Slow-running code for removing rows
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I have a set of code that works, but a part of it runs very slowly. Basically I'm trying to remove zero-only rows from two matrixes, and I'm able to tell that both have an empty row if the first element in one matrix is zero. Here's the actual code:
for a=1:numberofrows
if slopes(a,1)==0
slopes(a,:)=[];
switchingtimes(a,:)=[];
end
end
I'd try to do a nonzeros function and then a reshape function, but sometimes slopes have a zero or two at the end of a column.
Is there any way to speed up this code?
采纳的回答
Geoff Hayes
2019-3-1
Christian - what about using find as
slopes(find(slopes(:,1)==0), :) = []
We use find to return all indices of the first column of slopes whose element is a zero. We then remove all of these rows with the assignment to [] (like you have done already).
2 个评论
Christian F
2019-3-4
find is not required. Logical indexing is simpler and faster:
idx = slopes(:,1)==0;
slopes(idx,:) = []
switchingtimes(idx,:) = [];
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