Huffman encoding using cell array - matlab

Question

I have some problems in final part of Huffman encoding.
Currently I have my coding table in cell array
code =
{
[1,1] = 000
[1,2] = 001
[1,3] = 010
[1,4] = 011
[1,5] = 100
...
}
Where second index represent ascii character in my other cell array
huffman_tree =
{
[1,1] = A
[1,2] = B
[1,3] = C
[1,4] = D
[1,5] = E
...
}
I'm using following code for encoding input to output:
output= [];
for i=1:length(input)
x = findInArray(huffman_tree, input(i));
output= [output code(x)];
end
function [index] = findInArray(array, searched)
index = -1;
for i=1:length(array)
if array{i} == searched
index = i;
end
end
end
At this point my code is O(n^2) or even worse. I'm having problem with large input where
length(input) = 1000000
There must be some faster way to transform input with my coding table to output.

Answer 1

Because you're using cell arrays, that's going to be inherently slow so you have no choice but to iterate over each cell. However, I can provide some suggestions to help speed things up. What you can do is use strcmp to compare strings. I'm assuming that each character in your cell array is represented as a one character string. strcmp has the ability to take an individual string and compare itself to a cell array of strings. The output will be an array that is the same size as the cell array of strings and give you a logical true if the input string matches a position in the cell array and false otherwise.
Because your Huffman dictionary will contain a unique set of characters, you will only get one possible match per character. Therefore, we can use this logical array output to index the codebook directly to retrieve the corresponding code you want. Logical indexing works by supplying a logical vector that is the same length as the vector of interest and it retrieves those values whose corresponding positions are true. Therefore, if we only had one true value in the logical vector with the rest of the elements being false, this means that we would get just the corresponding element we desire and nothing else.
Therefore, we can change your code to do this. Note that I've changed the loop counter i to idx because it has actually been shown that using i as a loop counter slows down your code by a slight amount. See this post by Shai Bagon for more details: Using i and j as variables in Matlab. Also, I've changed the length call to numel... mainly because I don't like using length.... just a personal choice though.
output = [];
for idx = 1:numel(input)
output = [output code(strcmp(input(idx), huffman_tree))];
end
Give the above a whirl and see if it performs any faster. For one thing, this will escape using an additional for loop for searching for a match as strcmp is very efficiently implemented, so the above code won't be O(n^2), but could be slightly better than quadratic.