Changing duplicates in an array to zero?
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SO I have an array that looks like:
0 0 6 6 3 3
Basically my question is, how do you check for a duplicate value and if there is a duplicate existing, keep the first value, but set the second value to zero.
The result would be:
0 0 6 0 3 0
Thanks in advance, Dom
2 个评论
Joseph Cheng
2014-3-25
Are you just talking about consecutive duplicate values? such as [0 0 6 6 3 3 6 6 3 3] turns to [ 0 0 6 0 3 0 6 0 3 0]
or all duplicate values? such as [ 0 0 6 3 6 3] becomes [0 0 6 3 0 0].
Dominic Green
2014-3-25
采纳的回答
Joseph Cheng
2014-3-25
编辑：Joseph Cheng
2014-3-25
here is a simple thing i just threw together. Hopefully it helps. As you can see what i'm doing below is finding the differences using diff() which subtracts adjacent values. when offset by x(1) and you can see that the zero values in y line up with the duplicate values found in x. using the find function you can get the index of zero values when y is zero. then you can substitute the values (here i used z) for zero.
x = [1 1 6 6 3 3 2 3 4 6 6 3 3 2 2 2 3 3 3]
y = [ x(1) diff(x)]
y = find(y==0);
z = x
z(y)=0
if it is all duplicate values as i asked in my earlier comment. Still thinking about that one.
3 个评论
Dominic Green
2014-3-25
Dominic Green
2014-3-25
Joseph Cheng
2014-3-25
simpler way, you can also use [C,IA,IC] = unique(x,'stable')and it'll give you the index of the first occurrence in IA for that number. and then you can set the rest into 0
更多回答（1 个）
Jos (10584)
2014-3-25
x = [1 1 6 6 3 3 2 3 4 6 6 3 3 2 2 2 3 3 3]
y = zeros(size(x))
[~,i] = unique(x,'first')
y(i) = x(i)
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