Splitting Matrix based on another matrix

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?

[Rounak Saha Niloy]

NOTE: intersect may not work here as there can be repition of rows.

[Matt J]

Please give the intended result for the example you've posted.

[Rounak Saha Niloy]

A = [52.6460   10.9120    2.3890
    52.5690   10.9130    2.4030
    52.5700   10.9120    2.3890]; % unique in cx w.r.t popx

B = [52.5640   10.9110    2.3890
    52.5690   10.9130    2.3890
    52.5690   10.9120    2.3890] % not unique in cx w.r.t. popx; I need the index of these rows w.r.t. popx as well. these indixes will be used somewhere else.

C=[52.6460   10.9120    2.3890
    52.5690   10.9130    2.4030
    52.5700   10.9120    2.3890
    52.5640   10.9110    2.3890
    52.5690   10.9130    2.3890
    52.5690   10.9120    2.3890] % Combined matrix

[Matt J]

WHat does it mean to be "unique in cx with respect to popx"?

[Rounak Saha Niloy]

It means that, the rows which are in cx but not in popx.

Answer 1

[Rounak Saha Niloy]

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];

Answer 2

[Matt J]

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]

Nah! I am sorry, this is not what I want.

Answer 3

[Rounak Saha Niloy]

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];

Answer 4

[Matt J]

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]

[Rounak Saha Niloy]

well, this does not always work. If there is repitation, it fails. For example, let's say there are 3 rows which are common in popx and cx. Among these 3 rows, two rows are identical. Using setdiff/intersect function will give me the unique rows. But I want all three common rows.

[Rounak Saha Niloy]

And most importantly, I need the indices of rows of B with respect to popx.

[Matt J]

What is the meaning of " the indices of rows of B with respect to popx"? B are the remaining rows in cx that are not in A. They have nothing to do with popx.

[Rounak Saha Niloy]

It is true that "B are the remaining rows in cx that are not in A". But B is also subset of popx. Please look at the venn diagram. It may make sense.

[Matt J]

A0=setdiff(cx,popx,'rows');
B0=setdiff(cx,A0,'rows');

loc=ismember(cx,A0,'rows');
A=cx(loc,:);

loc=ismember(popx,B0,'rows');
B=popx(loc,:); %loc contains desired indices into popx

[Rounak Saha Niloy]

Okay, let me try this.

[Rounak Saha Niloy]

Will it be A0 in 2nd line?

[Matt J]

Yes, I fixed it.

[Rounak Saha Niloy]

I appreciate your effort, mate. But unfortunately, it does not work in all cases.

[Matt J]

Such as?

[Rounak Saha Niloy]

Try with these values-

popx=[   52.6440   10.8560    2.3890
   52.6450   10.9110    2.3890
   52.6470   10.9120    2.3890
   52.6470   10.9120    2.3890
   52.6470   10.9120    2.3890
   52.6460   10.9120    2.3890];

cx=[52.6470   10.8560    2.4330
   52.6450   10.9110    2.3890
   53.1770   10.7480    2.3890
   52.6440   10.7950    2.3890
   52.6470   10.9120    2.3890
   52.6470   10.9120    2.3890];

[Rounak Saha Niloy]

Here,

[A;B] should be equal to cx. (if the order of rows are different, there is no problem).

[Rounak Saha Niloy]

This is working untill I find some other extreme case-

archive = unique(popx,"rows","stable");
[A,ia]=setdiff(cx,popx,"rows","stable");
ia1=setdiff(1:size(cx,1),ia);
B0=cx(ia1,:);
ib=find((ismember(archive,B0,"rows"))~=0);
B=archive(ib,:);
C=[A;B];

[Matt J]

A0=setdiff(cx,popx,'rows');
B0=setdiff(cx,A0,'rows');

loc=ismember(cx,A0,'rows');
A=cx(loc,:);

loc=ismember(cx,B0,'rows');
B=cx(loc,:); %loc contains desired indices into popx

C=[A;B];

popxIndices=ismember(popx,C,'rows')

[Rounak Saha Niloy]

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]

[~,id]=ismember(B,archive,"rows")