Octave - 霍夫曼代码不起作用 - SIG 的所有元素必须是 [1,N] 范围内的整数 [英] Octave - Huffman code doesn't work - All elements of SIG must be integers in the range [1,N]

问题描述

我在 Octave 中使用 huffmandict 和 huffmanenco 时遇到问题.

这是我的错误:

<块引用>

error: huffmanenco: SIG 的所有元素必须是范围内的整数[1,N]

这是我的代码:

inputSig = [1 1 2 6 6 6 6 4 5 5];list_symb = [1 2 6 4 5];list_proba = [0.2, 0.1, 0.4, 0.1, 0.2];dict = huffmandict(list_symb,list_proba);代码 = huffmanenco(inputSig,dict);

我的字典是

dict ={[1,1] = 1[1,2] = 0 1[1,3] = 0 0 1[1,4] = 0 0 0 0[1,5] = 0 0 0 1}

所以我的错误在于这条线

code = huffmanenco(inputSig,dict);

因为我的 dict 的长度是 5，而我的 inputSig 的长度是 10.

如何在没有此错误的情况下进行霍夫曼编码?

但是，此代码似乎适用于 Matlab.

解决方案

你说

<块引用>

因为我的 dict 的长度是 5，而我的 inputSig 的长度是 10.

这不是您收到此错误的原因.来自文档:

<块引用>

一个限制是信号集必须严格属于范围 '[1,N]' 且 'N = length (dict)'.

换句话说，您的dict"仅包含 5 个单元格，但您的inputSig"包含 [1,6] 而不是 [1,5] 范围内的整数.

因此，您基本上必须在 [1,5] 范围内重新编码"/映射您的信号(即范围 [1,5] 将成为索引/标签，到您的实际符号数组).

例如

inputSig = [1 1 2 6 6 6 6 4 5 5];list_symb = unique( inputSig );list_proba = [0.2, 0.1, 0.4, 0.1, 0.2];dict = huffmandict( list_symb, list_proba );[~, idx] = ismember( inputSig, list_symb );代码 = huffmanenco( idx, dict )% 代码 = 0 1 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0

<小时>

附注.为了完成，下一个显而易见的问题是，鉴于整个实际交易品种与指数"业务，您如何对此进行解码.简单的;您使用解码输出(对应于索引)并将其用作 list_symb 向量的索引向量，从而检索原始符号.即:

deco = huffmandeco ( code, dict )% 装饰 = 1 1 2 5 5 5 5 3 4 4list_symb( 装饰 )% ans = 1 1 2 6 6 6 6 4 5 5

I have a problem in Octave using huffmandict and huffmanenco.

Here is my error :

error: huffmanenco: all elements of SIG must be integers in the range [1,N]

Here is my code :

inputSig = [1 1 2 6 6 6 6 4 5 5];
list_symb = [1 2 6 4 5];
list_proba = [0.2, 0.1, 0.4, 0.1, 0.2];
dict = huffmandict(list_symb,list_proba);
code = huffmanenco(inputSig,dict);

my dict is

dict =
{
 [1,1] =  1
 [1,2] = 0   1
 [1,3] = 0   0   1
 [1,4] = 0   0   0   0
 [1,5] = 0   0   0   1
}

So my error is with the line

code = huffmanenco(inputSig,dict);

because the lenght of my dict is 5 and my lenght of my inputSig is 10.

How can I do my huffman coding without this error?

However, this code seems to work on Matlab.

解决方案

You say

because the length of my dict is 5 and the length of my inputSig is 10.

That's not quite why you get this error. From the documentation:

A restriction is that a signal set must strictly belong in the range '[1,N]' with 'N = length (dict)'.

In other words, your 'dict' only contains 5 cells, but your 'inputSig' contains integers in the range [1,6] instead of [1,5].

Therefore, you basically have to 'recode' / map your signal in the range [1,5] (i.e. the range [1,5] will become indices/labels, to an array of your actual symbols).

E.g.

inputSig   = [1 1 2 6 6 6 6 4 5 5];
list_symb  = unique( inputSig );
list_proba = [0.2, 0.1, 0.4, 0.1, 0.2];
dict       = huffmandict( list_symb, list_proba );
[~, idx]   = ismember( inputSig, list_symb );
code       = huffmanenco( idx, dict )
% code = 0 1 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0

---

PS. For completion, the obvious next question is how do you decode this, given the whole 'actual symbol vs indices' business. Simple; you use the decoded output (which corresponds to indices) and use it as an index vector to the list_symb vector, thus retrieving the original symbols. I.e.:

deco = huffmandeco ( code, dict )
% deco = 1 1 2 5 5 5 5 3 4 4

list_symb( deco )
% ans  = 1 1 2 6 6 6 6 4 5 5

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