功能 – Lua中的LZW压缩

这是Lempel-Ziv-Welch压缩的伪代码.

pattern = get input character
 while ( not end-of-file ) {
     K = get input character
     if ( <<pattern,K>> is NOT in 
             the string table ){
         output the code for pattern
         add <<pattern,K>> to the string table
         pattern = K
     }
     else { pattern = <<pattern,K>> }
 }
 output the code for pattern
 output EOF_CODE

我试图在Lua中编写代码,但它并没有真正起作用.这是我在Python中使用LZW函数建模的代码,但是我在第8行得到了“尝试调用字符串值”错误.

function compress(uncompressed)

 local dict_size = 256
 local dictionary = {}

 w = ""
 result = {}
 for c in uncompressed do
  -- while c is in the function compress
     local wc = w + c
     if dictionary[wc] == true then
         w = wc
     else
         dictionary[w] = ""
         -- Add wc to the dictionary.
         dictionary[wc] = dict_size
         dict_size = dict_size + 1
         w = c
    end
 -- Output the code for w.
 if w then
   dictionary[w] = ""
 end
 end
 return dictionary
 end

 compressed = compress('TOBEORNottOBEORTOBEORNOT')
 print (compressed)

我真的想要一些帮助,让我的代码运行,或帮助我在Lua中编写LZW压缩.非常感谢！

解决方法

假设uncompressed是一个字符串,你需要使用这样的东西来迭代它：

for i = 1,#uncompressed do
  local c = string.sub(uncompressed,i,i)
  -- etc
end

第10行还有另一个问题; ..用于Lua中的字符串连接,因此该行应该是本地的wc = w .. c.

您可能还希望阅读this关于字符串连接的性能.简而言之,将每个元素保存在表中并使用table.concat()返回它通常更有效.