Equivalent of A&(~B) without using ~

Question

Yi-xiao Liu 2022-6-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1745490-equivalent-of-a-b-without-using

评论： Yi-xiao Liu 2022-6-22

I am looking for a way to generate the same output as A&(~B) w/o using the ~ operator. The problem is both A and B are very large sparse matrices with only a few non-zero elements (trues). Evaluating ~B will produce a temporary matrix that cannot fit in memory. What's the best way to circumvent this?

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

David Goodmanson 2022-6-22

在 MATLAB Online 中打开

Hi YL,

does

a-(a&b)

get it done?

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Yi-xiao Liu 2022-6-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1745490-equivalent-of-a-b-without-using#answer_991195

编辑：Yi-xiao Liu 2022-6-22

在 MATLAB Online 中打开

Hat tip to @David Goodmanson and @Stephen23. I found 4 solutions:

A=rand(5)<0.6;B=rand(5)>0.2;

ref=A&(~B);

Method1=A;

Method1(B)=false;

Method2=A-(A&B);

Method3=(A-B)>0;

Method4=xor(A,A&B);

all((Method1==ref)&(Method2==ref)&(Method3==ref)&(Method4==ref),"all")

ans = logical

1

rng("shuffle")

t1=nan(10,1);t2=nan(10,1);t3=nan(10,1);t4=nan(10,1);

for ii=1:10

idx=ceil(1e6*rand(2e6,2));

idx=unique(idx,"rows");

i=idx(:,1);j=idx(:,2);

A=sparse(i(1:1e6),j(1:1e6),true(1e6,1),1e6,1e6);

B=sparse(i((1e6+1):end),j((1e6+1):end),true(numel(i)-1e6,1),1e6,1e6);

tic

Method1=A;

Method1(B)=false;

t1(ii)=toc;

tic

Method2=A-(A&B);

t2(ii)=toc;

tic

Method3=(A-B)>0;

t3(ii)=toc;

tic

Method4=xor(A,A&B);

t4(ii)=toc;

end

t1=mean(t1);t2=mean(t2);t3=mean(t3);t4=mean(t4);

bar([t1,t2,t3,t4])

Personally I like the 4th one the most. It's fast, it's short, and it does not invoke type conversion in case you are short on Bytes.

4 个评论
显示 2更早的评论隐藏 2更早的评论

Paul 2022-6-22

编辑：Paul 2022-6-22

I thought there was a concern about memory consumption, so thought it worthwile to point out that Method2 requires 8x more memory, in addtion to the need to cast to logical if used later on as logical.

Yi-xiao Liu 2022-6-22

@Paul You are absolutely right. This is in fact (one of) the reason I prefer method 4 (see last line of my answer).

On conversion back to logical, my experience is that MATLAB will do the conversion for you if you try to use double as logical. there is no need to do it explicitly.

请先登录，再进行评论。