How to replace values?

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Ugur
Ugur on 16 Jul 2013
How can I replace values? I've a matrix of 3653x337 Let's consider I've
  • dates values
  • 0101 0
  • 0102 0
  • 0103 0
  • 0104 1
  • 0105 0
  • 0106 0
  • 0107 0 ....
I need a function to find all these "1"'s in my matrix and to replace it for two days before and after I've a "1".
Thanks!
  4 Comments
Cedric
Cedric on 16 Jul 2013
And why not to 0101 and 0107?
the cyclist
the cyclist on 16 Jul 2013
I think he means that if there is a 1 in the second column, put two more 1's before and two more after (all in the second column).

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Accepted Answer

the cyclist
the cyclist on 16 Jul 2013
Here is a simple way that should be easy to understand. I am sure there are slicker ways, but maybe this will suit your purpose.
oldIndex = find(values==1);
newIndex = unique([oldIndex-2; oldIndex-1; oldIndex; oldIndex+1; oldIndex+2]);
values(newIndex) = 1;
  3 Comments
Ugur
Ugur on 17 Jul 2013
it works very well!!!
Thanks!
Jos (10584)
Jos (10584) on 17 Jul 2013
This errors when values has 1's in the beginning or the end ... Moreover, you can skip unique.

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More Answers (1)

Jos (10584)
Jos (10584) on 17 Jul 2013
Here's a one-liner:
values = [0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1]
values(max(min(bsxfun(@plus,reshape(find(values==1),[],1),-2:2),numel(values)),1))=1
or in more readable format:
values = [0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1]
oldIndex = find(values==1)
newIndex = bsxfun(@plus, oldIndex(:), -2:2)
newIndex = max(newIndex,1)
newIndex = min(newIndex,numel(values))
values2 = values % work on a copy
values2(newIndex) = 1
Note that you do not need unique numbers for right-hand indexing
B = [0 0 0]
B([2 2 2 2 2 2]) = 1
  3 Comments
Jos (10584)
Jos (10584) on 17 Jul 2013
you should type numel with an lower case L instead of the digit 1 ...
Ugur
Ugur on 20 Jul 2013
: )
thank you a lot!!!

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