emp.x=[];
emp.info=[];
emp.fit=[];
pop=repmat(emp,1,1);
pop(1).x=[1 2 3 4];
pop(2).x=[3 2 1 4];
pop(3).x=[1 2 3 4];
pop(4).x=[2 1 3 4];
How to make the above structure (pop) unique?
The result would be like the following figure.

4 commentaires

Walter Roberson
Walter Roberson le 25 Juin 2020
Will there ever be a case where the x values are not row vectors? Will there ever be a case when the row vectors are not the same length? For example will any of the entries ever be empty?
Md. Asadujjaman
Md. Asadujjaman le 25 Juin 2020
X values are not a row vector.
Row vectors are the same length.
Here, the 'info' and 'fit' entities are empty. After making the structure (pop) unique, I will do some analysis and therefore the 'info' and 'fit' entities would not be empty.
It would be also helpful for if anyone can make the following structure (pop) unique.
emp.x=[];
pop=repmat(emp,1,1);
pop(1).x=[1 2 3 4];
pop(2).x=[3 2 1 4];
pop(3).x=[1 2 3 4];
pop(4).x=[2 1 3 4];
Result wouldbe as follows:
X values are not a row vector.
Row vectors are the same length.
Which row vectors are the same length as each other if they are not the X values? (We know they are not the X values because you said that the X values are not row vectors.)
Md. Asadujjaman
Md. Asadujjaman le 25 Juin 2020
sorry, X values are a row vector

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 Réponse acceptée

Rasul Khan
Rasul Khan le 25 Juin 2020
Modifié(e) : Rasul Khan le 25 Juin 2020
You can achieve it using this script.
m = [];
for i = 1 : numel(pop)
m = [m ; pop(i).x];
end
[~ , ia , ~] = unique(m , 'rows');
pop = pop(ia);

2 commentaires

m = vertcat(pop.x);
[~ , ia , ~] = unique(m , 'rows');
pop = pop(ia);
Md. Asadujjaman
Md. Asadujjaman le 25 Juin 2020
Thank you. It's working.
I got the result.

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