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Input or put a large array of numbers into a function

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Ben Nguyen
Ben Nguyen 2022 年 10 月 28 日
コメント済み: Ben Nguyen 2022 年 11 月 2 日
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
Ben Nguyen
Ben Nguyen 2022 年 10 月 28 日
n3 = combinator(6, 3, 'p', 'r');
n4 = combinator(6, 4, 'p', 'r');
n5 = combinator(6, 5, 'p', 'r');
Ben Nguyen
Ben Nguyen 2022 年 10 月 28 日
That's all I had for my code. My next step is to transfer all the outputs of n3, n4, n5 to Excel spreasheets manually. Since the combinator function output the combinations only in numbers, I will replace the numbers with letters in Excel manually.

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採用された回答

David Hill
David Hill 2022 年 10 月 28 日
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)
X = 120×3 char array
'col' 'clo' 'ocl' 'olc' 'lco' 'loc' 'aol' 'alo' 'oal' 'ola' 'lao' 'loa' 'tol' 'tlo' 'otl' 'olt' 'lto' 'lot' 'eol' 'elo' 'oel' 'ole' 'leo' 'loe' 'acl' 'alc' 'cal' 'cla' 'lac' 'lca' 'tcl' 'tlc' 'ctl' 'clt' 'ltc' 'lct' 'ecl' 'elc' 'cel' 'cle' 'lec' 'lce' 'tal' 'tla' 'atl' 'alt' 'lta' 'lat' 'eal' 'ela' 'ael' 'ale' 'lea' 'lae' 'etl' 'elt' 'tel' 'tle' 'let' 'lte' 'aco' 'aoc' 'cao' 'coa' 'oac' 'oca' 'tco' 'toc' 'cto' 'cot' 'otc' 'oct' 'eco' 'eoc' 'ceo' 'coe' 'oec' 'oce' 'tao' 'toa' 'ato' 'aot' 'ota' 'oat' 'eao' 'eoa' 'aeo' 'aoe' 'oea' 'oae' 'eto' 'eot' 'teo' 'toe' 'oet' 'ote' 'tac' 'tca' 'atc' 'act' 'cta' 'cat' 'eac' 'eca' 'aec' 'ace' 'cea' 'cae' 'etc' 'ect' 'tec' 'tce' 'cet' 'cte' 'eta' 'eat' 'tea' 'tae' 'aet' 'ate'
P=prod(prob(p),2)
P = 120×1
0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0072 0.0072 0.0072 0.0072
  6 件のコメント
David Hill
David Hill 2022 年 10 月 29 日
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)
X = 216×3 char array
'lll' 'oll' 'cll' 'all' 'tll' 'ell' 'lol' 'ool' 'col' 'aol' 'tol' 'eol' 'lcl' 'ocl' 'ccl' 'acl' 'tcl' 'ecl' 'lal' 'oal' 'cal' 'aal' 'tal' 'eal' 'ltl' 'otl' 'ctl' 'atl' 'ttl' 'etl' 'lel' 'oel' 'cel' 'ael' 'tel' 'eel' 'llo' 'olo' 'clo' 'alo' 'tlo' 'elo' 'loo' 'ooo' 'coo' 'aoo' 'too' 'eoo' 'lco' 'oco' 'cco' 'aco' 'tco' 'eco' 'lao' 'oao' 'cao' 'aao' 'tao' 'eao' 'lto' 'oto' 'cto' 'ato' 'tto' 'eto' 'leo' 'oeo' 'ceo' 'aeo' 'teo' 'eeo' 'llc' 'olc' 'clc' 'alc' 'tlc' 'elc' 'loc' 'ooc' 'coc' 'aoc' 'toc' 'eoc' 'lcc' 'occ' 'ccc' 'acc' 'tcc' 'ecc' 'lac' 'oac' 'cac' 'aac' 'tac' 'eac' 'ltc' 'otc' 'ctc' 'atc' 'ttc' 'etc' 'lec' 'oec' 'cec' 'aec' 'tec' 'eec' 'lla' 'ola' 'cla' 'ala' 'tla' 'ela' 'loa' 'ooa' 'coa' 'aoa' 'toa' 'eoa' 'lca' 'oca' 'cca' 'aca' 'tca' 'eca' 'laa' 'oaa' 'caa' 'aaa' 'taa' 'eaa' 'lta' 'ota' 'cta' 'ata' 'tta' 'eta' 'lea' 'oea' 'cea' 'aea' 'tea' 'eea' 'llt' 'olt' 'clt' 'alt' 'tlt' 'elt' 'lot' 'oot' 'cot' 'aot' 'tot' 'eot' 'lct' 'oct' 'cct' 'act' 'tct' 'ect' 'lat' 'oat' 'cat' 'aat' 'tat' 'eat' 'ltt' 'ott' 'ctt' 'att' 'ttt' 'ett' 'let' 'oet' 'cet' 'aet' 'tet' 'eet' 'lle' 'ole' 'cle' 'ale' 'tle' 'ele' 'loe' 'ooe' 'coe' 'aoe' 'toe' 'eoe' 'lce' 'oce' 'cce' 'ace' 'tce' 'ece' 'lae' 'oae' 'cae' 'aae' 'tae' 'eae' 'lte' 'ote' 'cte' 'ate' 'tte' 'ete' 'lee' 'oee' 'cee' 'aee' 'tee' 'eee'
p=prod(prob(idx),2)
p = 216×1
0.0010 0.0018 0.0002 0.0040 0.0008 0.0022 0.0018 0.0032 0.0004 0.0072
Ben Nguyen
Ben Nguyen 2022 年 11 月 2 日
David, you are a gods-sent. Based on your code (the first one without the for loop since I understood it better), I improvised my own. It worked. Thank you so much!!!

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