Index through datestring, selecting whole hours
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Hi,
i have the following problem. I have created a cell array consisting of two columns date and value, which looks like this (extract):
'02-10-2014 20:30' [10934]
'02-10-2014 21:00' [10195]
'02-10-2014 21:30' [ 9388]
'02-10-2014 22:00' [ 8808]
'02-10-2014 22:30' [ 8730]
I want to filter through the array and select only values that correspond to whole hours like:
'02-10-2014 21:00' [10195]
'02-10-2014 22:00' [ 8808]
Any suggestion please?
Many thanks
Stefan
Réponse acceptée
Plus de réponses (4)
Image Analyst
le 3 Oct 2014
Something like
counter = 1;
for row = 1 : size(ca, 1)
cellContents = ca{row, 1}; % Extract first column of this row.
if strfind(cellContents, ':00')
% Cell contains :00, so append it to the output cell array.
output_ca{counter, 1} = cellContents;
output_ca(counter, 2) = ca(row, 2);
counter = counter + 1;
end
end
That's one way at least, sort of the intuitive but brute force way. There are probably other ways that are more compact, like using cellfunc() or something. Note how I used braces and parentheses. If you don't know why, see the FAQ : http://matlab.wikia.com/wiki/FAQ#What_is_a_cell_array.3F. You may find it easier to use a structure array than a cell array.
Andrei Bobrov
le 3 Oct 2014
Modifié(e) : Andrei Bobrov
le 3 Oct 2014
Z = {'02-10-2014 20:30' [10934]
'02-10-2014 21:00' [10195]
'02-10-2014 21:30' [ 9388]
'02-10-2014 22:00' [ 8808]
'02-10-2014 22:30' [ 8730]};
[~,~,~,~,m] = datevec(Z(:,1),'dd-mm-yyyy HH:MM');
out = Z(m == 0,:);
or
out = Z(~cellfun('isempty',regexp(Z(:,1),':00')),:);
Matz Johansson Bergström
le 3 Oct 2014
Modifié(e) : Matz Johansson Bergström
le 3 Oct 2014
Another alternative where I just pick the last two characters from the first column of the cell:
%mock data:
cel{1,1} = '02-10-2014 21:00'; cel{1,2} = 3.2;
cel{2,1} = '02-10-2014 21:03'; cel{2,2} = 7.2;
mask = cellfun(@(x) strcmp(x(end-1:end), '00'), cel(:,1))
>> cel(mask,:)
ans =
'02-10-2014 21:00' [3.2000]
If you want to find out how many entries that are on the hour:
sum(mask)
Stefan Hristov
le 3 Oct 2014
0 votes
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