executing a loop most probably while
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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 个评论
Stephen23
2018-1-16
@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.
采纳的回答
Guillaume
2018-1-16
编辑:Guillaume
2018-1-16
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.
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