基百科，我知道步骤，我只需要帮助代码以及我应该如何继续，谢谢！

HuffmanNode来自这个嵌套类：

class HuffmanNode(object):

    def __init__(self, left=None, right=None, root=None):
        self.left = left
        self.right = right
        self.root = root

Answer 1

维基百科中的霍夫曼算法告诉您如何创建节点树，因此您的程序可以基于该算法或其他类似算法。这是一个Python程序，其中的注释显示了相应的维基百科算法步骤。测试数据是英文文本中字母表字母的频率。

创建节点树后，需要将其向下移动以将霍夫曼代码分配给数据集中的每个符号。由于这是作业，这一步取决于你，但递归算法是处理它的最简单和最自然的方法。它只有六行代码。

import queue

class HuffmanNode(object):
    def __init__(self, left=None, right=None, root=None):
        self.left = left
        self.right = right
        self.root = root     # Why?  Not needed for anything.
    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)
print(node)

# Recursively walk the tree down to the leaves,
#   assigning a code value to each symbol
def walk_tree(node, prefix="", code={}):
    return(code)

code = walk_tree(node)
for i in sorted(freq, reverse=True):
    print(i[1], '{:6.2f}'.format(i[0]), code[i[1]])

在字母数据上运行时，生成的霍夫曼代码为：

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

Answer 2

@Dave walk_tree缺少树处理代码

# Recursively walk the tree down to the leaves,
# assigning a code value to each symbol
def walk_tree(node, prefix="", code={}):
    if isinstance(node[1].left[1], HuffmanNode):
        walk_tree(node[1].left,prefix+"0", code)
    else:
        code[node[1].left[1]]=prefix+"0"
    if isinstance(node[1].right[1],HuffmanNode):
        walk_tree(node[1].right,prefix+"1", code)
    else:
        code[node[1].right[1]]=prefix+"1"
    return(code)

Answer 3

另一个解决方案是返回字典{label:code}和包含结果图的递归字典tree。输入vals采用字典{label:freq}的形式：

def assign_code(nodes, label, result, prefix = ''):    
    childs = nodes[label]     
    tree = {}
    if len(childs) == 2:
        tree['0'] = assign_code(nodes, childs[0], result, prefix+'0')
        tree['1'] = assign_code(nodes, childs[1], result, prefix+'1')     
        return tree
    else:
        result[label] = prefix
        return label

def Huffman_code(_vals):    
    vals = _vals.copy()
    nodes = {}
    for n in vals.keys(): # leafs initialization
        nodes[n] = []

    while len(vals) > 1: # binary tree creation
        s_vals = sorted(vals.items(), key=lambda x:x[1]) 
        a1 = s_vals[0][0]
        a2 = s_vals[1][0]
        vals[a1+a2] = vals.pop(a1) + vals.pop(a2)
        nodes[a1+a2] = [a1, a2]        
    code = {}
    root = a1+a2
    tree = {}
    tree = assign_code(nodes, root, code)   # assignment of the code for the given binary tree      
    return code, tree

这可以用作：

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') ]    
vals = {l:v for (v,l) in freq}
code, tree = Huffman_code(vals)

text = 'hello' # text to encode
encoded = ''.join([code[t] for t in text])
print('Encoded text:',encoded)

decoded = []
i = 0
while i < len(encoded): # decoding using the binary graph
    ch = encoded[i]  
    act = tree[ch]
    while not isinstance(act, str):
        i += 1
        ch = encoded[i]  
        act = act[ch]        
    decoded.append(act)          
    i += 1

print('Decoded text:',''.join(decoded))

可以使用Graphviz将树可视化为：

该图由以下脚本生成（需要Graphviz）：

def draw_tree(tree, prefix = ''):    
    if isinstance(tree, str):            
        descr = 'N%s [label="%s:%s", fontcolor=blue, fontsize=16, width=2, shape=box];\n'%(prefix, tree, prefix)
    else: # Node description
        descr = 'N%s [label="%s"];\n'%(prefix, prefix)
        for child in tree.keys():
            descr += draw_tree(tree[child], prefix = prefix+child)
            descr += 'N%s -> N%s;\n'%(prefix,prefix+child)
    return descr

import subprocess
with open('graph.dot','w') as f:
    f.write('digraph G {\n')
    f.write(draw_tree(tree))
    f.write('}') 
subprocess.call('dot -Tpng graph.dot -o graph.png', shell=True)

Answer 4

@Dave类HuffmanNode（对象）有一个微妙的bug。当两个频率相等时，抛出异常：例如，让

    freq = [ (200/3101, 'd'), (100/3101, 'e'), (100/3101, 'f') ]

然后你得到TypeError：unorderable类型：HuffmanNode（）＆lt; STR（）。 问题与PriorityQueue实现有关。我怀疑当元组的第一个元素比较相等时，PriorityQueue想要比较第二个元素，其中一个是python对象。您将 lt 方法添加到您的课程中，问题就解决了。

    def __lt__(self,other):
        return 0