How to replace values in a very large array

2022 8 3
1 个回答
回答已采纳
更新时间：2022 8 4
5 次查看（30 天）

0 个投票

Hi
I have an array idx_Laser which is very large (~1 billion values). Whenever a certain condition is fulfilled for another array tdcRelTime I want to replace the values in my idx_Laser array.
The trivial solution would probably be:
for xx = 1:length(tdcRelTime)
if tdcRelTime(xx) < 1.5e10
idx_Laser(idx_Laser==xx) = 0;
end
end
However, as both arrays are quite big, this takes extremely long.
Is there a solution for that problem that is computionally more efficient?
I thought about iterating only through a part of the idx_Laser array, as this is sorted by the size of the values, but I couldnt come up with a good solution.

3 个评论
显示 1更早的评论 隐藏 1更早的评论

dpb
dpb 2022-8-3
Can you hold the arrays in memory?
If that is the case, it's already vectorized for you --
ix=(tdcRelTime < 1.5e10);
idx_Laser(ix) = 0;
If the arrays are being held in virtual memory and you're paging to address them in the above code, that will be slow.
In that case, investigate tall arrays...
Jonathan Weber
Jonathan Weber 2022-8-4
Hey :)
Thanks for the comment.
Yes I can hold the arrays in memory. But
idx_Laser(ix) = 0;
is not really what I was searching for.
The Problem is in the line
idx_Laser(idx_Laser==xx) = 0;
Because sweeping through that array is what takes so long.
I will take a look at these tall arrays though.
Jonathan Weber
Jonathan Weber 2022-8-4
What I came up with is a bit confused, but the performance is not so bad.
As I mentioned the array idx_Laser is sorted and the longest bin has "noElements" values.
Maybe someone has a better idea:
for zz = 1:length(tdcRelTime)
if tdcRelTime(zz) < 1.5e10
idx = find(idx_Laser == zz, 1);
for xx = idx:idx + noElements
if idx_Laser(xx) == zz
idx_Laser(xx) = NaN;
end
end
end
end

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Mohammad Sami
Mohammad Sami 2022-8-4

1 个投票

You can modify what @dpb suggested as follows.
xx=find(tdcRelTime < 1.5e10);
idx_Laser(ismember(idx_Laser,xx)) = 0;

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Logical 的更多信息

产品

版本

R2020a

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by