Effacer les filtres
Effacer les filtres

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

1 vue (au cours des 30 derniers jours)
Talat
Talat le 3 Avr 2011
Clôturé : Walter Roberson le 3 Jan 2017
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);
  8 commentaires
Oleg Komarov
Oleg Komarov le 12 Avr 2011
And what is errors? Post a question!!
Walter Roberson
Walter Roberson le 12 Avr 2011
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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