# Delete Entries in Cell arrays with a certain criterea

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Maximilian comploj on 21 Oct 2016
Answered: Nick Counts on 5 Nov 2016
Hello I Have a cell :
A= [2011 12 3 8 52 36] [2019]
[2011 12 3 8 52 40] [2004]
[2011 12 3 8 52 52] [2004]
[2011 12 3 14 50 50] [2015]
the first row contains a Matrix with the date in yyyy mm dd HH MM SS
i need to delete the double entry for 2004 delete second or third row if the date is less distant than one day Thanks a lot

Nick Counts on 5 Nov 2016
This is messy, but here is an example. I'm sure there's a slick way to do this with logical indexing, but I can't quite wrap my head around it.
Sometimes a for-loop will get it done when slicker methods are elusive. I tried to comment the example, so hopefully it makes sense to you. I also added an additional row with [2004] in column 7 that is more than 1 day away, and it is correctly left alone.
A = [[2011 12 3 8 52 36] [2019];
[2011 12 3 8 52 40] [2004];
[2011 12 5 8 52 52] [2004];
[2011 12 3 14 50 50] [2015];
[2011 12 3 8 52 52] [2004]]
disp('Starting with these dates:')
datestr(datenum(deal(A(:,1:6))))
% Find unique values in column 7
uniqueValues = unique(A(:,7));
% how often does each unique value from column 7 occur?
freq = hist(A(:,7), uniqueValues);
% find index of column 7 values that appear more than once
nonUniqueInd = ismember(A(:,7),uniqueValues(freq>1));
% Make a list of values that are repeated in column 7
nonUniqueVals = A(nonUniqueInd,7)
% we will be building an index of rows to delete
indToRemove = [];
% Loop through each value that is repeated. Will run on the same
% value more than once to caputure the case when a column 7 value
% occurs multiple times on multiple days
for i = 1:numel(nonUniqueVals)
% try each non-unique value from column 7
% get the index of each occurrence of the current non-unique
% value in the original array
nuInd = find(A(:,7)==nonUniqueVals(i))
% find elapsed time from first occurrence of the repeated value
% in column 7
deltas = datenum(deal(A(nuInd(1),1:6)))-(datenum(deal(A(nuInd,1:6))))
% find any time changes that are 0 or greater than 1. This
% gives us the first occurrence and anthing on a different day
thisIndToRemove = nuInd(find( (deltas == 0 | deltas > 1)))
% Add indices to the running list. We could delete them here,
% but it's a little better to build a small array of indices then
% to keep resizing the original array, which could potential be
% very large.
indToRemove = vertcat(indToRemove, thisIndToRemove);
end
% Remove duplicates matching our conditions
A(indToRemove,:) = [];
disp('Here are the remaining rows');
datestr(datenum(deal(A(:,1:6))))