霍夫曼编码：如何用Python编写二进制数据

我已经尝试过使用struct模块的方法,如代码中注释掉的行所示,但没有成功.基本上,我有两个选择：我可以按代码编写二进制数据代码(我的代码是长度在3到13位之间的位序列),或将n个字符的整个字符串(在这种情况下为n = 25000)转换为二进制数据.但是我不知道如何实现这两种方法.码：

import heapq
import binascii
import struct

def createFrequencyTupleList(inputFile):
    frequencyDic = {}

    intputFile = open(inputFile, 'r')
    for line in intputFile:
        for char in line:
            if char in frequencyDic.keys():
                frequencyDic[char] += 1
            else:
                frequencyDic[char] = 1

    intputFile.close()
    tupleList = []
    for myKey in frequencyDic:
        tupleList.append((frequencyDic[myKey],myKey))
    return tupleList

def createHuffmanTree(frequencyList):
    heapq.heapify(frequencyList)
    n = len(frequencyList)
    for i in range(1,n):
        left = heapq.heappop(frequencyList)
        right = heapq.heappop(frequencyList)
        newNode = (left[0] + right[0], left, right)
        heapq.heappush(frequencyList, newNode)
    return frequencyList[0]

def printHuffmanTree(myTree, someCode,prefix=''):
    if len(myTree) == 2:
        someCode.append((myTree[1] + "@" + prefix))
    else:
        printHuffmanTree(myTree[1], someCode,prefix + '0')
        printHuffmanTree(myTree[2], someCode,prefix + '1')

def parseCode(char, myCode):
    for k in myCode:
        if char == k[0]:
            return k[2:]


if __name__ == '__main__':
    myList = createFrequencyTupleList('input')
    myHTree = createHuffmanTree(myList)
    myCode = []
    printHuffmanTree(myHTree, myCode)
    inputFile = open('input', 'r')
    outputFile = open('encoded_file2', "w+b")
    asciiString = ''
    n=0
    for line in inputFile:
        for char in line:
            #outputFile.write(parseCode(char, myCode))
            asciiString += parseCode(char, myCode)
            n += len(parseCode(char, myCode))
    #values = asciiString
    #print n
    #s = struct.Struct('25216s')
    #packed_data = s.pack(values)
    #print packed_data
    inputFile.close()
    #outputFile.write(packed_data)
    outputFile.close()

解决方法:

您正在寻找：

packed_data = ''.join(chr(int(asciiString[i:i+8], 2)) 
                         for i in range(0, len(asciiString), 8))

从asciiString一次将花费8位,将其解释为整数,然后输出相应的字节.

您的问题在于,这要求asciiString的长度为8位的倍数才能正常工作.如果不是,您将在最后几个实数之前插入零位.

因此,您需要将位数存储在最后一个字节中的某个位置,这样您就知道在将它们取回时会忽略这些位,而不是将它们解释为零.您可以尝试：

packed_data = chr(len(asciiString) % 8) + packed_data

然后,当您读回它：

packed_input = coded_file.read()
last_byte_length, packed_input, last_byte = (packed_input[0], 
                                             packed_input[1:-1], 
                                             packed_input[-1])
if not last_byte_length: last_byte_length = 8
ascii_input = ''.join(chain((bin(ord(byte))[2:].zfill(8) for byte in packed_input),
                      tuple(bin(ord(last_byte))[2:].zfill(last_byte_length),)))
# OR
# ascii_input = ''.join(chain(('{0:0=8b}'.format(byte) for byte in packed_input),
#                       tuple(('{0:0=' + str(last_byte_length) + '8b}').format(last_byte),)))

编辑：您需要从bin()返回的字符串中剥离’0b’,或者在2.6或更高版本上,最好使用我添加的新的替代版本,该版本使用字符串格式而不是bin(),切片和zfill() .

编辑：谢谢eryksun,很好用的链条可以避免制作ASCII字符串.另外,需要在bin()版本中调用ord(byte).

标签：huffman-code,binary,binary-data,python
来源： https://codeday.me/bug/20191102/1991087.html