How to make a structure unique?
显示 更早的评论
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 个评论
Walter Roberson
2020-6-25
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
2020-6-25
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:
Walter Roberson
2020-6-25
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
2020-6-25
采纳的回答
Rasul Khan
2020-6-25
编辑：Rasul Khan
2020-6-25
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 个评论
Walter Roberson
2020-6-25
m = vertcat(pop.x);
[~ , ia , ~] = unique(m , 'rows');
pop = pop(ia);
Md. Asadujjaman
2020-6-25
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