Find if two numbers occurs in the same sequence

Hello

I would like to know how to determine if numbers occur in the same sequence.

[3 1 5 0 0
 1 2 5 0 0
 1 3 4 1 5
 2 1 5 0 0]

For example in row 1 , we can see that 1 is followed by a 5. Now I need to determine if the same happens in row 2. But in this case 1 is followed by a 2 and then 5.

In that case I need to delete the second row and check with the third row.

When I compare row 1 and row 3 , though 1 is followed by a 5 , 3 should be followed by a 1 in row 1. But in row 3, 3 is followed by a 4, so need to delete this as well.

When you compare row 1 and row 4, since 1 is followed by a 5 in both cases I retain row 4 ( even though the first numbers are different).

So my final answer should be.

[3 1 5 0 0
 2 1 5 0 0]

If a row is retained , then the checking needs to start from that row on wards. That is row 4 will now be checked with row 5 and not row 1.

I hope I have explained it correctly, if not I will clear it out better if anyone has any questions.I am not sure how to proceed and any help would be appreciated. Thanks in advance

[Ced]

How big are the matrices in question? Do you know which numbers can be present?

The easiest (note: inefficient) way I can think of right now is to do it in two steps:

Step 1: Set up a matrix, where e.g. column 1 contains all occurring numbers, and column 2 the numbers that should come after them.

Step 2: iterating over column one, look for each occurrence of the value in your matrix, check if the next value matches your value from column 2, and act accordingly.

[Guillaume]

For your information I've turned a variation of this question into a cody problem.

Answer 1

[Guillaume]

Does this do what you want?

m = [3 1 5 0 0
     1 2 5 0 0
     1 3 4 1 5
     2 1 5 0 0];

out = m(1, :); %output matrix
comprow = 1;   %row that must be matched
for row = 2:size(m, 1) %iterates from row 2 onwards
   [~, match] = ismember(m(row, :), nonzeros(m(comprow, :))); %find positions of numbers in common, ignoring 0s
   %we now need to make sure that the elements that match are in sequence, that is that the
   %min of match (ignoring 0, unmmatched element) to the max of match are in sequence.
   %strfind also find sequences in numbers
   minpos = min(nonzeros(match));
   maxpos = max(match);
   if strfind(match, minpos:maxpos) %works with numbers as well
      out(end+1, :) = m(row, :); %match, then copy
      comprow = row; %use current row for further comparison
   end
end

[Manoj]

Hello Guillaume

Thank you so much it works perfectly. I hope its not too much to ask but can we implement one more condition into your code.

If two rows are identical Eg

    1 2 5 0 0
    1 2 5 0 0

Then we delete them as well, everything else is just perfect. Thank you so much once again

[Guillaume]

If you have indentical rows, do you want to remove all of them or keep just one.

The latter is just:

out = unique(out, 'rows', 'stable');

The former is a bit more complicated but should still involve unique.

[Manoj]

It is the later but this operation should be performed in the loop as well.

Eg

   1 2 5 0 0  row 1
   3 2 5 0 0  row 2
   3 2 5 0 0  row 3
   1 2 5 0 0  row 4

Then the answer would be

   1 2 5 0 0   row 1
   3 2 5 0 0   row 2
   1 2 5 0 0   row 4

What I am trying to say is that when we compare row 1 and row 2 , row 2 remains. Then when we compare row 2 and row 3, we delete row 3 since they are identical.

Next we compare row 2 and row 4 , and row 4 remains. Hope I have made myself clear

[Guillaume]

Alright, then you just change the if to:

if strfind(match, minpos:maxpos) && ~isequal(m(row, :), m(comprow, :))

[Manoj]

Thank you once again , works brilliant.