"def compress(S)" 函数使用 运行 长度编码进行图像压缩

Question

我需要编写一个名为 compress(S) 的函数，它将长度小于或等于 64 的二进制字符串 S 作为输入，将另一个二进制字符串 returns 作为输出。输出二进制字符串应该是输入字符串的 运行 长度编码。

运行-长度编码用8位字节的序列（称为"run-length sequence"）表示一幅图像：

每个字节的第一位代表图像中下一个将出现的位，0 或 1。 最后七位包含在图像中当前位置连续出现的那些位的二进制数。

>>>compress( 64*'0' )
'01000000'
>>> compress( '11111' )
'10000101'
>>> Stripes = '0'*16 + '1'*16 + '0'*16 + '1'*16
>>> compress(Stripes)
'00010000100100000001000010010000'

Answer 1

首先运行长度编码，定义为-

def rle(input_string):
            """ takes input as string and checks repeating bit
                return repeating bit with their counts.
            """
            count = 1
            prev = ''
            lst = []
            for character in input_string:
                if character != prev:
                    if prev:
                        entry = (prev,count)
                        lst.append(entry)
                        #print lst
                    count = 1
                    prev = character
                else:
                    count += 1
            else:
                entry = (character,count)
                lst.append(entry)

            return lst

生成元组列表。

输入：print rle('1111010')

输出：[('1', 4), ('0', 1), ('1', 1), ('0', 1)]

 ----------------------------------

现在使用这个列表来制作字典，将重复计数进行二进制转换，并将其格式化为 7 位长。最后添加字典的相应键和值，使总数字保持 8 位。

def new_dict(S):
            """ input of rle(S) i.e., tuples of bit and repeating counts
                output dict as bit as key and value as counts with binary conversion.
            """  

            dict=rle(S)
            new_dict = []
            temp = []
            for k,v in dict:
                temp = k + "%07d" % int(bin(v)[2:])
                new_dict.append(temp)
            return new_dict

输入：打印new_dict('1111010')

输出：['10000100', '00000001', '10000001', '00000001']

---

现在用 compress(S) 函数压缩这个二进制字符串。

        def compress(S):
            """ takes a binary string S of length less than or equal to 64 as input and returns another binary string as output. 
            """
            l = new_dict(S)
            return''.join(str(elem) for elem in l)

打印压缩（'1111010'）
打印压缩( 64*'0' )

输出：'10000100000000011000000100000001'

输出：'01000000'

打印压缩（'11111'）

输出：'10000101'

条纹 = '0'*16 + '1'*16 + '0'*16 + '1'*16

打印压缩（条纹）

输出：'00010000100100000001000010010000'