功能 – Lua中的LZW压缩
发布时间:2020-12-15 00:23:11 所属栏目:大数据 来源:网络整理
导读: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 } el
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. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |