Change a value after a maximum five-consecutive column of zero
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Jalu Naradi
el 9 de Mayo de 2018
Comentada: Jalu Naradi
el 15 de Mayo de 2018
I have a matrix [3000,1000] which has only 0, 1, and -1. If there is a -1 in a row followed by a (maximum) consecutive five-columns of zeros then followed by 1, this 1 value should change to 2.
I am looking for the fastest way to do it in Matlab. Is there anyway without a loop? if not, what is the best way?
For example
A = [1 0 0 -1 0 0 0 0 1 0 0 -1 0 0 1;
0 -1 0 0 0 0 0 0 1 -1 0 1 0 0 -1];
% I would like to change matrix A as follow
A = [1 0 0 -1 0 0 0 0 2 0 0 -1 0 0 2;
0 -1 0 0 0 0 0 0 1 -1 0 2 0 0 -1 ];
Thanks in advance,
3 comentarios
Respuesta aceptada
Guillaume
el 9 de Mayo de 2018
I don't think this can be done more efficiently than with a loop over the rows:
for row = 1:size(A, 1)
col = find(A(row, :));
tochange = diff(col) <= 5 & A(row, col(1:end-1)) == -1 & A(row, col(2:end)) == 1;
A(row, col([false, tochange])) = 2;
end
4 comentarios
Guillaume
el 14 de Mayo de 2018
The result of diff is a vector with one less element than col. tochange(1) actually corresponds to col(2|, so I just prepend false to vector (since the 1st element is never going to have to be changed anyway).
Más respuestas (2)
Jan
el 9 de Mayo de 2018
Editada: Jan
el 9 de Mayo de 2018
A = [1 0 0 -1 0 0 0 0 1 0 0 -1 0 0 1; ...
0 -1 0 0 0 0 0 1 -1 0 1 0 0 1 0];
[s1, s2] = size(A);
Av = reshape(A.', 1, []); % Convert it to 1 vector
for k = 0:5
index = strfind(Av, [-1, zeros(1, k), 1]);
index = index(mod(index, s2) < s2 - k); % Omit matches at end of row
Av(index + k + 1) = 2;
end
B = reshape(Av, s2, []).';
Ameer Hamza
el 9 de Mayo de 2018
This is an approach to modify an identify an arbitrary pattern and modify a value, Using for loop and comparing them by converting them to char array.
A = [1 0 0 -1 0 0 0 0 1 0 0 -1 0 0 1;
0 -1 0 0 0 0 0 1 -1 0 1 0 0 1 0];
pattern = [-1 0 0 0 0 1];
strA = num2str(A+1); % +1 is added to remove negative signs for easy manipulation as strings
strPattern = strrep(num2str(pattern+1), ' ', '');
for i=1:size(A,1)
index = strfind( strrep(strA(i,:), ' ', ''), strPattern) + length(pattern) -1;
if ~isempty(index)
A(i, index) = 2;
end
end
2 comentarios
Guillaume
el 9 de Mayo de 2018
Note that you don't need to convert numbers to char to use strfind. Despite not being actually documented, strfind works just as well for detecting patterns of numbers
strfind(A(i, :), [-1 0 0 0 0 0 1])
The problem here is that several patterns are acceptable, [-1 1], [-1 0 1], ..., [-1 0 0 0 0 0 1], so a pattern search is not particularly useful. I guess a regexp would work (which does requires a conversion to char) but I'm not convinced the extra complexity and time taken by the regex would be better than the simple for loop I've detailed.
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