Splitting Matrix based on another matrix

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Rounak Saha Niloy
Rounak Saha Niloy am 19 Okt. 2022
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 Kommentare
Matt J
Matt J am 19 Okt. 2022
WHat does it mean to be "unique in cx with respect to popx"?
Rounak Saha Niloy
Rounak Saha Niloy am 19 Okt. 2022
It means that, the rows which are in cx but not in popx.

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Akzeptierte Antwort

Rounak Saha Niloy
Rounak Saha Niloy am 20 Okt. 2022
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];

Weitere Antworten (3)

Matt J
Matt J am 19 Okt. 2022
Bearbeitet: Matt J am 19 Okt. 2022
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 am 19 Okt. 2022
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 am 19 Okt. 2022
Bearbeitet: Matt J am 19 Okt. 2022
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 Kommentare
Rounak Saha Niloy
Rounak Saha Niloy am 19 Okt. 2022
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,:)
Rounak Saha Niloy
Rounak Saha Niloy am 19 Okt. 2022
[~,id]=ismember(B,archive,"rows")

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