Input or put a large array of numbers into a function
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I'm fairly new to MATLAB. My assignment is to pick a word, break it down into each letter, generate all the possibilities with replacement of all the letters, assign each letter with a probability, calculate the their probabilities in 3rd, 4th, and 5th order extenstion, and use the built-in huffmandict function to generate the binary equivalence of each symbol.
In my case, I picked the word 'locate', so each letter will be 'l' 'o' 'c' 'a' 't' 'e'.
3rd order extension means combination of 3 letters. For instance, 'loc' 'loa' 'lot' etc... The same goes for 4th and 5th order. So 3rd order will have 216 possible combinations, 4th order will have 1296 and 5th will have 7776 combinations.
From my thought process to approach this problem, here is what I have so far:
I used the COMBINATOR function that I found to generate all the possible combinations
n3 = combinator(6, 3, 'p', 'r'); % 3rd order possible combinations
n4 = combinator(6, 4, 'p', 'r'); % 4th order possible combinations
n5 = combinator(6, 5, 'p', 'r'); % 5th order possible combinations
then I move the outputs of n3, n4, and n5 to Excel, replace the number with the letters (1 = l , 2 = o , 3 = c , 4 = a, 5 = t , 6 = e) , then replace them again with assigned probabilities (l = 0.1 , o = 0.18 , c =0.02 , a = 0.4 , t = 0.08 , e = 0.22), calcalate the probabilities for each symbol by the product of all the letter in that symbol. For instance, 3rd order, 'loc', will be 0.1*0.18*0.02 = 0.00036. I'm currently stuck at how to move back all 216, 1296 or 7776 probabilties to MATLAB to use the huffmandict function. The function requires 2 arguments, numbers of symbols and their corresponding probabilities.
I hope this makes sense. Any advice, help, tip or guidance will be appreciated. I know all of these can be done on MATLAB, but I'm not fluent in the programming language. Main reason that I involved Excel.
3 件のコメント
the cyclist
2022 年 10 月 28 日
Can you upload the code? You can use the paper clip icon in the INSERT section of the toolbar. Reading your description is not as useful as seeing the code.
Ben Nguyen
2022 年 10 月 28 日
Ben Nguyen
2022 年 10 月 28 日
採用された回答
x='locate';prob=[.1 .18 .02 .4 .08 .22];
p=[];order=3;
n=nchoosek(1:length(x),order);
for k=1:size(n,1)
p=[p;perms(n(k,:))];
end
X=x(p)
P=prod(prob(p),2)
6 件のコメント
Ben Nguyen
2022 年 10 月 28 日
Walter Roberson
2022 年 10 月 29 日
but the code uses perms() to ensure that all of the permutations are generated.
Ben Nguyen
2022 年 10 月 29 日
x='locate';prob=[.1 .18 .02 .4 .08 .22];
[a,b,c]=ndgrid(1:length(x));%for order 3, increase for higher orders
[d,e,f,g]=ndgrid(1:length(x));%for order 4
idx=[a(:),b(:),c(:)];
idx2=[d(:),e(:),f(:),g(:)];
w=x(idx)
s=sum(prod(prob(idx),2))%sum of all probs
w2=x(idx2)
s2=sum(prod(prob(idx2),2))
Even better
x='locate';prob=[.1 .18 .02 .4 .08 .22];
order=3;
b=cell(1,order);
[b{:}] = ndgrid(1:length(x));
idx=[];
for k=1:length(b)
idx=[idx,b{k}(:)];
end
X=x(idx)
p=prod(prob(idx),2)
Ben Nguyen
2022 年 11 月 2 日
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