Fast way to delete the last n rows of a matrix

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Hi,
I would like to be able to delete the last n rows of a matrix. I have found that this can be done in the following way. For example, suppose I want to delete the last two rows of a matrix:
tst=[1 1 1; 2 2 2; 3 3 3];
lastn=2;
tic
tst((end-(lastn-1)):end,:)=[];
toc
This works, but it seems quite expensive: "Elapsed time is 0.001152 seconds." In my code, I will be deleting rows many hundreds of thousands of times. Do you know if there is a faster way to delete the last n rows of a matrix?
Thank you.
Andrew DeYoung
Carnegie Mellon University

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Sean de Wolski
Sean de Wolski il 20 Mag 2011
tst = tst(1:end-lastn,:); %glass half full
No guarantees it'll be faster. The typical goal here is just minimize the number of matrix resizing operations you have to do.
  3 Commenti
Hafsa Arif
Hafsa Arif il 7 Mar 2020
is this making a new one? informmation will be lost or get arranged in the matrix?
Walter Roberson
Walter Roberson il 7 Mar 2020
Newtst = tst(1:end-lastn,:); %glass half full
If you want to keep the old matrix. With Sean's version to write to tst then the last rows would be thrown away.

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Walter Roberson
Walter Roberson il 20 Mag 2011
Andrew, could you transpose so that you are deleting columns instead of rows? If you do that then the portion that is left behind is contiguous in memory, and it should be faster to work with that than to have to copy over each partial column as you delete rows.
  3 Commenti
Walter Roberson
Walter Roberson il 20 Mag 2011
Correct. MATLAB arranges memory in columns and the first entry for the next column is stored in memory directly after the last entry for this column. Therefor if you are keeping entire columns then the block of memory that needs to be copied to the new array is arranged sequentially in memory and can just be described as "starting address" and "number of items to copy". Especially for larger blocks, there are often ways to optimize that at the machine level, such as by copying full system "pages" of memory using DMA.
Andrew
Andrew il 20 Mag 2011
Thank you so much. I really appreciate your taking the time to explain this to me. I will be able to use in the future what you have taught me.

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