selecting elements from union of multiple sets, (edited again , need immediately )

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Talat 2011-4-3

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https://ww2.mathworks.cn/matlabcentral/answers/4626-selecting-elements-from-union-of-multiple-sets-edited-again-need-immediately

关闭： Walter Roberson 2017-1-3

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I am editing it again to elaborate my problem...code given below gives sets which are 'H', 'RE', 'LE' and 'SH' , 'LN' and 'RN'......now i need a set of elements of 'errors' that exist B/W [(LN, LM )U(RM, RN)] ...here 'errors' must be between LN(excluding) and LM(excluding), and similarly between RM(excluding) and RN(excluding) and there's union....

   LM = 0;
   RM = 1; 
   for i=1:size(errors, 1)
       for j=1:size(errors, 2)
           if (errors(i, j)== LM)         % Errors equal to LM
               H(Hindex)=errors(i,j);
               Hindex=Hindex+1;
           if (errors(i, j) < LM)      % Errors less than LM
               LE(LEindex)=errors(i,j);
               LEindex=LEindex+1;
           elseif (errors(i, j)== RM)     % Errors equal to RM
               SH(SHindex)=errors(i,j);
               SHindex=SHindex+1;
           elseif (errors(i, j)> RM)      % Errors greater than RM
               RE(REindex)=errors(i, j);
               REindex=REindex+1;
           else EX(EXindex)=errors(i, j); % Remaining Errors
                EXindex=EXindex+1;
           end
       end
   end    
   %%selecting minimum occurrence of numbers LN and RN from histrogram of LE and RE respectively
   % MINIMUMS OF LE
   uniqueLX = unique(LE);
   L_elements = hist(LE, uniqueLX);
   min_count_of_each_LE = min(L_elements);
   index_to_all_min=(L_elements == min_count_of_each_LE);
   LN = uniqueLX(index_to_all_min);
   %MINIMUMS OF RE
   uniqueRX=unique(RE);
   R_elements = hist(RE, uniqueRX);
    min_count_of_each_RE = min(R_elements);
   index_to_all_minRE=(R_elements == min_count_of_each_RE);
   RN = uniqueRX(index_to_all_minRE);

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Oleg Komarov 2011-4-12

And what is errors? Post a question!!

Walter Roberson 2011-4-12

Note: Talat had multiple threads going on topics and did not clearly indicate whether previous problems had been solved or not. This particular thread showed up as new again because I edited it to improve the formatting and grammar.

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