# Removing rows - duplicates based on a condition

3 views (last 30 days)
Nick on 1 Dec 2019
Commented: Turlough Hughes on 1 Dec 2019
Hi,
I have a table with a number of columns and would like to delete some rows based on some conditions.
Here is an example:
A = table([1;2;3;3;3;3;4;4;4], [ 0;0;0;0;1;1;1;1;1], [1999;1999;1999;2000;2001;2002;2000;2001;2004]);
A.Properties.VariableNames = {'ID' 'Size' 'Date'};
The condidions are:
If ID is the same and within that ID the size = 1 then if there are more than one rows, keep the row that contains the earliest Date for that ID.
So in this case the new table would look like this:
B = table([1;2;3;3;3;4], [ 0;0;0;0;1;1], [1999;1999;1999;2000;2001;2000]);
B.Properties.VariableNames = {'ID' 'Size' 'Date'};
Thanks!

Rik on 1 Dec 2019
I can't come up with anything other than a solution involving a nested loop, which will have horrible performance on larger sets of data. How large do you expect your real data to be?
Nick on 1 Dec 2019
Hi Rik,
It is a prety big data set. 3million rows x 50 columns.

dpb on 1 Dec 2019
Edited: dpb on 1 Dec 2019
[~,ia]=unique(A(A.Size==1,1),'first');
B=[A(A.Size==0,:);A([ia+find(A.Size==0,1,'last')],:)];
returns
>> B
B =
6×3 table
ID Size Date
__ ____ ____
1 0 1999
2 0 1999
3 0 1999
3 0 2000
3 1 2001
4 1 2000
>>
I'm guessing the selection on Size==1 is only artificial given no duplicates by inspection. Logic works more simply without having to not subset that grouping (the mess about the complicated indexing expression for B)
The solution relies on the Date field being sorted so unique returns the first/lowest date...if possibly not, then sort first.

#### 1 Comment

Turlough Hughes on 1 Dec 2019
The above also assumes that A.Size is sorted, if this is not the case in your data you should first sort the Size and then the Date where Size=1 as follows:
A = table([1;2;3;3;3;3;4;4;4;5;5;5;5], [ 0;0;0;0;1;1;1;1;1;0;0;1;1], ...
[1999;1999;1999;2000;2001;2002;2000;2001;2004;2000;2005;2005;2001], ...
'VariableNames',{'ID','Size','Date'}) % sample data to illustrate
T=sortrows(A,2); % Sort the size first.
T2=sortrows(T(find(T.Size),:),3) % then where Size=1, sort Date
A2=[A(A.Size==0,:);T2];
Then you just sub A2 in for A into dbp's solution.