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Change a value after a maximum five-consecutive column of zero

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Jalu Naradi
Jalu Naradi 2018-5-9
评论： Jalu Naradi 2018-5-15
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,
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Jalu Naradi
Jalu Naradi 2018-5-9
Sorry, it was my mistake. I edit the example so that there are 6 consecutive zeros now before 1

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采纳的回答

Guillaume
Guillaume 2018-5-9
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
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Guillaume
Guillaume 2018-5-14
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).

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更多回答（2 个）

Jan
Jan 2018-5-9
编辑：Jan 2018-5-9
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
Ameer Hamza 2018-5-9
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 个评论
Guillaume
Guillaume 2018-5-9
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.
Jalu Naradi
Jalu Naradi 2018-5-10
Thank you Ameer for your answer. As Guillaume mentioned above, there are several patterns that acceptable here.

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