pythonはハフマン符号化を実現

一、問題
二叉木の構造を利用してHuffman木を符号化し、最短符号化二、解決を実現する

  1 #  
  2 class TreeNode:
  3     def __init__(self, data):
  4         """
  5         :data is a tuple the first element is value and the second is priority
  6         :param data:
  7         """
  8         self.value = data[0]
  9         self.priority = data[1]
 10         self.left_child = None
 11         self.right_child = None
 12         self.code = ""
 13 
 14 
 15 #  
 16 def create_node_queue(codes):
 17     queue = []
 18     for code in codes:
 19         queue.append(TreeNode(code))
 20     return queue
 21 
 22 
 23 #  
 24 def add_queue(queue, node_new):
 25     if len(queue) == 0:
 26         return [node_new]
 27     for i in range(len(queue)):
 28         if queue[i].priority >= node_new.priority:
 29             return queue[:i] + [node_new] + queue[i:]
 30     return queue + [node_new]
 31 
 32 
 33 #  
 34 class NodeQueue:
 35     def __init__(self, code):
 36         self.queue = create_node_queue(code)
 37         self.size = len(self.queue)
 38 
 39     def add_node(self, node):
 40         self.queue = add_queue(self.queue, node)
 41         self.size += 1
 42 
 43     def pop_node(self):
 44         self.size -= 1
 45         return self.queue.pop(0)
 46 
 47 
 48 #    
 49 def frequent_char(string_s):
 50     store_d = {}
 51     for c in string_s:
 52         if c not in store_d:
 53             store_d[c] = 1
 54         else:
 55             store_d[c] += 1
 56     return sorted(store_d.items(), key=lambda x: x[1])
 57 
 58 
 59 #  Huffman 
 60 def create_huffman_tree(node_queue):
 61     while node_queue.size != 1:
 62         node1 = node_queue.pop_node()
 63         node2 = node_queue.pop_node()
 64         r_1 = TreeNode([None, node1.priority + node2.priority])
 65         r_1.left_child = node1
 66         r_1.right_child = node2
 67         node_queue.add_node(r_1)
 68     return node_queue.pop_node()
 69 
 70 
 71 code_dict1 = {}
 72 code_dict2 = {}
 73 
 74 
 75 #  Huffman Huffman 
 76 def huffman_code_dict(head, x):
 77     # global code_dict, code_list
 78     if head:
 79         huffman_code_dict(head.left_child, x + "0")
 80         head.code += x
 81         if head.value:
 82             code_dict2[head.code] = head.value
 83             code_dict1[head.value] = head.code
 84         huffman_code_dict(head.right_child, x + "1")
 85 
 86 
 87 #  
 88 def trans_encode(string_s):
 89     # global code_dict1
 90     trans_code = ""
 91     for c in string_s:
 92         trans_code += code_dict1[c]
 93     return trans_code
 94 
 95 
 96 #  
 97 def trans_decode(string_s):
 98     # global code_dict1
 99     code = ""
100     answer = ""
101     for c in string_s:
102         code += c
103         if code in code_dict2:
104             answer += code_dict2[code]
105             code = ""
106     return answer

 
三、総括ハフマンツリーの符号化形式でデータの圧縮が可能であるため、ハフマンの応用も広範である.ここにメモしておくと便利です.
転載先:https://www.cnblogs.com/future-dream/p/10801934.html