How to replace values?
显示 更早的评论
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 个评论
Azzi Abdelmalek
2013-7-16
This is not clear
Ugur
2013-7-16
Cedric
2013-7-16
And why not to 0101 and 0107?
the cyclist
2013-7-16
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).
采纳的回答
the cyclist
2013-7-16
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 个评论
Azzi Abdelmalek
2013-7-17
newIndex = unique([oldIndex-2; oldIndex-1;oldIndex+1; oldIndex+2]);
without oldIndex
Ugur
2013-7-17
Jos (10584)
2013-7-17
This errors when values has 1's in the beginning or the end ... Moreover, you can skip unique.
更多回答（1 个）
Jos (10584)
2013-7-17
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 个评论
Ugur
2013-7-17
Jos (10584)
2013-7-17
you should type numel with an lower case L instead of the digit 1 ...
Ugur
2013-7-20
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