executing a loop most probably while

1 vue (au cours des 30 derniers jours)
Busy Bee
Busy Bee le 16 Jan 2018
Modifié(e) : Guillaume le 16 Jan 2018
I have created a function which groups a vast data. Here, for simplicity, I have used 20 data points. group(x,y) creates the required group (g1) and the next 'for' loop removes the elements of this group to create a new dataset. Again the group(x,y) creates the next group(g2) and the initial dataset is modified. I want to create a loop which will execute this till the dataset has zero elements,i.e. all the elements are grouped with group numbers,g1,g2 etc. I have tried to use while loop but I cannot change the group names, hence the error. Any help will be appreciated.
x = gallery('uniformdata',20,1,1);
y = gallery('uniformdata',20,1,1);
group(x,y);
g1=ans
for i=1:length(g1)
x(any(x==g1(i,1),2),:)=[];
y(any(y==g1(i,2),2),:)=[];
end
group(x,y);
g2=ans
for i=1:length(g2)
x(any(x==g2(i,1),2),:)=[];
y(any(y==g2(i,2),2),:)=[];
end
  3 commentaires
Stephen23
Stephen23 le 16 Jan 2018
@Busy Bee:
  • Why do you not return any output from group ?
  • Do NOT use ans as a variable name.
  • Do not try to magically generate variable names g1, g2, etc:
Just use simpler and much more efficient indexing.
Busy Bee
Busy Bee le 16 Jan 2018
this is the answer i want
g1 =
0.1996 0.0736
0.4428 0.0015
0.8407 0.0159
0.9102 0.3281
0.9528 0.4493
0.9539 0.9641
g2 =
0.3759 0.1164
0.5982 0.0765
0.7041 0.2088
0.8368 0.2609
0.8986 0.5779
g3 =
0.3068 0.4931
0.5187 0.3469
0.5383 0.4436
0.7153 0.6566
g4 =
0.4290 0.4946
0.5253 0.8889
g5 =
0.0222 0.9105
0.3031 0.9285
g6 =
0.0345 0.9509

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Guillaume
Guillaume le 16 Jan 2018
Modifié(e) : Guillaume le 16 Jan 2018
First, some important notes
group(x, y);
g = ans;
Do not do this! This is just a more complicated and confusing way of simply doing
g = group(x, y);
Secondly, you create x and y as column vectors, but then have
x(any(x == g(i,1), 2), :) = [];
y(any(x == g(i,1), 2), :) = [];
which is designed for x and y matrices. If x and y are indeed vector then the above can be reduced simply to
x(x == g(i, 1)) = [];
y(y == g(i, 2)) = [];
In addition the i loop can be entirely removed, for x, y matrices:
x(any(ismember(x, g(:, 1)), 2)) == [];
y(any(ismember(y, g(:, 2)), 2)) == [];
and for x, y vectors simply:
x(ismember(x, g(:, 1))) = [];
y(ismember(y, g(:, 2))) = [];
Now to answer your question, first do not number variables, see Stephen's comment. Instead create a cell array that can easily be indexed:
x = gallery('uniformdata',20,1,1);
y = gallery('uniformdata',20,1,1);
g = {}; %better variable name required
count = 0;
while ~isempty(x)
count = count + 1;
g{count} = group(x, y);
x(ismember(x, g{count}(:, 1))) = [];
y(ismember(y, g{count}(:, 2))) = [];
end
g{i} is the ith group.
  1 commentaire
Busy Bee
Busy Bee le 16 Jan 2018
Modifié(e) : Busy Bee le 16 Jan 2018
Thank you so much!!!!!

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