Splitting Matrix based on another matrix

2022 10 19
4 个回答
回答已采纳
更新时间：2022 10 20
1 次查看（30 天）

0 个投票

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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Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
NOTE: intersect may not work here as there can be repition of rows.
Matt J
Matt J 2022-10-19
Please give the intended result for the example you've posted.
Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
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
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

0 个投票

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:

0 个投票

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

0 个投票

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

0 个投票

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
编辑：Rounak Saha Niloy 2022-10-19
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
Rounak Saha Niloy 2022-10-19
And most importantly, I need the indices of rows of B with respect to popx.
Matt J
Matt J 2022-10-19
编辑：Matt J 2022-10-19
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
Rounak Saha Niloy 2022-10-19
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
Matt J 2022-10-19
编辑：Matt J 2022-10-19
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
Rounak Saha Niloy 2022-10-19
I appreciate your effort, mate. But unfortunately, it does not work in all cases.
Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
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
Rounak Saha Niloy 2022-10-19
Here,
[A;B] should be equal to cx. (if the order of rows are different, there is no problem).
Rounak Saha Niloy
Rounak Saha Niloy 2022-10-19
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
Matt J 2022-10-19
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
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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