Splitting Matrix based on another matrix

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Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
回答： Rounak Saha Niloy 2022-10-20
I have two matrices as follows.
popx=[52.647 10.912 2.389
52.564 10.911 2.389
52.569 10.912 2.389
52.569 10.912 2.389
52.569 10.913 2.389
52.569 10.913 2.389];
cx=[52.646 10.912 2.389
52.564 10.911 2.389
52.569 10.913 2.403
52.570 10.912 2.389
52.569 10.913 2.389
52.569 10.912 2.389];
Now, I want to split cx into two matrirces as per the following-
  1. rows which are UNIQUE with respect to popx.
  2. rows wich are NOT UNIQUE with respect to cx.
Finally, I will again merge these two matrices which will be equivalent to cx (i do not mind if the order of rows are diffferent).
How can I do this?
  5 个评论
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Matt J
Matt J 2022-10-19
WHat does it mean to be "unique in cx with respect to popx"?
Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
It means that, the rows which are in cx but not in popx.

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采纳的回答

Rounak Saha Niloy
Rounak Saha Niloy 2022-10-20
I would like to thank Matt J for helping me solve the problem. My sincere gratitude to him.
Meanwhile, this is the way how I have solved the issue-
archive = unique(popx,"rows","stable");
ia=find(~ismember(cx,popx,'rows')==1);
A = cx(ia,:); % A is the matrix of rows which are present in cx but not present in popx
ia1=setdiff(1:size(cx,1),ia);
B0=cx(ia1,:);
if size(B0,2)~=0
[~,ib]=ismember(B0,archive,"rows");
end
B=archive(ib,:);
C = [A;B];

更多回答（3 个）

Matt J
Matt J 2022-10-19
编辑：Matt J 2022-10-19
Ran in:
This might be what you want.
popx=[52.647 10.912 2.389
52.564 10.911 2.389
52.569 10.912 2.389
52.569 10.912 2.389
52.569 10.913 2.389
52.569 10.913 2.389];
cx=[52.646 10.912 2.389
52.564 10.911 2.389
52.569 10.913 2.403
52.570 10.912 2.389
52.569 10.913 2.389
52.569 10.912 2.389];
[~,~,Gp]=unique(popx,'rows');
[~,~,Gc]=unique(cx,'rows');
Hp=histcounts(Gp,1:max(Gp)+1);
Hc=histcounts(Gc,1:max(Gc)+1);
crit1=(Hp==1);
crit2=(Hc>1);
M1=cx(crit1(Gp),:)
M1 = 2×3
52.6460 10.9120 2.3890 52.5640 10.9110 2.3890
M2=cx(crit2(Gc),:)
M2 = 0×3 empty double matrix
result=[M1,M2]
result = 2×3
52.6460 10.9120 2.3890 52.5640 10.9110 2.3890
Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
This can be a way, I think. Need to think about the extreme cases though.
[A,ia]=setdiff(cx,popx,"rows","stable");
ia1=setdiff(1:size(cx,1),ia);
ic=find((ismember(popx,cx,"rows"))~=0);
if size(ia1,2)==size(unique(popx(ic,:),"rows"),1)
[~,~,ib]=intersect(cx,popx,'rows','stable');
else
ib=find((ismember(popx,cx,"rows"))~=0);
end
B=popx(ib,:);
C=[A;B];
Matt J
Matt J 2022-10-19
编辑：Matt J 2022-10-19
popx=[52.647 10.912 2.389
52.564 10.911 2.389
52.569 10.912 2.389
52.569 10.912 2.389
52.569 10.913 2.389
52.569 10.913 2.389];
cx=[52.646 10.912 2.389
52.564 10.911 2.389
52.569 10.913 2.403
52.570 10.912 2.389
52.569 10.913 2.389
52.569 10.912 2.389];
[A,ia]=setdiff(cx,popx,'rows')
B=setdiff(cx,A,'rows')
C=[A;B]
  16 个评论
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Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
Mate, Please help me solve the following. Then my whole problem will be solved. I will comment it once it is solved.
B= [ 52.6450 10.9110 2.3890
52.6470 10.9120 2.3890
52.6470 10.9120 2.3890]
archive = [ 52.6440 10.8560 2.3890
52.6450 10.9110 2.3890
52.6470 10.9120 2.3890
52.6460 10.9120 2.3890]
how can I find the corresponding row index of archive for each row of B?
i.e. I want the answer as-
id = [2 3 3]
It means that,
B(1,:)=archive(2,:)
B(2,:)=archive(3,:)
B(3,:)=archive(3,:)

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