Finding repeating values in an array

2020 4 25
3 个回答
回答已采纳
更新时间：2020 4 27
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0 个投票

Hello all,
I have 2 arrays
A = [ 0.3 0.6 1 0.6 0.3]
B = [ 3 2 3 6 11 ]
I need to find the position of same elements in B and then the max value of the elements on the corresponding position in A.
In this case the number 3 is repeated in B on positions 1 and 3 so the corresponding values in A are 0.3 and 1 => max( 0.3 , 1 ) = 1
The end resault should be:
A1 = [ 0.6 1 0.6 0.3 ]
B1 = [ 2 3 6 11 ]
Any help is appreciated

3 个评论
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andrea
andrea 2020-4-25
maybe :
pos_in_b = B == 3 ;
max ( A ( pos_in_b) )
dpb
dpb 2020-4-25
The find part is easy enough, the logic of how to build the A1, B1 vectors from A,B and the lookups escapes me entirely, though...???
Vladimir Kostic
Vladimir Kostic 2020-4-25
编辑：Vladimir Kostic 2020-4-25
dpb
If you look at the elements of A as the numerator and elements of B as the denominator i need to find fractions with the same denominator and use the one with the higher numerator.
In my example i have
0.3 0.6 1 0.6 0.3
3 2 3 6 11
So there are 2 fractions with the denominator of 3 (it isn't predetermined that it is 3 just happened to be this way, it can be any number and I can be more that just 1 number that repeats) and the numerators of those 2 fractions are 0.3 and 1, where the higher number is 1 ( 1>0.3). So i need to erase the fraction with the lower numerator.
0.6 1 0.6 0.3
2 3 6 11
Hope this clears it up

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 采纳的回答

aleksa markovic
aleksa markovic 2020-4-25
编辑：aleksa markovic 2020-4-25

0 个投票

You ca try something like this:
Xa = [3 2 3 6 11];
mua = [.3 .6 1 .6 .3];
tmpX = [];
tmpmu = [];
for i = 1:size(Xa,2)
if(sum(tmpX == Xa(i)) > 0)
tmpmu(tmpX == Xa(i)) = max(mua(i),tmpmu(tmpX == Xa(i)));
else
tmpX = [tmpX Xa(i)];
tmpmu = [tmpmu mua(i)];
end
end
Xa = tmpX;
mua = tmpmu;

更多回答（2 个）

Rik
Rik 2020-4-25
编辑：Rik 2020-4-27

0 个投票

You can use the outputs of the unique function to achieve this.
A = [ 0.3 0.6 1 0.6 0.3];
B = [ 3 2 3 6 11];
[B1,~,ind]=unique(B);
A1=accumarray([ones(numel(A),1) ind],A(:),[],@max);
A1
B1

2 个评论
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Vladimir Kostic
Vladimir Kostic 2020-4-25
Could you explain further into detail, I'm still pretty new at MATLAB
Rik
Rik 2020-4-27
A bit late, but here you go, no loops required.

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andrea
andrea 2020-4-25

0 个投票

maybe i do not understand the problem but anyway
[val, ind] = min ( A ( pos_in_b) )
A(ind) = []

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