Checking a matrix for duplicates in specific row, removing the respective columns

So, I try to explain this.
I have a two-row matrix of values, [x;y], f.e.
x=[1 2 2 3 4 5 6 6 7]
y=[1 2 3 4 5 6 7 7 8]
being merged into the matrix:
d= [1 2 2 3 4 5 6 6 7;
1 2 3 4 5 6 7 7 8]
I then want to check this matrix for repeats in the first row & remove the respective columns, while saving the x values of the columns being removed as a seperate vector. So Output should be something like:
d=[1 3 4 5 7;
1 3 4 5 8]
x_cut=[2 6]
It should be noted that this would have to scan for multiple repeats of different values, as shown above.
Thank you.
Have a great day & stay safe
Claudius Appel

 Réponse acceptée

Bruno Luong
Bruno Luong le 6 Août 2020
Modifié(e) : Bruno Luong le 6 Août 2020
d= [1 2 2 3 4 5 6 6 7;
1 2 3 4 5 6 7 7 8]
dd = diff([nan,d(1,:),nan])==0;
remove = dd(1:end-1) | dd(2:end)
x_cut = unique(d(1,remove),'stable')
d(:,remove) = []

3 commentaires

This works nicely, but I'd have a short follow up question, that is admittetly not within the original question, because I didn't realise it was necessary:
If I, instead of just conserve x_cut, wanted to cut and preserve the pairs, i.e. having as output:
cut=[2 2 6 6;
2 3 7 7]
I am having trouble figuring out how excactly that would be done.
I was experimenting with extracting y_cut as is done with x_cut, but that leaves me with the problem that I don't know how to merge those into the right vectors again:
x_cut = unique(d(1,remove),'stable')
% yields
x_cut =
2 6
y_cut=unique(d(2,remove),'stable')
% would yield
y_cut =
2 3 7
I wouldn't know how to merge these together to form:
cut=[2 2 6 6;
2 3 7 7]
d= [1 2 2 3 4 5 6 6 7;
1 2 3 4 5 6 7 7 8]
dd = diff([nan,d(1,:),nan])==0;
remove = dd(1:end-1) | dd(2:end);
cut = d(:,remove)
keep = d(:,~remove) % rename keep to d if you like
Or if yoy want to get cut from x_cut
dd = diff([nan,d(1,:),nan])==0;
remove = dd(1:end-1) | dd(2:end)
x_cut = unique(d(1,remove),'stable')
cut = d(:,ismember(d(1,:), x_cut))
d(:,remove) = []
Thank you.
That works perfectly.
Have a nice day & stay healthy.

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