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fast delete of rows in a table

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Francesco Giuseppe Fornari
Francesco Giuseppe Fornari 2020 年 5 月 27 日
hi
I've got a table with 120k rows, and I need to delete some rows if there is a specific condition.
I wrote this:
for i=1:(size(db,1)-1)
if db.INTERVENTION(i)==1 && isequal(db.SETTLEMENTDATE(i-1),db.SETTLEMENTDATE(i)) && isequal(db.DUID(i-1),db.DUID(i))
db(i-1,:)=[];
end
end
This code takes a long time to execute.
Is there a faster way?
thanks!

採用された回答

Image Analyst
Image Analyst 2020 年 5 月 27 日
Try this:
numRows = size(db,1)-1;
rowsToDelete = false(numRows, 1);
for k = 2 : numRows % Has to start at 2, right? Since you're using k-1. Don't use i, the imaginary variable, as a loop index.
if db.INTERVENTION(k)==1 && isequal(db.SETTLEMENTDATE(k-1),db.SETTLEMENTDATE(k)) && isequal(db.DUID(k-1),db.DUID(k))
rowsToDelete(k) = true;
end
end
db(rowsToDelete, :) = []; % Remove these rows.
% Or equivalently
%db = db(~rowsToDelete, :); % Extract everything BUT the rows to delete.
  3 件のコメント
Image Analyst
Image Analyst 2020 年 5 月 27 日
I think the reason being is that when you removed a single row from the table, it had to rebuild the table each time. If you give it a vector, it has to rebuild the table only once.
Francesco Giuseppe Fornari
Francesco Giuseppe Fornari 2020 年 5 月 27 日
yes, it makes a lot of sense ;)

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その他の回答 (1 件)

darova
darova 2020 年 5 月 27 日
Write indices and delete outside the for loop
ind = logical(db*0);
for i=1:(size(db,1)-1)
if db.INTERVENTION(i)==1 && isequal(db.SETTLEMENTDATE(i-1),db.SETTLEMENTDATE(i)) && isequal(db.DUID(i-1),db.DUID(i))
ind(i,:) = true;
end
end
db(ind) = [];
  2 件のコメント
Francesco Giuseppe Fornari
Francesco Giuseppe Fornari 2020 年 5 月 27 日
seems ok, but I don't get the inizialization of ind. Line
ind = logical(db*0);
gives "Undefined operator '*' for input arguments of type 'table'"
Did I got wrong?
Thanks anyway!
darova
darova 2020 年 5 月 27 日
See Image Analyst answer below. It's correct

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