Replacing row in matrix with previous row depending on condition
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Hi
Say I have a matrix:
A=[2,5,3;5,8,2;1,-2,5]
If a value in any of the rows are -2 the whole row should be replaced by the previous row.
So the result would be:
A=[2,5,3;5,8,2;5,8,1]
The matrix consists of 1 million rows, so I'm looking for the fastest method.
0 件のコメント
採用された回答
the cyclist
2020 年 6 月 19 日
Here is one way:
while any(A(:)==-2)
rowToReplace = find(any(A==-2,2));
A(rowToReplace,:) = A(rowToReplace-1,:);
end
This solution could probably made faster by using only logical indices, without the find command, but I felt lazy.
2 件のコメント
the cyclist
2020 年 6 月 19 日
This is a little faster, in limited testing:
while any(A(:)==-2)
rowToReplace = any(A==-2,2);
A(rowToReplace,:) = A([rowToReplace(2:end); false],:);
end
The optimized version might depend on the pattern of -2's in the array.
その他の回答 (1 件)
David Hill
2020 年 6 月 19 日
If you have consecutive rows containing -2, you will have to repeat until all the -2 rows have been replaced if that is your goal
[a,~]=find(A==-2);
a=unique(a);
A(a,:)=A(a-1,:);
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