Remove rows with consecutive numbers

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sena
sena 2019-1-3
编辑： Bruno Luong 2019-1-4
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];

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Luna
Luna 2019-1-3
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
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sena
sena 2019-1-3
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.

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更多回答（2 个）

Luna
Luna 2019-1-3
Hello Sena,
Try this below:
newA = A(~all((diff(A')' == 1)')',:);
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sena
sena 2019-1-3
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
Luna 2019-1-3
Please look at the function I have given in another answer.
Write your k -> any number of consecutive you wish and your A matrix.

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Bruno Luong
Bruno Luong 2019-1-3
编辑：Bruno Luong 2019-1-4
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
  1 个评论
Luna
Luna 2019-1-3
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.

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