Going back from cumsum for a matrix

1 次查看（过去 30 天）

显示更早的评论

valentino dardanoni 2022-10-21

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1832003-going-back-from-cumsum-for-a-matrix

评论： valentino dardanoni 2022-10-21

采纳的回答： David Hill

Suppose I cumsum a matrix, say A=rand(3,3); B=cumsum(A).

Knowing B, how to I get back to A, in a reasonably efficient way, for a rather large B?

Thanks!

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

valentino dardanoni 2022-10-21

Thankyou David (and Walter). It works perfectly in my application.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

David Hill 2022-10-21

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1832003-going-back-from-cumsum-for-a-matrix#answer_1080203

在 MATLAB Online 中打开

A=round(rand(100,100),4);
B=cumsum(A);
a=round([B(1,:);diff(B)],4);
isequal(A,a)
ans = logical
   1

Key points here are the use of diff(), the duplication of the first entry, and the rounding or other way of comparing with tolerance for the cross-check (since you would need to deal with round-off errors.)

请先登录，再进行评论。

类别

MATLAB

在 Help Center 和 File Exchange 中查找有关 MATLAB 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by