taking second mode from strings

I want to write something that takes the most common letter from a cell array of it does not appear in another string

 newguess=char(mode(+[dict{:}]));  %gives the most common letter out of a cell array of words

Say pastguesses='eas' and newguess='e' I want to return the second most common letter until newguess doesn't appear in pastguesses.

I tried something like

 if newguess==pastguesses
%take next most common newguess until newguess doesn't appear in pastguesses
end

Answer 1

[Thorsten]

You can compress your dictionary to a single string of letters and then remove the most common letter from the string. Then find the second most common letter in this string. Note that I used double instead of + to convert from char to double, because it is easier to understand:

 dict = {'a' 'b' 'ba' 'c' 'hallo'};
 letters = [dict{:}];
 mostcommon = char(mode(double(letters)))
 newletters = strrep(letters, mostcommon, '')
 secondmostcommon = char(mode(double(newletters)))

[Max]

 newguess=char(mode(+[dict{:}]));  %gives the most common letter out of a cell array of words

I mean is it possible to take the highest appearing letter that is not in pastguesses from example if pastguesses='eai' and newguess='e' I ignore newguess and take the second most common letter so newguess='a' but newguess='a' appears in pastguesses so we take the next most common letter so newguess='i' but it appears in pastguesses so we take the next most common letter say newguess='o' 'o' doesnt appear in pastguesses so we stop at newguess='o'

The newguess is generated from a list of words and pastguesses keeps getting updated in my code. so for example say

dict= {'aa';'aal';'aalii';'aam';'aani';'aardvark';'aardwolf';'aaron';'aaronic';'aaronical';'aaronite';'aaronitic';'aaru';'ab';'aba';'ababdeh';'ababua';'abac';'abaca';'abacate'}

newguess= %is the highest occuring letter in dict that does not appear in pastguesses.

Thanks for the help.

[Thorsten]

You remove the letters in passguesses from the letters in your dictionary, and then determine the mode of the remaining letters:

   dict= {'aa';'aal';'aalii';'aam';'aani';'aardvark';'aardwolf';'aaron';'aaronic';'aaronical';'aaronite';'aaronitic';'aaru';'ab';'aba';'ababdeh';'ababua';'abac';'abaca';'abacate'}
     letters = [dict{:}];
     pastguesses='eas' 

     for i=1:length(pastguesses)
       letters = strrep(letters, pastguesses(i), '');
     end
     mostcommonnotinpastguesses = char(mode(double(letters)))