Remove rows with consecutive numbers

If I have a matrix such as:

A = [1 2 3 4 5

3 5 7 9 11

1 1 4 5 7

3 5 6 6 9

1 4 10 15 16];

I would like to remove all the rows from matrix A that contain more than 4 consecutive numbers. In matrix A we can see that first row contains 5 consecutive numbers [1 2 3 4 5], so this row should be removed from matrix A yielding:

A = [3 5 7 9 11

1 1 4 5 7

3 5 6 6 9

1 4 10 15 16];

Answer 1

[Luna]

I wrote down a function your A (matrix), k (number of consecutive numbers) and result is newA.

Call that function in your workspace with defining A and k.

function newA = filterArray(A,k)

pattern = ones(1,k-1);
idx = contains(cellstr(num2str((diff(A')' == 1))),num2str(pattern));
newA = A(~idx,:);

end

[Luna]

for example:

k = 3;
A = [..];

pattern = ones(1,k-1);
idx = contains(cellstr(num2str((diff(A')' == 1))),num2str(pattern));
newA = A(~idx,:);

%% OR 

myNewResult = filterArray(A,3);

[sena]

It works perfectly! I just checked it with my large matrix and it looks all good.

Thank you, Luna, very much. You have been a great help.

[Luna]

Your welcome :)

Answer 2

[Luna]

Hello Sena,

Try this below:

newA = A(~all((diff(A')' == 1)')',:);

[sena]

Hello Luna,

Thank you for answering my question. Your suggestion works absolutely fine if the entire row contains consecutive numbers. However, I need a condition that if there are more than let's say 3 consecutive numbers than that entire row needs to be removed from matrix A.

So for example, if matrix A is as follows:

A = [1 2 3 4 10

3 4 5 9 11

1 2 4 5 7

3 5 6 6 9

1 4 10 15 16];

Then given the condition that all rows with 3 or more consecutive numbers are removed the resulting matrix A should look like:

A = [1 2 4 5 7

3 5 6 6 9

1 4 10 15 16];

Hopefully, I'm clearer this time around and thank you once again for your help.

[Luna]

Ups sorry it was my mistake i didn't read the question well.

Here it checks the sum of the colums in diff of A which are greater than 3 gets deleted.

newA = A(~(sum((diff(A')' == 1)') >= 3)',:)

[Luna]

Actually I realized that, in the question you said it will be more than 4 consecutive numbers that means 5 consecutive numbers. First code I shared does the exact thing you said. That means all row:

Now you say 3 or more numbers.

[sena]

It works really well with example I used in my question and comment, however when I tried to apply it to my actual problem it only works for the first row and then it stops there. For example, I used augmented matrix A as:

A = [1 2 3 4 10 2 5

3 4 5 9 11 3 8

1 2 4 5 7 3 1

3 5 6 6 9 1 4

1 4 10 15 16 8 7];

But new matrix A looks like:

newA = [3 4 5 9 11 3 8

1 2 4 5 7 3 1

3 5 6 6 9 1 4

1 4 10 15 16 8 7]

and it still contains the second row from original matrix A which satisfies condition of having 3 consecutive numbers.

Additionally, since the proposed formula uses syntex 'sum' there might be cases which satisfy sum condition but are not actully 3 consecutive numbers. For example,

[3 4 7 10 11 4 5]

which gives us logicals

[1 0 1 1]

so this will be equal to 3 but it does not contain 3 consecutive numbers.

[Luna]

Please look at the function I have given in another answer.

Write your k -> any number of consecutive you wish and your A matrix.

Answer 3

[Bruno Luong]

A = [1     2     3     4     5;
    3     5     7     9    11;
    1     1     4     5     7;
    3     5     6     6     9;
    1     4    10    15    16];
% Remove all rows of A contain at least n consecustive numbers
n = 3;
D = diff(A,1,2);
A(arrayfun(@(i) ~isempty(strfind(D(i,:), ones(1,n))),1:size(D,1)),:) = []

Result:

A =

     3     5     7     9    11
     1     1     4     5     7
     3     5     6     6     9
     1     4    10    15    16

[Luna]

When A is like below:

A = [1 2 3 4 10

3 4 5 9 11

1 2 4 5 7

3 5 6 6 9

1 4 10 15 16];

result should be like:

A = [1 2 4 5 7

3 5 6 6 9

1 4 10 15 16];

So she asks for 3 or more consecutive numbers as I understand.