Creating a loop to use all columns in a dataset array

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Giorgos
Giorgos 2011 年 7 月 26 日
Hello , i have a data set with 10 columns.The first column contains the names of the variables and the 9 next contain the data for the variables across 9 periods. I want , for each period ,to take the data which are lower than the data's median .I use the following command for the first period: s1=data(data.mv1<median(data.mv1),{'Name','mv1',}), where mv1 is the header of the first period's column and s1 a new dataset which contains only the variables i want. My question is , can i write a (for) loop that will automatically do this for the whole 9 periods, thus giving me s1,s2,...,s9?
  1 件のコメント
Jan
Jan 2011 年 7 月 26 日
There must be one or two typos in "s1=data(data.mv1<median(data.mv1),{'Name','mv1',})"
Please fix them by editing your question.

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採用された回答

Oleg Komarov
Oleg Komarov 2011 年 7 月 26 日
data = dataset({('a':'z').','names'},{rand(26,2),'mv1','mv2'});
EDIT
% Retrieve column names/variables (from second onwards)
varnames = data.Properties.VarNames(2:end);
nV = numel(varnames);
% Preallocate
C = cell(1,nV);
% Loop per each column (variable)
for v = 1:nV
idx = data.(varnames{v}) < median(data.(varnames{v}));
C{v} = data.(varnames{v})(idx);
C{v} = data(idx,[1 v]);
end
  5 件のコメント
Oleg Komarov
Oleg Komarov 2011 年 7 月 27 日
Sry for the distraction, it should be v (why v+1?).
Why do you want it to be extracted to the workspace, I suggest to keep it in the cell array, easier to reference. Read http://matlab.wikia.com/wiki/FAQ#How_can_I_create_variables_A1.2C_A2.2C....2CA10_in_a_loop.3F
Giorgos
Giorgos 2011 年 7 月 27 日
If i use v ,the first dataset doesn't show the first data but instead repeates the names of the variables. Agreed, it's much easier if i keep them in the cell ,it came to me right after i posted it unfortunately. Thanks again for all your help.

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その他の回答 (2 件)

Jan
Jan 2011 年 7 月 26 日
It would be easy, if you do not use symbols like 's1' and 'mv1', which have an index inside the name. Better use and index as index: s{1}, s{2}, ... and mv{1}, mv{2}, ...

Titus Edelhofer
Titus Edelhofer 2011 年 7 月 26 日
Hi,
I guess this leads to the question, how to get to data.mv1 where mv1 is given in a variable?
Note, that data.mv1 is the same as data.('mv1'). So if you have e.g.
header = {'mv1', ...};
then you could do
for i=1:length(header)
col = data.(header{i});
% do your median thing
dataNew = data(col<median(col));
end
Hope, this helps,
Titus
  2 件のコメント
Giorgos
Giorgos 2011 年 7 月 26 日
Is your <header> a cell array? When trying to run your code i get a 'Dataset array subscripts must be two-dimensional.' error. Also , shouldn't there be a suscript to dataNew like dataNew{i}?
Titus Edelhofer
Titus Edelhofer 2011 年 7 月 26 日
Yes, header is a cell array containing the names. And yes, there should be some subscript depending on what further you want to do with the reduced data ...

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